Mahesh

Students can go through AP Inter 1st Year Physics Notes 8th Lesson Oscillations will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 8th Lesson Oscillations

→ If a body repeats its motion at regular intervals of time, the motion is said to be periodic.

→ If a body moves to and fro about a fixed point in its path and if the acceleration is proportional to the displacement of the body from a fixed point and directed towards the fixed point, then the motion of the body is called simple harmonic motion.

→ One complete to and fro motion of a body is called an oscillation or vibration.

→ The time required for one oscillation of a body is called its period of oscillation.

→ The maximum displacement of a vibrating body from the rest position is called its amplitude.

→ The number of vibrations made by a body in unit time is called its frequency.

→ The phase of vibration of a particle is the state of motion related to the time with reference to the average position of rest.

→ The force constant of a system is equal to the force to be applied on the particle to cause unit displacement.

→ The time taken for one complete oscillation is known as the time period of simple harmonic motion given by T = \(\frac{2 \pi}{\omega}\)

→ The number of oscillations per second is known as the frequency (i.e., υ = \(\frac{1}{T}\) )

→ The velocity of the particle in SHM varies with displacement ‘y’ given by v = ω \(\sqrt{A^2-y^2}\)

→ The velocity is equal to zero at the extreme position and maximum at the mean position. v_max = Aω.

→ The acceleration of the pan Me varies with displacement as, a = -ω²y. The acceleration is zero at the mean position and maximum at the extreme position, a_max = Aω².

→ A simple harmonic motion with amplitude ‘A’ and angular frequency ‘ω’ may be represented as y = A sin (ωt ± Φ₀) or y = A cos (ωt ± Φ₀).

→ A simple pendulum of length ‘l’ makes simple harmonic oscillations with small amplitudes. The period of oscillation is given by T = 2π \(\sqrt{\frac{l}{g}}\).

→ The time period of a loaded spring is T = 2π \(\sqrt{\frac{m}{K}}\) where K is force constant.

→ The particle velocity and acceleration during SHM as a function of time are given by
v(t) = -ωA sin (ωt + Φ)
a(t) = -ω²A cos (ωt + Φ) = -ω²x(t)

→ The damped simple harmonic motion is not strictly simple harmonic.

→ In an ideal case of zero damping, the amplitude of SHM at resonance is infinite.

→ Under forced oscillation, phase of harmonic motion of the particle differs from the phase of the driving force.

→ The phenomenon of increase in amplitude when the driving force is close of the natural frequency of the oscillator is called resonance.

Students can go through AP Inter 1st Year Physics Notes 7th Lesson Systems of Particles and Rotational Motion will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 7th Lesson Systems of Particles and Rotational Motion

→ The angle through which the radius vector rotates in a given time is called angular displacement.

→ The rate of angular displacement of a body is called angular velocity w = \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\)

→ The rate of change of angular velocity of a body is called angular acceleration, a = \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\)

→ A body which does not undergo any change in its shape and volume by the application of force is called a rigid body.

→ A pair of equal, unlike, parallel and non-collinear forces acting on a rigid body constitute a couple.

→ Uniform circular motion is the motion of a particle that moves on the circumference of a circle with constant angular velocity or constant linear speed.

→ Centripetal acceleration (a_c) is the acceleration of a particle moving on the circumference of a circle. Its magnitude is \(\frac{V^2}{r}\) (or) rω² (or) vw. It is directed towards the centre of the circle and variable.

→ Centripetal force (F_c) is the force required by a particle to perform circular motion. Its magnitude is \(\frac{M v^2}{r}\) (or) Mrω². It is also directed towards the centre of the circle and variable. It is a real force like gravitational force, electrostatic force, force of friction etc. It is to be supplied to the particle by an external agency.

→ Safe maximum speed on an unbanked road is v = \(\sqrt{r g \mu}\).

→ A body performing horizontal circular motion has same speed at all points. When a stone tied to one end of a string is revolved in a horizontal circle,
The tension on the string = Centripetal force = \(\frac{M v^2}{r}\) = mrω²

→ When a body performs rotational motion, each particle moves on a circle. The centers of all such circles are on a line called axis of rotation.

→ Torque or moment of force about a point is defined as the product of the force and the perpendicular distance of the point of application of the force from the point. In vector from, t = r × F.

→ Moment of inertia of a rigid body about an axis is defined as the sum of the products of the masses of different particles, supposed to be constituting the body and the square of their respective perpendicular distances from the axis of rotation.

→ Moment of inertia of a point mass m is I = mr², where r is the perpendicular distance of the point mass from the axis of rotation.

→ Radius of gyration of a rotating body is the distance between the axis of rotation and a point at which the whole mass of the body can be supposed to be concentrated so that its moment of inertia would be the same with the actual distribution of mass. Radius of gyration,
K = \(\sqrt{\frac{1}{M}}=\sqrt{\sum_{i=1}^n \frac{m_i r_i^2}{\text { Total mass }}}\)

→ Both moment of inertia and radius of gyration depend upon the position of axis of rotation and the distribution of mass about the axis of rotation. But moment of inertia depends on mass also.

→ Parallel axes theorem states that the moment of inertia of a rigid body about any axis is equal to the moment of inertia about a parallel axis through its center of mass plus the product of the mass the body and the square of the distance between the two parallel axes. Moment of inertia of a rigid body, I = I_G + Mr², where I_G is the moment of inertia of the body about a parallel axis through its center of mass and r is the distance between the parallel axis.

→ Perpendicular axes theorem states that the moment of inertia of plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about two axis perpendicular to each other, in its own plane and intersecting each other at the point, where the axis perpendicular to the plane passes.

→ Torque is the cause of angular acceleration. Relation between torque, t and angular acceleration, a is t = la.

→ Angular momentum of a particle about a point is the product of linear momentum of the particle and the perpendicular distance of the line of motion of the particle from the point. In vector form, L = r × p.
Its SI unit is kg m² s^-1 (or)Js. Its dimensional formula is [M L²T^-1].

→ Angular momentum, L = mvr, where m is mass of a particle, v is the velocity and r is the perpendicular distance.

→ Angular momentum is expressed in terms of w as L = Iω.

→ Relation between τ and L is τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}\) and τ and α is t = Iα.

→ Principle of conservation of angular momentum : When there is no resultant external torque on a rotating system, the angular momentum of the system remains constant both in magnitude and direction.

→ When there is no resultant external torque, L is constant i.e., Iω = constant (or) ω is inversely proportional to I.

→ When an object tied to one end of a string is revolved in a vertical circle, its velocity changes due to gravity.

→ The tension on the string at the highest point of a vertical circle = \(\frac{\mathrm{Mv}_2^2}{\dot{\mathrm{r}}}\) – Mg and at the lowest point, tension = \(\frac{\mathrm{Mv}_1^2}{\dot{\mathrm{r}}}\) + Mg. Tension at any position, T = \(\frac{M v^2}{r}\) – Mg cos θ.

→ The minimum velocity at the highest point, v₂ = \(\sqrt{r g}\) and at the lowest point, v₁ = \(\sqrt{5r g}\).

→ Centre of mass : Def: It is a point where the entire mass of a body or a system is supposed to be concentrated

→ Co-ordinates of Centre of mass :
X_CM = \(\frac{\Sigma m_i x_i}{\Sigma m_i}\); Y_CM = \(\frac{\Sigma m_i y_i}{\Sigma m_i}\); Z_CM = \(\frac{\Sigma m_i z_i}{\Sigma m_i}\)

→ Co-ordinates of velocity of C.M : V_CM = \(\frac{\Sigma m_i v_i}{\Sigma m_i}\)

→ Internal forces do not change the position (or) velocity of centre of mass.

→ Centre of mass obeys Newton’s laws of motion.

→ Matter may (or) may not be present at the centre of mass.

→ The linear momentum of,the centre of mass of a system is the sum of the linear momenta of all the particles comprising the system
Mv_c = Σ m_iv_i or P_c = P₁ + P₂ + …………….. P_n
M is the mass of the entire system, v_c in the velocity of the CM of the system. P_c in the momentum of centre of mass.

→ The acceleration imparted to the system to the external force F is equal to the acceleration of the centre of mass of the system.
a_c = \(\frac{\Sigma m_i z_i}{\Sigma m_i}\)

→ If a shell explodes in. mid air as it moves, the fragments of the shell move in different parabolic paths. but the centre of mass of the shell continues to move in the same parabolic path as the shell, as a single piece. would have moved.

Students can go through AP Inter 1st Year Physics Notes 6th Lesson Work, Energy and Power will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 6th Lesson Work, Energy and Power

→ The scalar product (or) dot product of any two vectors A and B denoted as A.B = AB cos θ.

→ The dot product of A and B is a scalar.

→ Dot product obey’s commutative law. \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}\)

→ Dot product obeys distributive law \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)

→ The work done is a scalar quantity. It can be positive (or) negative.

→ Work done by the friction (or) viscous force on a moving body is negative.

→ The work – energy theorem states that the change in K.E of a body is the work done by the net force on the body.
K_f – K_i = W_Net

→ The work-energy theorem is not independent of Newtons second law.

→ The work-energy theorem holds in all internal forces.

→ The P.E of a body subjected to a conservative force is always undetermined upto a constant.

→ Work done by friction over a closed path is not zero and no potential energy can be associated with friction.

→ The energy possessed by a body by virtue of its position (or) state is called potential energy. V(h) = mgh, if h is taken as variable.

→ The energy possessed by a body by virtue of its motion is called kinetic energy
K = \(\frac{1}{2}\)mv².

→ Work done by a variable force W = \(\int_{x_1}^{x_1} F(x)\)

→ The total mechanical energy of a system observed if the forces, doing work on it, are conservative.

→ The spring force is a conservative force.

→ Eiitstein mass-energy relation is E = mC²
Where C = 3 × 10⁸ m/s = Velocity of light

→ The rate of doing work is called power, P_av = \(\frac{W}{t}\)

→ Another unit of power is horse – power (HP) 1 H.P = 746 W.

→ Collision is an interaction between two (or) more bodies in which sudden changes of momenta take place.

→ In an elastic collision both K.E. and linear momentum are conserved.

→ A collision in which only momentum.is conserved but not the K.E is called inelastic collision.

→ Coefficient of restitution (e) = \(=\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\)

Students can go through AP Inter 1st Year Physics Notes 5th Lesson Laws of Motion will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 5th Lesson Laws of Motion

→ Newton’s First Law ot Morion : “Every body continues to be in its state of rest or of uniform motion in a straightline unless compelled by an external force to change that state”.

→ Newton’s Second Law of Motion: “The rate of change of momentum of a body is directly proportional to the external force applied and takes place in the same direction in which the external force is acting”,

→ Newton’s Third Law of Motion : “To every action there is an equal an opposite reaction.

→ Two objects connected by a cable, one object having horizontal motion and the other having vertical motion
Tension T = \(\frac{ m_1 m_2 }{m_1+m_2}\) g

→ Two objects of unequal suspended by a cable passing over a pulley (At wood’s machine) In this case Tension T = \(\frac{2 m_1 m_2 }{m_1+m_2}\) g

→ Resultant force F_R of two forces F₁ and F₂ acting on a body simultaneously making an angle θ with each other is F_R = \(\sqrt{\mathrm{F}_1^2+\mathrm{F}_2^2+2 \mathrm{~F}_1 \mathrm{~F}_2 \cos \theta}\)

→ When the lift moves up with an acceleration ‘a’ then reaction force R = mg(1 + \(\frac{\mathrm{a}}{\mathrm{g}}\))

→ When the lift moves downwards with an acceleration ‘a1 then Reaction force R = mg (1 – \(\frac{\mathrm{a}}{\mathrm{g}}\)).

→ When the lift is stationary or moves with uniform velocity then acceleration is zero and net force is zero.

→ Impulse ( I ) is the product of force and time of action of force and is equal to change in momentum of the body. I = mv – mu

→ When the resultant external force acting on a system is zero, the total momentum of the system remains constant. This is called law of conservation of linear momentum.

→ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

→ If a body displaces by the application of force, work is said to be done by the force.

→ Impulse = Force × time duration = Change in momentum.

→ Mass is the measure of inertia.

→ Momentum of a body is defined to be the product of its mass m and velocity v.
p = mv

→ Friction : It is the force which always opposes, the relative motion between two surfaces in contact. It acts parallel to the surfaces and opposite to the direction of motion.

→ Three kinds of friction :
a) Static friction: It is the frictional force acting, when there is no relative motion between the surfaces of the bodies. It is always equal to the applied force. Static friction is a self adjusting force. The maximum value of the static friction is called “limiting friction”, F_s.
b) Kinetic friction (F_k) : It is the frictional force present, when a body slides over the other body.

→ The resistance encountered by a body in static friction while tending to move under the action of an external force is. called static friction. It is always equal and opposite to the applied force till the static friction reaches a maximum value.

→ The resistance encountered by a sliding body on a surface is kinetic friction. The applied force required to keep the body moving with uniform velocity, under friction is numerically equal to the value of kinetic friction. It is in the opposite direction of relative velocity between the . bodies in contact.

→ The resistance encountered by a rolling body on a surface is known as rolling friction.

→ Normal reaction is the resultant contact force acting on a body placed on a rigid surface perpendicular to the plane of contact. It is equal to mg on a horizontal surface and mg cos 0 on an inclined plane where m is the mass of the body and 0 is the angle of inclination of the inclined plane.

→ Laws of friction : i) The frictional force is independent of the area of contact, ii) The frictional force is directly proportional to the normal reaction.

→ In case of static friction, the frictional force is limiting friction. Coefficient of static friction μ_s = \(\frac{\mathrm{f}_{\mathrm{L}}}{\mathrm{N}}\)

→ In case of dynamic friction, the frictional force is kinetic friction. Coefficient of dynamic friction m_k = \(\frac{f_k}{N}\)

→ Laws of rolling friction :

1. The smaller the area of contact the lesser will be the rolling friction,
2. The larger the radius of the rolling body, the lesser will be the rolling friction,
3. The rolling friction is directly proportional to the normal reaction.
   Coefficient of rolling friction, μ_r = \(\frac{\mathrm{f}_{\mathrm{r}}}{\mathrm{N}}\)

→ The angle made by the resultant of the normal reaction and the limiting friction with normal reaction is called angle of friction. Coefficient of static friction μ_s = tan θ.

→ Acceleration or a body on a rough horizontal force, a = \(\frac{P-f_k}{m}=\frac{p-\mu_k m g}{m}\) where P is the applied force and m is the mass of the body.

→ Angle of repose is defined as the angle of inclination of a plane with respect to horizontal for which the body will be in limiting equilibrium on the inclined plane. If a is the angle of repose μ_s = tan α.

→ When a body slides down an inclined plane of angle of inclination (θ) greater than the angle of repose (α) i.e., θ > α, the acceleration of the body, a = g (sin θ – μ_k cos θ).

→ When a body starting from rest slides down the inclined plane of length l, its final velocity v = \(\sqrt{\sqrt{2 g l(\sin \theta}-\mu_{\mathrm{k}} \cos \theta}\) and time taken to slide down t = \(\sqrt{\frac{2 l}{g\left(\sin \theta-\mu_k \cos \theta\right)}}\)

→ When a body is to be moved up a rough inclined plane with uniform velocity the force to be applied F = mg (sin θ + m_k cos θ).

→ The acceleration, velocity and time taken by a body sliding along & smooth horizontal surface can be obtained by putting μ_k = 0. i.e., a = g sin θ, v = \(\sqrt{2 g l \sin \theta}\) and t = \(\sqrt{\frac{2 l}{g \sin \theta}}\). The force to move up the plane with uniform velocity is F = mg sin θ.

→ Pulling is easier than pushing. The net pul ing force is given by
P = F(cos θ + μ_k sin θ) – μ_R mg and the net pushing force is given by P’ = F(cos θ – μ_R sin θ) – μ_R mg.

→ Pulling force F = \(\frac{W \sin \phi}{\cos (\theta-\phi)}\) and poshing mg. F’ = \(\frac{W \sin \phi}{\cos (\theta-\phi)}\) where W is the weight of body, Φ is the angle of friction and θ is the angle made by F with the horizontal.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties

Very Short Answer Questions

Question 1.
What is the difference in the approach between the Mendeleev’s periodic law and the modern periodic law ?
Answer:

• According to Mendeleev the physical and chemical properties of elements are periodic functions of their atomic weights.
• According to modem periodic law the physical and chemical properties of elements are periodic functions of their atomic numbers.

Question 2.
In terms of period and group, where would you locate the element with Z = 114 ?
Answer:
Element Z = 114 is present in 7^th period and IVA group (Group – 14)

Question 3.
Write the atomic number of the element, present in the third period and seven¬teenth group of the periodic table.
Answer:
The Element present in 3^rd period and Group – 17 (VIIA group) is chlorine (Cl). It’s atomic number is 17.

Question 4.
Which element do you think would have been named by
a) Lawrence Berkeley Laboratory .
b) Seaborg’s group
Answer:
a) Lawrence Berekeley Laboratory – Lanthanide
b) Seaborg’s group – Actinide (Transuranic element).

Question 5.
Why do elements in the same group have similar physical and chemical properties ?
Answer:
Elements in the same group have same no. of valency shell electrons and have similar outer elec¬tronic configuration so these have similar physical and chemical properties.

Question 6.
What are representative elements ? Give their valence shell configuration.
Answer:

• Representative elements are s and p-block elements except zero group.
• These have general electronic configuration ns^1-2np^1-5.

Question 7.
Justify the position of f-block elements in the periodic table.
Answer:
The two series of elements lanthanides and actinides have been grouped separately and placed at the bottom of the periodic table, though they belong to the sixth and seventh periods of third group (III B).

The justification for assigning one place to these elements has been given on the basis of their similar properties. The properties are so similar that elements from Ce to Lu can be considered as equivalent to one element. In case these elements are assigned different positions (i.e.,) arranged in order of their increasing atomic numbers, the symmetry of the whole arrangement would be disrupted. The same explanation can be given in the case of actinides.

Question 8.
An element ‘X’ has atomic number 34. Give its position in the periodic table.
Answer:
The element X with atomic no (z) = 34 has electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. Hence the element is present in 4,h period and 16th group (VIA group).

Question 9.
What factors impart characteristic properties to the transition elements ?
Answer:
Transition elements exhibits characteristic properties because

• The differentiating electron enters into penultimate d-subshell
• These elements have small size.
• These possess high effective nuclear charge.

Question 10.
Give the outer shells configuration of d-block and f-block elements.
Answer:

• The outer shell electronic configuration of d-block – elements is ns^1-2 (n-1)d^1-10
• The outer shell electronic configuration of f-block – elements is ns²(n-1)d^{0 (or) 1} (n – 2) f^1-14

Question 11.
State and give one example for Dobereiner’s law of triads and Newland’s law of octaves.
Answer:
1) Dobereiner law :

• According to Dobereiner in a traid (3 – elements) the atomic weight of the middle element is the arithmatic mean of the other two elements.
  

2) Newland’s law of octaves : According to Newlands, the elements arranged in the increasing order of atomic weights noticed that every eight element had properties similar to the first element. This relationship was just like every eight note that resembles the first in octaves of music.

Question 12.
Name the anomalous pairs of elements in the Mendaleev’s periodic table.
Answer:
In Mendeleev’s periodic table anamalous pairs are the elements whose atomic weights increasing order is reversed.
Eg:

Question 13.
How does atomic radius vary in a period and in a group ? How do you explain the variation ?
Answer:
In a period : Atomic radius decreases generally from left to right in a period.
Reason : In periods electrons are entered into same subshells.
In a group : Atomic radius increases generally from top to bottom in a group.
Reason : In groups electrons are entered into new subshells.

Question 14.
Among N^-3; O^-2, F^–, Na⁺, Mg⁺² and Al⁺³
a. What is common in them ?
b. Arrange them in the increasing ionic radii.
Answer:
Given ions are
N^-3, O^-2, F^–, Na⁺, Mg⁺² and Al⁺³.
a) The above ions have same number of electrons (All have 10 electrons). So these are called iso electronic species.
b) The increasing order of ionic radii among above ions is
Al⁺³ < Mg⁺² < Na⁺ < F^– < O^-2 < N^-3
Reason : – In case of iso electronic species as the nuclear charge increases ionic radii decreases.

Question 15.
What is the significance of the term isolated gaseous atom while defining the ionization enthalpy.
Hint: Requirement for comparison.
Answer:

• Isolated gaseous atom’s ionisation enthalpy is taken as reference value and it is required to compare this values to various ions of this elements and to compare this values with various elements.

Question 16.
Energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.
Answer:
Given that the energy of the electron in the ground state for hydrogen atom = – 2.18 × 10^-18 J.
For 1 mole of atoms is given by – 2.18 × 10^-18 J × 6.023 × 10²³
= -13.13 × 10⁵ J/Mole
∴ Ionisation enthalpy of hydrogen atom = 13.13 × 10⁵ J/Mole.

Question 17.
Ionization enthalpy, (IE₁) of O is less than that of N — explain.
Answer:

• Oxygen has electronic configuration 1s² 2s² 2p⁴
  
• Nitrogen has electronic configuration 1s² 2s² 2p³
  
• Nitrogen has half filled shell and is stable so more amount of energy is required to remove an electron, than in oxygen.
  Hence IE, of ‘O’ is less than that of ‘N’.

Question 18.
Which in each pair of elements has a more negative electron gain enthalpy?
a. O or Fb. F or Cl
Answer:
a)

has more negative electron gain enthalpy than that of

b)

has more negative electron gain enthalpy than that of

Question 19.
What are the major differences between metals and non-metals ?
Answer:
Metals

• These are generally in solid form (Except Hg)
• These are good conductors of heat and electricity.
• These have high m.pts and b.pts.
• Generally these are electropositive.
• These forms more ionic compounds.

Nón metals

• These may be solids (or) gases (or) liquids.
• These are not good conductors of heat and electricity.
• These have low m.pts and b.pts.
• Generally these are electronegative.
• These forms more covalent compounds.

Question 20.
Use the periodic table to identify elements.
a. With 5 electrons in the outer subshell
b. Would tend to lose two electrons
c. Would tend to gain two electrons.
Answer:
a) The elements possessing 5 electrons in the outer most shell are group 15 (V_A) elements.

• General outer electronic configuration is ns² np³ Eg. : N, P, As………

b) The elements tend to lose two electrons are Group — II elements.

• General outer electronic configuration is ns² Eg: Mg, Ca, Sr etc.

c) The elements tend to gain two electrons are Group – VI_A elements (16^th group).

• General outer electronic configuration is ns² np⁴ Eg : O, S, Se ………

Question 21.
Give the outer electronic configuration of s, p, d and f – block elements.
Answer:

Question 22.
Write the increasing order of the metallic character among the elements B, Al, Mg and K.
Answer:
Given elements are B, Al, Mg and K
The increasing order of metallic character is

Question 23.
Write the correct increasing order of non – metallic character for B, C, N, F and Si.
Answer:
Given elements are B, C, N. F and Si
The increasing order of non- metallic character is

Question 24.
Write the correct increasing order of chemical reactivity in terms of oxidizing property for N, O, Fand Cl.
Answer:
The correct increasing order of chemical reactivity in terms of oxidizing property for N, O, F and Cl is F > 0 > Cl > N.

Question 25.
What is electronegativity ? How is this useful in understanding the nature of elements?
Answer:
Electronegativity : The tendency of an element (or) atom to attract the shared pair of electrons towards itself in a molecule is called electronegativity.

• On the basis of electronegativity values nature of elements can be predicted. Higher electro-negativity values indicates that element is non metal and lower values indicates that the element is a metal.
• On the basis of electronegativity values bond nature also predicted (Ionic/covalent).

Question 26.
What is screening effect? How is it related to lE?
Answer:
The decrease of nuclear attraction on outer most shell electrons due to presence of inner energy electrons is called screening effect.

• As the screening effect increases LE. values decreases.

Question 27.
How are electronegativity and metallic & non-metallic characters related?
Answer:

• Greater the electronegativity values of an element indicates that non metallic nature and low metallic nature of that element.
• Lower the electronegativity value of an element indicates that low non metallic nature and high metallic nature of that element.

Electronegativity ∝ Non metallic nature
Electronegativity ∝ \(\frac{1}{\text { Metallic Nature }}\)

Question 28.
What is the valency possible to Arsenic with respect to oxygen and hydrogen?
Answer:

• The valency of Arsenic with respect to hydrogen is ‘3’
  Eg : AsH₃
• The valency of Arsenic with respect to oxygen is ‘5’
  Eg : As₂O₅

Question 29.
What is an amphoteric oxide? Give the formula of an amphoteric oxide formed by an element of group – 13.
Answer:
The oxide which contains both acidic as well s basic nature is called amphoteric oxide.

• The oxides reacts with both acids and bases and forms salts.
  Eq : Al₂O₃ is one of the amphoteric oxide formed by the Group — 13 element Aluminium.

Question 30.
Name the most electronegative element. Is it also having the highest electron gain enthalpy? Why or Why not?
Answer:
The most electronegative element is fluorine (F).

• It doesnot have high electron gain énthalpy.

Reasons : —

• Due to small size
• Due to high inter electronic repulsions.
• Chlorine has high electron gain enthalpý.

Question 31.
What is diagonal relation? Give one pair of elements, that have this relation.
Answer:
On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship. e.g. : (Li, Mg); (Be, Al); (B, Si)

Question 32.
How does the nature of oxides vary in the third period?
Answer:
In 3^rd period from left to right the oxide nature varies from high basic nature to high acidic nature.

• Basic nature gradually decreases and acidic nature gradually increases.
• 

Question 33.
Radii of iron atom and its ions follow Fe > Fe²⁺ > Fe³⁺ – explain.
Answer:
When the positive charge on the ion increases, the effective nuclear charge on the outer electrons increases.
Hence the ionic size decreases in the order Fe > F²⁺ > F³⁺.

Question 34.
IE₂ > IE₁ for a given element — why?
Answer:
IE₂ > IE₁ for a given element

Reason : —

• IE₁ means minimum amount of energy required to remove an electron from isolated neutral
  gaseous atom.
• IE₂ means minimum amount of energy required to remove an electron from uni positive ion.
• In case of unipositive ion nuclear attraction increases on outer most electrons than in isolated gaseous atom. So more amount of energy needed to remove an electron from unipositive ion.
  Hence IE₂ > → IE₁.

Question 35.
What is Ianthanide contraction? Give one of its consequences.
Answer:
Lanthanide contraction : Slow decrease in size of the atoms or ions among the lanthanides is known as lanthanide contraction.

Consequences due to lanthanide contraction:

1. Due to lanthanide contraction, the crystall structure and other properties of the elements become very closely similar.
2. Due to this, it becomes difficult to separate lanthanides from a mixture.

Question 36.
What is the atomic number of the element, having maximum number of unpaired 2p electrons ? To which group does it belong ?
Answer:

• The atomic number of the element having maximum no.of unpaired 2p electrons is 7′ (Z = 7)
• Element is nitrogen.
• Electronic configuration is 1s² 2s² 2p³ (3 unpaired 2p electrons)

Question 37.
Sodium is strongly metallic, while chlorine is strongly non-metallic – explain.
Answer:
Sodium is an alkali metal and it is present in group – I and it has the ability to lose the valency electron readily.

• It has high electropositive nature. So it has metallic nature.
• Chlorine is a halogen and it is placed in Group – 17 and it has the ability to gain the electron readily.
• It has high electronegative nature. So it has non metallic nature.

Question 38.
Why are zero group elements called noble gases or inert gases ?
Answer:

• Zero group elements has general outer electronic configuration ns² np⁶ (except for He).
• These contains stable octet configuration. So these are stable and chemically inert. Hence these are called inert gases.
• These elements neither lose nor gain electrons. Hence these are called ‘noble gases’.

Question 39.
Select in each pair, the one having lower ionization energy and explain the reason,
a. I and I^–
b. Br and K
c. Li and Li⁺
d. Ba and Sr
e. O and S
f. Be and B
g. N and O
Answer:
a) I^– has lower ionisation energy than I because of increase of size \(\mathrm{I}^{\ominus}\) ion than ‘I’.
b) K has lower ionisation energy than ‘Br’ because of low electronegative value of K (0.8) than ‘Br’ (2.8).
c) ‘Li” has lower ionisation energy than Li+ because of large size of ‘Li’ than Li⁺.
d) ‘S’ has lower ionisation energy than ‘O’ because of large size of ‘S’ than ‘O’.
e) ‘B’ has lower ionisation energy than ‘Be’ because ‘Be’ has completely filled electronic configuration (1s² 2s²).
f) ‘O’ has lower ionisation energy than ‘N’ because ‘N’ has half filled electronic configuration (1s² 2s² 2p³).

Question 40.
IE₁ of O < IE₁ of N but IE₂ of O > IE₂ of N – Explain.
Answer:

• ‘N’ has half filled electronic configuration (1s² 2s² 2p³)
  So IE₁ of O < IE₁ of ‘N’.
• O⁺ ion has half filled electronic configuration (1s² 2s² 2p³)
  So IE₂ of O > IE₁ of N.

Question 41.
Na⁺ has higher value of ionization energy than Ne, though both have same electronic configuration – Explain.
Answer:
Na⁺ has higher value of I.E. than Ne, though both have same electronic configuration.

Reason : –

• Both have electronic configuration 1s² 2s² 2p⁶
• In case of Na⁺ ion effective nuclear charge increases and size decreases than in ‘Ne’.

Question 42.
Which in each pair of elements has a more electronegative gain enthalpy ? Explain.
a. N or O
b. F or Cl
Answer:
a) Oxygen has high electronegative gain enthalpy than Nitrogen because ‘N’ has stable half filled electron configuration.
b) Chlorine (- 349 KJ /mole) has high electronegative gain enthalpy than Fluorine (- 328 KJ/mole) because ‘F’ has small size and more inter electronic repulsions.

Question 43.
Electron affinity of chlorine is more than that of fluorine – explain.
Answer:
Chlorine (- 349 KJ / mole) has high electronegative gain enthalpy than Fluorine (- 328 KJ/mole) because ‘F’ has small size and more inter electronic repulsions.

Question 44.
Which in each has higher electron affinity ?
a. F or Cl^–
b. O or O-
c. Na⁺ or F
d. F or F^–
Answer:
a) Fluorine has high electron affinity than Cl^– ion because of inert gas configuration of Cl^– ion.
b) Oxygen has high electron affinity than O^– because O^– has positive of 2nd electron affinity.
c) F has high electron affinity than Na⁺ because Na⁺ has inert gas configuration.
d) F has high electron affinity than F^– because F^– has inert gas configuration.

Question 45.
Arrange the following in order of increasing ionic radius :
a. Cl^–, P^-3, S^-2, F^–
b. Al⁺³, Mg⁺², Na⁺, O^-2, F^–
c. Na⁺, Mg⁺², K⁺
Answer:
a) The increasing order of ionic radius is F^– < Cl^– < S^-2 < P^-3
b) The increasing order of ionic radius is Al⁺³ < Mg⁺² < Na⁺ < F^– < O^-2
c) The increasing order of ionic radius is Mg⁺² < Na⁺ < K⁺

Question 46.
Mg⁺² is smaller than O^-2 in size, though both have same electronic configuration –
explain.
Answer:
Mg⁺² and O^-2 ions are iso electronic species.
In case of iso electronic species nuclear charge increases size of ion decreases. So Mg⁺² has small size than O^-2.

Question 47.
Among the elements B, Al, C and Si
a. Which has the highest first ionization enthalpy ?
b. Which has the most negative electron gain enthalpy ?
c. Which has the largest atomic radius ?
d. Which has the most metallic character ?
Answer:
a) Highest I.E. is possessed by the element carbon
b) Most negative gain enthalpy is for carbon (- 122 KJ/mole)
c) Large atomic radius is for Al (1.43 A)
d) Most metallic nature having element is ‘Al’.

Question 48.
Consider the elements N, P, O and S and arrange them in order of ;
a. Increasing first ionization enthalpy
b. Increasing negative electron gain enthalpy
c. Increasing non-metallic character
Answer:
a) Increasing first Ionisation energy order is S < P < O < N.
b) Increasing negative electron gain enthalpy order is N < P < O < S. .
c) Increasing non metallic nature order is P < N < S < O.

Question 49.
Arrange in given order :
a. Increasing EA :O, S and Se
b. Increasing IE₁ : Na, K and Rb
c. Increasing radius : I^–, I⁺ and I
d. Increasing electronegativity : F, Cl, Br, I
e. Increasing EA : F, Cl, Br, I
f. Increasing radius : Fe, Fe⁺², Fe⁺³
Answer:
a) Increasing order of electron affinity is O < Se < S.
b) Increasing order of IE₁ is Rb < K < Na.
c) Increasing order of radius is I⁺ < I < I^–
d) Increasing order of electronegativity is I < Br < F < Cl
e) Increasing order of electron affinity is I < Br < F < Cl
f) Increasing order of radius is Fe⁺³ < Fe⁺² < Fe

Question 50.
a. Name the element with highest ionization enthalpy
b. Name the family with highest value of ionization enthalpy.
c. Which element possesses highest electron affinity ?
d. Name unknown elements at the time of Mendeleef
e. Name any two typical elements.
Answer:
a) Highest I.E₁ possessing element is ‘Helium’.
b) The family that possess highest values of I.E is Noble gases (or) inert gases.
c) Highest electron affinity element is ‘Chlorine’.
d) Unknown elements at the time of mendeleef are Germanium (Eka silicon). Scandium (Eka Alu-minium), Gallium (Eka Boron).
e) 3^rd period elements are called typical elements
Eg : Al, Si, P, Na, Mg etc

Question 51.
a. Name any two bridge elements.
b. Name two pairs showing diagonal relationship.
c. Name two transition elements.
d. Name two rare earths.
e. Name two transuranic elements.
Answer:
a) Bridge elements 2nd period elements are called Bridge elements Eg : Be, B.
b) i) ‘Li’ diagonally relates with ‘Mg’,
ii) ‘Be diagonally relates with ‘Al’.
c) Scandium, Titanium, Vanadium, Chromium, etc., are examples of transition elements.
d) Lanthanides are called rare earths
Eg : Cerium (Ce), Prasodimium (Pr), Promethium (Pm).
e) Neptunium (Np), Californium (Cf), Fermium (Fm) are examples of transuranic elements.

Question 52.
On the basis of quantum numbers, justify that the 6^th period of the periodic table should have 32 elements.
Answer:
6^th period contains the subshells 6s, 4f, 5d, 6p
6s can accomodate two electrons (2 elements)
4f can accomodate 14 electrons (14 elements)
5d can accomodate 10 electrons (10 elements)
6p can accomodate 6 electrons (6 elements)

Total no.of electrons can accomodation 6^th period are 2 + 14 + 10 + 6 = 32
∴ 6^th period of periodic table contains 32 elements.

Question 53.
How did Moseley’s work on atomic numbers show that atomic number is a fundamental property better than atomic weight ?
Answer:
Mosley’s equation is

where v = frequency, Z = atomic number a, b = constants
A plot of \(\sqrt{v}\) against ‘Z’ gives a straight line.
However, no such relationship was obtained when the plot was drawn between frequency and the atomic mass. The atomic number of the elements, according to Mosley, stands for serial numbers of the elements in the periodic table. As the atomic number of the elements increase, the wavelengths of characteristic X – rays decrease. Mosley concluded that there is a fundamental quantity in an atom which increases in regular steps with increasing atomic number. The correlation between X – ray spectra and atomic number indicated that an element is characterized by its atomic number and not by atomic mass.

Question 54.
State modern periodic law. How many groups and periods are present in the long form of the periodic table ?
Answer:
Modern periodic law : – The physical and chemical properties of elements are periodic functions of their atomic numbers.

• In modern periodic table 7 periods and 18 groups are present.

Question 55.
Why f-block elements are placed below the main table ?
Answer:
In Lanthanides 4f – orbitals and in Actinides 5f – orbitals are filled. Since these elements have the same electronic configuration in the ultimate and penultimate shells they have similar properties. Hence they were placed at the bottom of the periodic table though they belongs to the sixth and seventh periods of IIIB groups.

Question 56.
Mention the number of elements present in each of the periods in the long form periodic table.
Answer:

Question 57.
Give the outer orbit general electronic configuration of
a. Noble gases
b. Representative elements
c. Transition elements
d. Inner transition elements
Answer:
Type of elements – General electronic configurations
a) Noble gases – ns² np⁶ (except ‘He’ which has 1s²)
b) Representative elements – ns^1-2 np^0-5
c) Transition elements – (n – 1)d^1-10ns^1-2
d) Inner transition elements – (n – 2) f^1-14(n – 1)d^{0, 1} ns²

Question 58.
Give any four characteristic properties of transition elements.
Answer:
Characteristic properties of elements:
a) They exhibit more than one oxidation state.
b) Most of the elements and their ions exhibit colour.
c) These elements and their compounds are good catalysis for various chemical processes.
d) They and their ions exhibit paramagnetic properties.
e) They form useful alloys.

Question 59.
What are rare earths and transuranic elements?
Answer:

1. Lanthanides are rare earths. In these elements the differentiating electron enters into 4f – orbital.
2. The elements present after Uranium are called Transurariic elements. All of these are radioactive and synthetic elements.

Question 60.
What is isoelectronic series? Name a series that will be isoelectronic with each of the following atoms or ions.
a. F^–
b. Ar
c. He
d. Rb⁺
Answer:
The species containing same no.of electrons are called Iso electronic species and this series is called Isoelectronic series.
a) F^– relating series
N^-3, O^-2, F^–, Ne, Na⁺, Mg⁺², Al⁺³
b) Ar relating series P^-3, S^-2, Cl^–, Ar, K⁺, Ca⁺²
c) ‘He’ relating series H^–, He, Li⁺, Be⁺²
d) Rb⁺ relating series
As^-3, Se^-2, Br^–, Kr, Rb⁺, Sr⁺²

Question 61.
Explain why cation is smaller and anion is larger in radii than their parent atoms.
Answer:

• Cation means a positively charged species which is formed by an atom (or) element when an electron is lost.
  M → M⁺ + \(\mathrm{e}^\Theta\)
• Cation has high effective nuclear charge and decrease in size observed.
  Hence cation has smaller radii.
• Anion means a negatively charged species which is formed by an atom (or) element when an electron is gained.
  M + e^– → \(\mathrm{M}^{\Theta}\).
• Anion has very low effective nuclear charge and increase in size observed.
  Hence anion has larger radii.

Question 62.
Arrange the second period elements in the increasing order of their first ionization enthalpies. Explain why Be has higher IE₁ than B.
Answer:
The increasing order of the I.E.s of 2^nd period elements are as follows.

• Due to presence of incompletely filled p – orbitals in Boron, its IE value is less.
• Due to presence of completely filled s² configuration in ‘Be’, it has higher IE value.

Question 63.
IE₁ of Na is less than that of Mg but IE₂ of Na is higher than that of Mg – explain.
Answer:
IE₁ of Na is less than that of ‘Mg’
Reason :-
Na — has electronic configuration [Ne] 3¹
Mg — has electronic configuration [Ne] 3s²
Mg has completely filled configuration so Mg has more IE₁ than Na.
IE₂ of Na is higher than that of Mg

• 
• Na⁺ has stable inert gas configuration so IE₂ of Na is very high
  
• By the lose of one electron from Mg⁺ ion forms Mg⁺² ion which is more stable so low amount of energy is required.
  ∴ IE₂ of Na is higher than Mg.

Question 64.
What are the various factors due to which the IE of the main group elements tends to decrease down a group ?
Answer:
The factors influencing on IE are

1. Atomic radius
2. Nuclear charge
3. Screening effect
4. Half filled, completely filled electronic configurations
5. peretrating power
   • In main group elements in a group IE decrease from top to bottom.

Reason: –
In groups from top to bottom size of elements increases hence IE values decreases.

Question 65.
The first ionization enthalpy values (in KJ mol^-1) of group 13 elements are :

How do you explain this deviation from the general trend ?
Answer:
The given IE, values (in KJ / mole) of group 13 are as follow

• In genaral in a group IE values decrease down a group but from the above values we observe that there is no smooth decrease down the group.
• The decrease from B to Al is due to increase of size.
• The observed discontinuity in the IE values of Al and Ga, and between In and Tl are due to inability of d- and f – electrons, which have low screening effect to compensate the increase in nuclear charge.

Question 66.
Would you expect the second electron gain enthalpy of oxygen as positive, more negative or less negative than the first ? Justify.
Answer:
2^nd gain enthalpy means energy released when an electron is added to uni negative ion.
O_(g) + e^– → \(\mathrm{O}_{(g)}^{-}\) + 141 KJ/mole
O_(g) + e^– → \(O_{(g)}^{-(2)}\) – 780 KJ/mole

• 2^nd gain enthalpy of oxygen is positive because O^– ion doesnot accept an electron readily and the entering electron have to over come the repulsive force.

Question 67.
What is the basic difference between the electron gain enthalpy and electropositivity?
Answer:

• Electron gain enthalpy means energy released when an electron is added to isolated neutral gaseous atom.
• The tendency to lose the electrons by an element is called electropositivity.
• Electron gain enthalpy is the measure of electronegativity.
• Electron gain enthalpy and electropositivity are inversely related.

Question 68.
Would you expect IE₁ for two isotopes of the same element to be the same or different ? Justify.
Answer:

• Isotopes means existance of same elements with different mass no.s
• The isotope with higher mass no. have low I.E value than the normal isotope.
• This is due to the less nuclear attraction on valency electrons in case of heavier nuclide.
• But overall, the IE values of the isotopes are nearly same and the difference in IE values is negligible.

Question 69.
Increasing order of reactivity among group-1 elements is Li < Na < K < Rb < Cs, where as among group-17 elements it is F > Cl > Br > I- explain.
Answer:
a) Increasing order of reactivity among group -1 elements is Li < Na < K < Rb < Cs

Explanation: –

• Group -1 elements are Alkali metals.
• Group -1 elements have the tendency to lose the electrons
• Group -1 elements forms ionic bonds readily by losing electrons
• Group – 1 elements good reducing agents.
• Electro positive character increases from top to bottom this is due to increase of size.

b) Increasing order of reactivity among group -17 elements is F > Cl > Br > I

Explanation: –

• These are halogens (Group – 17)
• These have high electronegativity due to small size.
• These have the tendency to gain electrons.
• In a group from top to bottom electronegativity decrease, due to increase of size.
• These are oxidising agents and forms ionic bonds by gaining electrons.

Question 70.
Assign the position of the element having outer electronic configuration.
a. ns²np⁴ for n = 3
b. (n – 1)d²ns² for n = 4
Answer:
a) ns²np⁴ for n = 3 .

• 3s²3p⁴ → element is sulphur
• Sulphur belongs to VIA group (Group – 16) and 3^rd period in periodic table,

b) (n – 1 )d² ns² for n = 4
3d² 4s² → element is Titanium

• Titanium belongs to IVB group (Group – 4) and 4^th period in the periodic table.

Question 71.
Predict the formulae of the stable binary compounds that would be formed by the combination of the following pairs of elements.
a. Li and O
b. Mg and N
c. Al and I
d. Si and O
e. p and Cl,
f. Element with atomic number 30 and Cl
Answer:
a) Stable binary compound formed between Li and O is Li₂O (Lithiumoxide).
b) Stable binary compound formed between Mg and N is Mg₃N₂ (Magnesium nitride).
c) Stable binary compound formed between Al and I is AlI₃ (Aluminium Iodide).
d) Stable binary compound formed between Si and I is SiO₂ (Silicondioxide).
e) Stable binary compound formed between P and Cl is PCl₃ and PCl₅ (Phosphorous trichloride and phosphorous pentain chloride).
f) Stable binary compound formed between element with At. NO – 30 and Cl is ZnCl₂ (Zinc chloride) [At. No – 30 – (Zn)].

Question 72.
Write a note on the variation of metallic nature in a group or in a period.
Answer:
Metals shows electropositive nature (i.e.,) loss of electrons and form positive ions.
Non – metals shows electronegative nature (i.e.,) gain of electrons and form negative ions.
Periodicity:
a) Down the group : Going down a group of the periodic table, the tendency to form positive ions, increases. That means there is an increase in metallic nature as the size of atom increases down the group.
b) Along a period : From left to right in a period, size of atom decreases. So there is decrease in metallic nature.

Question 73.
How does the covalent radius increase in group- 7 ?
Answer:

• Covalent radius increases in a group from top to bottom
• The increase of covalent radius in VIIA group elements as follows.
  

Question 74.
Which element of 3rd period has the highest IE₁ ? Explain the variation of IE₁ in this period.
Answer:

• In periods IE values increase from left to right
• Among 3^rd period elements Argon (Ar) [Z = 18] possess highest IE,sub>1 value.
• IE₁ values of 3^rd period elements given below.
  

Exceptions:

• IE, of ‘Mg’ is higher than ‘Al’ because ‘Mg’ has completely filled ‘s’ subshells.
• IE, of ‘p’ is higher than ‘s’ because ‘p’ has half filled ‘p’ subshells.

Question 75.
What is valency of an element ? How does it vary with respect to hydrogen in the third period?
Answer:
Valency : The combining capacity of an element with another element is called valency.

The number of hydrogen atoms (or) chlorine atoms (or) double the number of oxygen atoms, with which one atom of the element combine is also called valency.
∴ Valency = no. of hydrogens = no. of chlorine atoms
= 2 × no. of oxygen atoms present in the molecule.
e.g.:

Periodicity of valency:
1) Each period starts with valency ‘1’ and ends in ‘0’.

2) In a group valency is either equal to the group number (i.e., upto 4^th group) or is equal to (8 – group number) (i.e., from 5^th group onwards).
Significance :
Valency of an element is useful in writing the formulae of compounds.

Question 76.
What is diagonal relationship ? Give a pair of elements having diagonal relationship. Why do they show this relation ?
Answer:
On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship, e.g. : (Li, Mg); (Be, Al); (B, Si)

Diagonal relationship is due to similar sizes of atoms (or of ions) and similar electronegativities of the representative elements.
Diagonally similar elements posses the same polarizing power.
Polarizing power = \(\frac{\text { (ionic charge) }}{\text { (ionic radius) }^2}\)

Question 77.
What is lanthanide contraction ? What are its consequences ?
Answer:
Lanthanide contraction : Slow decrease in size of the atoms or ions among the lanthanides is known as lanthanide contraction.

Consequences due to lanthanide contraction:

1. Due to lanthanide contraction, the crystall structure and other properties of the elements become very closely similar.
2. Due to this, it becomes difficult to separate lanthanides from a mixture.

Question 78.
The first IP of lithium is 5.41 eV and electron affinity of Cl is – 3.61 eV. Calculate ΔH in kJ mol^-1 for the reaction : Li_(g) + Cl_(g) → \(\mathrm{Li}_{(\mathrm{g})}{ }^{+}\) + \(\mathrm{Cl}{ }_{(g)}^{-}\)
Answer:
Given reaction is

ΔH = ΔH₁ + ΔH₂
= 5.41 – 3.61
= 1.8 ev
= 1.8 × 9.65 × 10⁴ J/mole
= 17.37 × 10⁴ J/Mole
= 173.7 KJ/mole

Question 79.
How many Cl atoms can you ionize in the process Cl → Cl⁺ + e by the energy liberated for the process Cl + e → Cl^– for one Avogadro number of atoms. Given IP = 13.0 eV and EA = 3.60 eV. Avogadro number = 6 × 10²³
Answer:
Given
Cl_(g) + e^– → \(\mathrm{Cl}_{(\mathrm{g})}^{-}\) ΔH = – 3.6ev
1 – atom → Electron affinity = 3.6 ev
6.23 × 10²³ atoms → ?
6.23 × 10²³ × 3.6 = 21.6828 × 10²³
∴ For one avagadro no.of Cl atoms electron affinity = 21.6828 × 10²³ eV
Given 13 eV can ionise 1 atom of Cl
21.6828 × 10²³ eV ionise -?
\(\frac{21.6828 \times 10^{23}}{13}\) = 1.667 × 10²³ eV

Question 80.
The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2g of chlorine atoms is completely converted to Cl^– ions in the gaseous state ? (1 e V = 23.06 kcal)
Answer:
Given electron affinity of Cl = 3.7 ev
Cl + e^– → Cl^– ΔH = – 3.7ev
35.5 gms of CZ contains 6.023 × 10²³ atoms
2 gms of Cl contains ?
= \(\frac{2 \times 6.023 \times 10^{23}}{35.5}\) atoms
6.23 × 10²³ atoms can liberate 3.7 eV
\(\frac{2 \times 6.023 \times 10^{23}}{35.5}\) atom can liberate?
= \(\frac{2 \times 6.023 \times 10^{23} \times 3.7}{35.5 \times 6.023 \times 10^{23}}\) = \(\frac{2 \times 3.7}{35.5}\) = \(\frac{7.4}{35.5}\)
= 0.2084 eV
= 0.2084 × 23.06 kcal/mole
= 4.81 k.cal/mole

Long Answer Questions

Question 1.
Discuss the classification of elements by Mendaleev’
Answer:
The periodic classification of elements based on “atomic weights” was done by “Lothar Meyer” and “Mendeleev” independently.
Mendeleev’s periodic law : “The physical and chemical properties of elements and their compounds are a periodic function of their atomic weights”.

Mendeleev arranged the 65 elements in a periodic table. He did not blindly follow the atomic weight but gave more importance to their chemical properties in arranging them in the table.
Explanation of the periodic law : When the elements are arranged in the increasing order of their atomic weights, elements with similar properties appear again and again, at regular intervals. This is called, periodicity of properties.

Mendeleev’s table : Mendeleev introduced a periodic table containing the known 65 elements. In this table, while arranging the elements, he gave importance only to their atomic weights, but also to their physical and chemical properties. This table was defective in some responses. Then he introduced another table, after rectifying the defects of that table. lt is called, “short form of periodic table”. He named the horizontal rows as ‘periods’ and the vertical columns, as ‘groups’. It has in all ‘9’ groups, I to VIII and a ‘O’ group. The first ‘7’ groups were divided into A and B sub groups. There are ‘7’ periods in the table. The VIII group contains three triods, namely, (Fe, Co, Ni), (Ru, Rh, Pd) and (Os, Ir, Pt).

Mendeleev’s observations :

1. When the elements are arranged according to their atomic weights, they exhibit periodicity of properties.
2. Elements with similar chemical properties have nearly equal atomic weights: Iron (55.85),. Cobalt (58.94) and Nickel (58.69).
3. The group number corresponds to the valency of element in that group.
4. Most widely distributed elements like H,C, O, N, Si, S etc., have relatively low atomic weights.
5. The atomic weight of an element may be corrected if the atomic weights of the adjacent elements are known. The properties of an element are the average properties of the neighbouring elements.
6. The atomic weights of beryllium. Indium, Uranium etc. were corrected, based on this observation.

Merits of Mendeleev’s table:

1. Actually it formed the basis for the development of other modern periodic tables.
2. Mendeleeffs left some vacant spaces in his periodic table, for the unknown elements. But the predicted the properties of those elements. Later on, when these elements were discovered, they exactly fitted into those vacant places having properties, predicted by Mendeleev.
   Ex : E_ka – boron (Scandium), E_Ka – Silicon (germanium)
   E_Ka – aluminium (gallium) etc.
3. ‘O’ group elements were not known at the time of Mendeleeff. Later when they were discovered, they found a proper place in that table under ‘0’ group of elements. Similarly, the radioactive elements.
4. In case of these pairs of elements Tellurium – Iodine Argon – Potassium and Cobalt – Nickel, there is a reversal of the trend. The first element has higher atomic weight than the second one. These are called anomalous pairs.
   However, based on their atomic numbers, and chemical properties, this arrangement proves quite justified.

Draw-backs of Mendeleev’s periodic table :

1. Dissimilar elements were placed in the same group.
   Ex : The coinage mentals Cu, Ag and Au are placed along with the alkali metals K, Rb, Cs etc. in the I group. The only common property among them is that they are all univalent (Valency = 1).
2. The 14 rare earths having different atomic weights are kept in the same place.
3. Hydrogen could not be given a proper place, as it resembles alkali metals and halogens in its properties.

Question 2.
From a study of properties of neighbouring elements, the properties of an unknown element can be predicted – Justify with an example.
Answer:
From a study of adjacent elements and their compounds, it is possible to predict the characteristics of certain elements. These predictions were found to be very accurate. These predicted properties helped the future scientists in the discovery of unknown elements, e.g. : E_Ka Aluminium (E_Ka Al) (now known as Gallium); E_Ka Silicon (E_Ka Si) (now known as Germanium); E_Ka Boron (E_Ka B) (now known as Scandium).

Illustration : Following table shows a comparison of the properties predicted by’Mendeleeff’ for the elements and those found experimentally after their discovery.

Question 3.
Discuss the construction of long form periodic table.
Answer:
The elements are arranged in the long form of the periodic table in the increasing order of atomic numbers. ‘Neils Bohr’ constructed the long form of the periodic table based on electronic configuration of elements.
The important features of the long form of the periodic table are :
It consists seven horizontal rows called periods and 18 vertical columns which are classified into 16 groups only.

Periods : Every period starts with an alkali metal and ends with an inert gas. The first period consists of two elements only (H, He) and is called very short period. Second period consists 8 elements (Li to Ne) and is called first short period. The third period consists (Na to Ar) 8 elements and is called second short period.

Fourth period contains 18 elements (K to Kr) and is called first long period. Fifth period is the second long period with 18 elements (Rb to Xe).
Sixth period is the longest period with 32 elements which starts with Cs and ends with Rn. This period includes 14 lanthanides.
Seventh period is an incomplete period with 20 radioactive elements.
Groups : There are 16 groups in the long form of the periodic table (in transition elements three vertical columns are fused and designated as VIII group). These groups are IA, IIA, NIB, IVB, VB, VIB, VIIB, VIII, IB, MB, IMA, IVA, VA, VIA, VIIA and zero group.

The elements of IA, IIA, IIIA, IVA, VA, VIA, VIIA groups are called representative elements or normal elements. Elements of IB, MB, NIB, IVB, VB, VIB, VIIB and VIII groups have their ultimate and penultimate shell incomplete. These are called transition elements. MB elements have (n – 1) d¹⁰ ns² outermost electronic configuration.

Zero group elements have stable electronic configuration. These elements are called inert gases, noble gases. These elements have been grouped at the extreme right of the periodic table.
In this long periods have been expanded and short periods are broken to accommodate the transitional elements in the middle of the long period.

Lanthanides and actinides have been grouped separately and placed at the bottom of the periodic table.

Question 4.
Discuss the relation between the number of electrons filled into the sub energy levels of an orbit and the maximum number of elements present in a period.
Answer:
Elements have been accommodated in these periods according to the following scheme.
1st period : The first main energy shell (K – shell) is completed. As the maximum capacity of K- shell is of 2 electrons, it consists of only two elements, hydrogen (1s¹)and helium (1s²).
2nd period : The second main energy shell is completed (i.e.,) 2s and 2p are completed. It includes eight elements from Li (2s¹) to Ne (2s²2p⁶).
3rd period : The 3s and 3p energy shells are completed. It includes also eight elements from Na(3s¹) to Ar (3s² 3p⁶).
4th period : The 4s, 3d and 3p energy shells are completed. It includes 18 elements from K(4s¹) to Kr (3d¹⁰ 4s²4p⁶). It includes two s – block elements, ten d – block elements and six p – block – elements.
5th period : The 5s, 4d and 5p energy shells are completed. It includes 18 elements from Rb (5s) to Xe (4d¹⁰ 5s¹ 5p⁶). 1st also includes two s – block elements, ten d – block elements and six p – block elements.
6th period : The 6s, 4f, 5d and 6p energy shells are completed, (i.e.,) it includes 32 elements from Cs (6s¹) to Rn (4f¹⁴ 5d¹⁰ 6s²6p⁶). It consists of two s – block elements, ten d – block elements, six p – block elements and fourteen f – block elements.
As this period can accommodate only 18 elements in the table, 14 members of 4f – series are separately accommodated in a horizontal row below the periodic table.
7th period : This period at present consists 26 elements. The 7s, 5f and 6d are completed (i.e.,) twenty six elements. Seven of the elements from atomic numbers 106 to 112 have recently been reported.

Question 5.
Write an essay on s, p, d and f block elements .
Answer:
According to the electronic configuration of elements, the elements have been classified into four blocks. The basis for this classification is the entry of the differentiating electron into the subshell. They are classified into s, p, d and f blocks.
‘s’ block elements : If the differentiating electron enters into ‘s’ orbital, the elements belongs to ‘s’ block.

In every group there are two ‘s’ block elements. As an ‘s’ orbital can have a maximum of two electrons, ‘s’ block has two groups IA and IIA.
‘p‘ block elements : If the differentiating electron enters into ‘p’ orbital, the elements belongs to ‘p’ block.

‘p’ block contains six elements in each period. They are IIIA to VIIA and zero group elements. The electronic configuration of ‘p’ block elements varies from ns²np¹ to ns²np⁶.

‘d‘ block elements : It the differentiating electron enters into (n – 1) d – orbitals the elements belongs to’d’ block. These elements are in between ‘s’ and ‘p’ blocks. These elements are also known as transition elements. In these elements n and (n – 1) shells are incompletely filled. The general electronic configuration of’d’ block elements is (n – 1) d^1-10 ns^1-2. This block consists of IIIB to VIIB, VIII, IB and IIB groups.

‘f’ block elements : If the differentiating electron enters into ‘f’ orbitals of antipenultimate shell (n – 2) of atoms of the elements belongs to ‘f block. They are in sixth and seventh periods in the form of two series with 14 elements each. They are known as lanthanides and actinides and are arranged at the bottom of the periodic table. The general electronic configuration is (n – 2)f^1-14 (n – 1)d^{0 – 1} ns².

In these shells the last three shells (ultimate, penultimate and anti penultimate) are incompletely filled. Lanthanides belongs to 4f series. It contains Ce to Lu. Actinides belong to 5f series. It contains Th to Lr.

Advantages of this kind of classification :

As a result of this classification of elements were placed in correct positions in the periodic table. It shows a gradual gradation in physical and chemical properties of elements. The metallic nature gradually decreases and non – metallic nature gradually increases from’s1′ block to ‘p1 block. This classification gave a special place for radioactive elements.

Question 6.
Relate the electronic configuration of elements and their properties in the Classification of elements.
Answer:
The chemical properties of all elements depends upon the electronic configuration. Upon the basis of complete and incomplete electron shells and chemical properties, the elements are classified into four types. (Type -1, Type – II, Type – III and Type – IV).

Type – I(inert gas elements): All the elements with an electronic configuration ns² np⁶ including He belongs to this type, nth shell of those elements are completely filled.
The elements show chemical inertness due to completely filled shells and hence they have extra stability. Because of their stability, they are chemically inactive.
e.g. : He, Ne, Ar, Kr, Xe and Rn.

Type – II (Representative elements) : Except inert gases, the remaining elements of s and p – blocks are called representative elements. All the elements with an electronic configuration ns¹ to ns² np⁵ excluding. He comes under this type. These are atoms in which all except the outermost shells (n^th) are complete. Elements of this type enter into chemical combination by loosing, gaining or sharing electrons to get stable inert gas configuration. Many of the metals, all non – metals and metalloids come under this type. Chemically these elements are reactive.

Type – III (Transition elements): All the elements with an electronic configuration (n – 1) d^{1 – 9} ns^1or2 belongs to this type. Atoms in which the two outermost shells are incomplete. These elements show variable oxidation states, form complex ions and coloured ions. The electronic configuration of d – block elements is (n – 1) d^{1 – 10} ns^{1 – 2}. Small size, high nuclear charge and unpaired’d’ orbitals impart characteristic properties to be transition elements.

Type- IV (Inner transition elements): All the elements with an electronic configuration (n – 2) f^{1 – 14} (n -1) d^{0, 1} ns² belongs to this type. Atoms in which three outermost shells are incomplete. Lanthanides and Actinides belong to this type.

Question 7.
What is a periodic property ? How the following properties vary in a group and in a period ? Explain
a) Atomic radius
b) Electron gain enthalpy.
Answer:
Recurrence of similar properties of elements at definite regular intervals with increasing atomic number i.e., according to their electronic configurations is known as periodicity. Any property which is periodic in nature is called periodic property.
a) Atomic radius : The atomic radius decreases from left to right in a period. With an increase in the atomic number in a period the nuclear charge increases. As a result the effective nuclear charge over the outermost electrons increases, due to this the orbitals are pulled closer to the nucleus causing in a decrease in the atomic radius.

The atomic radius increases from top to the bottom in a group because – with an increase in the atomic number the electrons are added to new shells resulting an increase in the number of inner shells. Hence atomic radius increases from top to bottom in a group.

b) Electron gain enthalpy : Electron gain enthalpy increases in a period from left to right because the size of the atom decreases and the nature of the element changes from metallic to non – metallic nature when we move from left to right in a period.

Electron gain enthalpy decreases from top to bottom in a group because there is an increase in the atomic size. But the second element has greater electron gain enthalpy than the first element.
e.g. : Chlorine has more electron affinity value (- 348 kJ mol^-1) than Fluorine (-333 kJ mol^-1). It is because fluorine atom is smaller in size than chlorine atom. There is repulsion between the incoming electron and electrons already present in fluorine atom i.e., due to stronger inter electronic repulsions.

Question 8.
What is a periodic property ? How the following properties vary in a group and in a period ?
Explain
a. IE
b. EN
Answer:
a) IE : In groups and periods of the periodic table the ionization enthalpy values are periodically change depend upon the electronic configuration and size of elements.
In a group of elements ionization energy decreases from top to bottom because atomic radius increases.
In general, in a period the atomic size decreases. Because of this, the ionization energy increases across a period.

b) Electro negativity : Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases.
In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

Question 9.
Write a note on
a. Atomic radius
b. Metallic radius
c. Covalent radius
Answer:
a) Crystal Radius (Atomic radius or Metallic radius): The term is used for metal atoms. A metallic crystal contains metal atoms in close packing. These metal atoms are considered spherical. They are supposed to touch each other in the crystal.
The crystal radius is half the internuclear distance between two adjacent atoms.
e.g.: The internuclear distance between two adjacent sodium atoms in a crystal of sodium metal is 3.72 A. So crystal radius of sodium is \(\frac{3.72}{2}\) = 1.86 A
For potassium it is 2.31 A.

b) Covalent radius : It is used generally for non – metals. Covalent radius is half the equilibrium distance between the nuclei of two atoms with a covalent bond.
Covalent radii of two atoms can be added to give internuclear distance between them.
e.g. : Covalent radius of H is 0.37A and for chlorine it is 0.99A. Hence internulear distance between H and Cl in HCl is 1.36A.

c) Vander Waals radius (Collision radius): The Vander Waals radius is half the equilibrium distance between the nuclei of two atoms bound by Vander Waals forces.
It is used for molecular substances in solid state and for inert gases.
e.g. : The Vander Waals radius for hydrogen is 1.2 A and that of chlorine is 1.80 A.
The Vander Waals radius of an atom is 40% larger than its covalent radius.

Question 10.
Define IE₁ and IE₂. Why is IE₂ > IE₁ for a given atom? Discuss the factors that effect IE of an element. (A.P. Mar. ’16)(T.S. Mar. ’13)
Answer:
1) Ionization energy is the amount of energy required to remove the most loosely held electron from isolated a neutral gaseous atom to convert it into gaseous ion. It is also known as first ionization energy because it is the energy required to remove the first electron from the atom.
It is denoted as I₁, and is expressed in electron volts per atom, kilo calories (or) kilo joules per mole.
M_(g) + I₁ → \(M_{(g)^{+}}\) + e^–
I₁ is first ionization potential.

2) The energy required to remove another electron from the unipositive ion is called the second ionization energy. It is denoted as I₂.
\(M_{(g)}^{+}\) + I₂ → \(\mathrm{M}_{(\mathrm{g})}{ }^{2+}\) + e^–

3) The second ionization potential is greater than the first ionization potential. On removing an electron from an atom, the unipositive ion formed will have more effective nuclear charge than the number of electrons. As a result the effective nuclear charge increases over the outermost electrons. Hence more energy is required to remove the second electron. This shows that the second ionization potential is greater than the first ionization potential.
For sodium, I₁ is 5.1 eV and I₂ is 47.3 eV.
I₁ < I₂ < I₃ ….. I_n

Factors affecting ionization potential:

1. Atomic radius : As the size of the atom increases the distance between the nucleus and the outermost electrons increases. So the effective nuclear charge on the outermost electrons decreases. In such a case the energy required to remove the electrons also decreases. This shows that with an increase in atomic radius the ionization energy decreases.

2. Nuclear charge : As the positive charge of the nucleus increases its attraction increases over the electrons. So it becomes more difficult to remove the electrons. This shows that the ionization energy increases as the nuclear charge increases.

3. Screening effect or shielding effect: In multielectron atoms, valence electrons are attracted by the nucleus as well as repelled by electrons of inner shells. The electrons present in the inner shells screen the electrons present in the outermost orbit from the nucleus. As the number of electrons in the inner orbits increases, the screening effect increases. This reduces the effective nuclear charge over the outermost electrons. It is called screening or shielding effect. With the increase of screening effect the ionization potential decreases. Screening efficiency of the orbitals falls off in the order s > p > d > f.

(Magnitude of screening effect) ∝ \(\frac{1}{(\text { Ionization enthalpy) }}\)
TREND IN A GROUP : The ionisation potential decreases in a group, gradually from top to bottom as the size of the elements increases down a group.
TREND IN A PERIOD : In a period from left to right I.P. value increases as the size of the elements decreases along the period.

Question 11.
How do the following properties change in group-1 and in the third period ? Explain with example.
a. Atomic radius
b. IE
c. EA
d. Nature of oxides
Answer:
a) Atomic radius

• In Group – 1 atomic radius from Li to Cs increases
  
  → In 3rd period from Na to Cl atomic radius decreases.
  

b) Ionization energy : In a group ionization energy values decreases with an increase in the size of the atom. In IA group Li is the element with highest ionization potential and Cs is the element with lowest ionization potential.

In third period the ionization potential increases from Na to Mg and then decreases at Al and increases upto P and decreases in case of S and then increases upto argon, i.e., Al has a lower value than Mg and S has a lower value than P, due to stable electronic configurations of Mg and P Among these elements argon has highest ionization potential.

c) Electron affinity :

• In 3^rd period E.A > from ‘Si’ to ‘P’ decreases and P to Cl increases.
  ‘Mg’ and ‘Ar’ has positive values.
• In 1^st group from Li to Cs electron affinity decreases due to increase of size.

d) Nature of oxides of elements : All IA group elements are alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions which changes fed litmus blue.
e.g. : Na₂O, CaO , MgO etc.
Na₂O + H₂O → 2 NaOH
CaO + H₂O → Ca(OH)₂
MgO + H₂O → Mg (OH)₂
The basic nature of these oxides increases from top to bottom in the group. In a period basic properties decreases and acidic properties increases.

Question 12.
Define electron gain enthalpy. How it varies in a group and in a period? Why is the electron gain enthalpy of O or F is less negative than that of the succeeding element in the group ?
Answer:

1. Electron affinity is the amount of energy released when an electron is added to a neutral gaseous atom in its ground state. It is known as first Electron affinity EA,sub>1. It has -ve value.
   X_(g) + e → \(\mathrm{X}_{(\mathrm{g})}{ }^{-}\) + EA₁
2. Energy can be absorbed when an other electron is added to uni negative ion. It is because to overcome the repulsion between negative ion and electron. Hence second electron affinity, EA₂ has +ve value.
   \(X_{(g)}{ }^{-}\) + e → \(X_{(g)}^{2-}\) + EA₂
3. It can be indirectly determined from Born —Haber cycle. Its units are kJ/mole.
4. Trend in a group : Generally EA decreases down the group. In a group the second element has higher EA value than the first member.
   e.g. : EA of Cl is more than that of F
   Fluorine possess lower EA value than chlorine. It is due to inner electron repulsions in Fluorine.
5. Trend in a period : Generally EA value increases across a period, but some irregularities can be observed.

a) I A group elements possess low EA values than the corresponding III A elements, e.g. : Be has EA value zero, it is due to completely filled 2s² orbital.
b) VA elements have low EA values than that of VI A elements.
e.g. : Due to presence of half filled p – orbitals [2s² 2p³], nitrogen has lower EA value than that of oxygen.
c) For inert gases EA value is zero.
d) The element with the highest EA value is chlorine.

• The electron gain enthalpy of O and F is less negative than the succeeding element in the group because. These have small size and inter electronic repulsions are high in these elements
  O → 141 KJ / mole and S → 200 kJ/Mole
  F → 328 KJ / mole and Cl → 349 kJ/Mole

Question 13.
a. What is electronegativity ?
b. How does it vary in a group and in a period ?
Answer:
a) Electronegativity : ‘The tendency of the atom of an element to attract the shared electron pair(s) more towards itself in a hetronuclear diatomic molecule or in a polar covalent bond.” Measuring electronegativity Pauling scale : Pauling scale is based on the values of bond energy. The bond energy of a compound A – B is the average of bond energies of A – A and B – B molecules.
E_{A – B} = \(\frac{1}{2}\left(E_{A-A}+E_{B-B}\right)\)
But the experimental value of E_{A – B} is found to exceed the theoretical value. The difference is ∆.
∴ ∆ = E’_{A – B} – E_{A – B}
∆ indicates the polarity of the covalent bond. It is measured in k. cal. mol^-1.
Pauling gave the relation X_A – X_B = 0.208 × \(\sqrt{\Delta}\)
In S.l. units X_A – X_B = 0.1017 \(\sqrt{\Delta}\) where ∆ is measured in kJ/mole
X_A and X_B are the electronegativities of A and B. Pauling arbitrarily fixed 2.1 as the electronegativity value of Flydrogen and calculated the electronegativities of other elements. On Pauling scale Fluorine has the highest EN value 4.0.

From the values of the electronegativity of elements, the nature of the chemical bond formed can be understood. If two bonded atoms differ by 1.70 or more ion their EN values, the bond between them would be either 50% or more than 50% ionic in nature. Similarly if the difference in the EN values of the atoms is less than 1.70, the bond formed is more than 50% covalent in nature.

b) Variation in a group and period : Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases. In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

Question 14.
Explain the following
a. Valency
b. Diagonal relation
c. Variation of nature of oxides in the Group -1
Answer:
a) Valency : The combining capacity of an element with another element is called valency. The number of hydrogen atoms (or) chlorine atoms (or) double the number of oxygen atoms, with which one atom of the element combine is also called valency.
∴ Valency = no. of hydrogens
= no. of chlorine atoms
= 2 × no. of oxygen atoms present in the molecule

Periodicity of valency :
1) Each period starts with valency’11 and ends in ‘O’.
e.g.:

2) In a group valency is either equal to the group number (i.e., upto 4^th group) or is equal to (8 – group number) (i.e., from 5^th group onwards).
Significance :
Valency of an element is useful in writing the formulae of compounds.

b) Diagonal relation : On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship,
e.g. : (Li, Mg); (Be, Al); (B, Si)

c) Nature of oxides of elements : All IA group elements are alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions which changes red litmus blue.
e.g. : Na₂O, CaO , MgO etc.
CaO + H₂O → Ca(OH)₂
MgO + H₂O → Mg (OH)₂
The basic nature of these oxides increases from top to bottom in the group. In a period basic properties decreases and acidic properties increases.

Solved Problems

Question 1.
What would be the IUPAC name and symbol for the element with atomic number 120?
Solution:
The roots for 1, 2 and 0 are un, bi and nil, respectively. Hence, the symbol and the name respectively are Ubn and unbinilium.

Question 2.
How would you justify the presence of 18 elements in the 5t” period of the Periodic Table ?
Solution:
When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5^th period.

Question 3.
The elements Z = 117 and 120 have not yet been discovered. In which family / group would you place these elements and also give the electronic configuration in each case.
Solution:
we see from that element with Z = 117, would belong to the halogen family (Group 17) and the electronic configuration would be [Rn] 5f¹⁴6d¹⁰7s²7p⁵. The element with Z = 120, will be placed in Group 2 (alkaline earth metals), and will have the electronic configuration [Uuo]8s²

Question 4.
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P.
Solution:
Metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is : P < Si < Be < Mg < Na.

Question 5.
Which of the following species will have the largest and the smallest size ?
Mg, Mg²⁺, Al, Al³⁺.
Solution:
Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic species, the one with the larger positive nuclear charge will have a smaller radius.
Hence the largest species is Mg; the smallest one is Al³⁺.

Question 6.
The first ionization enthalpy (∆_iH) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol^-1. Predict whether the first ∆_iH value for Al will be more close to 575 or 760 kJ mol^-1 ? Justify your answer.
Solution:
It will be more close to 575 kJ mol^-1. The value for Al should be lower than that of Mg because of effective shielding of 3p electrons from the nucleus by 3s-electrons.

Question 7.
Which of the following will have the most negative electron gain enthalpy and which the least negative ? P, S, Cl, F. Explain your answer.
Solution:
Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hence the element with most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorous.

Question 8.
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;
(a) silicon and bromine
(b) aluminium and sulphur.
Solution:
(a) Silicon is group 14 element with a valence of 4; bromine belongs to the halogen family with a valency of 1. Hence the formula of the compound formed would be SiBr₄.
(b) Aluminium belongs to group 13 with a valence of 3; sulphur belongs to group 16 elements with a valence of 2. Hence, the formula of the compound formed would be Al₂S₂.

Question 9.
Are the oxidation state and covalency of Al in [AICI(H₂O)₅]²⁺ same ?
Solution:
No. The oxidation state of Al is +3 and the covalency is 6.

Question 10.
Show by a chemical reaction with water that Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
Solution:
Na₂O with water forms a strong base Whereas Cl₂O₇ forms strong acid.
Na₂O + H₂O → 2NaOH
Cl₂O₇ + H₂O → 2HClO₄
Their basic or acidic nature can be qualitatively tested with litmus paper.

Students can go through AP Inter 1st Year Physics Notes 4th Lesson Motion in a Plane will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 4th Lesson Motion in a Plane

→ A physical quantity having only magnitude but no direction is called a scalar.

→ A physical quantity having magnitude as well as direction and obeys law of addition of vectors is called a vector.

→ A vector of zero magnitude is called null vector for) zero vector.

→ Vectors situated in the same plane are called coplanar vectors.

→ A vector with magnitude one is called unit vector.

→ A vector ‘a1 can be represented in unit vector form as a = a_xi + a_yj + a_zk. The magnitude of
\(|\vec{a}|=\sqrt{a_x^2+a_y^2+a_z^2}\)

→ Resultant of any two vectors is determined by parallelogram law of vectors.
R = \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}\), tan α = \(\frac{Q \sin \theta}{P+Q \cos \theta}\)
→ Relative velocity is defined as the velocity of one body w.r. to other body.

→ Relative velocity of A w.r. to B = | V_A – V_B|

→ Relative velocity of B w.r. to A = | V_B – V_A|

→ Dot product of two vectors is given by \(\vec{a} \cdot \vec{b}\) = ab cos θ

→ Cross product of two vectors \(\vec{p}\) and \(\vec{Q}\) is given by \(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{Q}}=\mathrm{PQ} \sin \theta \hat{\mathrm{n}}\)
∴ \(|\vec{P} \times \vec{Q}|=\left|\begin{array}{ccc}
i & 1 & k \\
P_x & P_y & P_z \\
Q_x & Q_y & Q_z
\end{array}\right|\)

→ Two vectors A and B are said to be equal if and only if, they have the same magnitude and the same direction.

→ The equation of motion for a projectile is given by y = (tan θ₀)x – \(\frac{\mathrm{g}}{2\left(\mathrm{v}_0 \cos \theta_0\right)^2}\) x² After t seconds (v₀) remains the same and final velocity is given by v₀ = sin θ – gt

→ The time taken by the projectile to reach the maximum height is given by t_a = \(\frac{v_0 \sin \theta}{g}\)
The time of descent is equal to the time of ascent time of flight is given by T = \(\frac{2 v_0 \sin \theta}{g}\)

→ Maximum height reached by the projectile (H) = \(\frac{\left(v_0 \sin \theta_0\right)^2}{2 g}\)

→ For angles of projection (45° + α) and (45° – α) the range is the same for a given initial velocity.

→ Horizontal range is given by R = \(\frac{v_0^2 \sin 2 \theta_0}{g}\)

→ Maximum Range (R_max) = \(\frac{v_0^2}{g}\)

→ The velocity of the projectile at any instant t is given by v = \(\sqrt{v_x^2+v_y^2}\) where v_x = υ₀ cos θ, v_y = v₀ sin θ – gt

→ The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant.

→ The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion.

→ The position vector of an object in x – y plane is given by

Students can go through AP Inter 1st Year Physics Notes 3rd Lesson Motion in a Straight Line will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 3rd Lesson Motion in a Straight Line

→ Path length is defined as the total length of the path traversed by an object.

→ Displacement is the change in position; ∆x = x₂ – x₁

→ Average velocity is the displacement divided by the time interval in which the displacement occurs. υ = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)

→ Instantaneous velocity is defined’as the limit of the average velocity as the time interval At becomes infinitesimany small.

→ Average acceleration is the change in velocity divided by the time interval during which the change occurs, \(\overline{\mathrm{a}}=\frac{\Delta v}{\Delta \mathrm{t}}\)

→ Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero.

→ Average speed is the ratio of total path length traversed and the corresponding time interval.

→ Equations of motion are
υ = υ₀ + at
x = υ₀t + \(\frac{1}{2}\) at²
υ² = υ²₀ + 2ax

→ Height of a tower is – h = ut – \(\frac{1}{2}\) gt²

→ An object is released near the surface of the earth is accelerated under the influence of gravity. The object is said to be in free fall.

→ Acceleration due to gravity (g) is positive along +y direction and g is negative along – y direction

→ g value in downward motion is negative a = -g = -9.8 m/s²

1. v = 0 – gt = -9.8 t
2. y = 0 – \(\frac{1}{2}\) gt² = – 4.9 t²
3. υ² = 0 – 2 gy = -19.6y

→ Relative velocity is the velocity one body w.r.to velocity of another body.

→ Relative velocity of A w.r.to B = \(\left|\vec{V}_A-\vec{V}_R\right|\)

→ Relative velocity of B w.r.to A = \(\left|\vec{V}_B-\vec{V}_A\right|\)

→ Dimensional formula of velocity = [LT^-1].

→ Dimensional formula of acceleration = [LT^-2].

Students can go through AP Inter 1st Year Physics Notes 2nd Lesson Units and Measurements will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 2nd Lesson Units and Measurements

→ The accuracy depends on the errors and also on the precision of measuring instrument.

→ There will be certain amount of uncertainty inherent in the measurement of physical quantities by any instrument. This uncertainty is called the error.

→ When the result of a series of measurements are in error by the same amount. Such an error is called constant error.

→ Systematic errors are those errors that tend to be always in one direction.

→ Systematic errors with a constant magnitude are called constant errors.

→ The different types of systematic errors are

1. Imperfection in experimental technique or procedure
2. Environmental errors,
3. Personal errors or Observational errors.

→ When the random errors could be eliminated we say the measurements are precise.

→ When all types of errors are eliminated then the measurements are accurate.

→ When two quantities are added or subtracted, the (maximum) absolute error in the result (in both cases) will be the sum of the absolute errors in the two quantities.

→ When two quantities are multiplied or divided, the (maximum) relative error in the result (in both cases) will be the sum of the relative errors in the two quantities.

→ The digits of a number that are definitely known plus one more digit that is estimated are called significant digits or significant figures.

→ The process of emitting the non-significant digits and retaining only the desired number of significant digits incorporating the required modifications to the last significant digit is called “rounding off the number”.

→ A physical quantity which is independent of any other physical quantity is called a fundamental quantity, e.g. : length, mass, time, temperature, strength of current, intensity of light and quantity of matter.

→ The units of fundamental quantities are called fundamental units.

→ The physical quantities which can be expressed in terms of the fundamental quantities are called derived physical quantities, e.g.: Volume, Velocity, Work etc.

→ The units of derived physical quantities are called the derived units.

→ S.I. System : It consists seven fundamental physical quantities. They are :

1. length
2. mass,
3. time,
4. thermodynamic temperature,
5. intensity of light,
6. strength of electric current,
7. quantity of matter.

→ The power to which the fundamental quantities are to be raised to obtain one unit of the physical quantity is called the “dimensions” of that physical quantity.

→ The derived quantity can be expressed as M^a L^b T^c which is called the dimensional formula. The powers a, b, c are called dimensions.

→ Uses of dimensional equations :
a) To change one system of units into another.
b) To derive a relation connecting different physical quantities.
c) To check the correctness of an equation for a physical quantity.

→ Limitations of dimensional method :

1. The proportionality constant in an equation cannot be obtained by dimensional method.
2. The equations involving trigonometrical or exponential functions cannot be derived.
3. The equations containing more than three physical quantities cannot be derived.
4. If the formula of a physical quantity is represented by the sum of some physical quantities dimensional method cannot be used to derive that formula.
5. Some proportionality quantities possess units. In such cases dimensional method cannot be used to analyse it.

→ The accuracy of a measurement of any physical quantity made by any measuring instrument is a measure of how close the measured value is to the true value of the quantity.

→ Large distances such a$ the distance of stars (or) of a planet from the earth can be measured using parallax method.

→ The smallest value that can be measured by the measuring instrument is called its least count.

→ A vernier callipers is used for lengths to an accuracy of 10^-4 m.

→ A screw gauge and a spherometer can be used to measure lengths as less as to 10^-5 m.

→ 1 Å = 10^-10 m = 10^-8 cm.

→ 1 light year = 9.46 × 1^-15 m
1 parsec = 3.08 × 10¹⁶ m.

→ 1 a.m.u — 1.66 × 10^-27 kg.

→ 1 Fermi = 10^-15 m.

Formulae

a_mean = true value = \(\frac{1}{n} \sum_{i=1}^n a_i\)

Relative error = \(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}\)

Percentage error = δa = \(\left(\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}} \times 100\right) \%\)

Relative error in multiplication \(\frac{\Delta x}{x}=\frac{\Delta a}{a}+\frac{\Delta b}{b}\)

Relative error in division \(\frac{\Delta x}{x}=\frac{\Delta a}{a}+\frac{\Delta b}{b}\)

Maximum relative error in (x – aⁿ) = \(\frac{\Delta x}{x}=n\left(\frac{\Delta a}{a}\right)\)

Students can go through AP Inter 1st Year Physics Notes 1st Lesson Physical World will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 1st Lesson Physical World

→ Physics deals with the study of the basic laws of nature and their manifestation in difference phenomena.

→ Basic forces in nature are

1. Gravitational force
2. Electromagnetic force
3. Strong nuclear force
4. Weak nuclear force.

→ The Raman effect deals with scattering of light by molecules of a medium, when they are excited to vibrational energy levels.

→ Conservation of energy, momentum, angular momentum, charge, etc are considered to be fundamental laws in Physics.

→ A conservation law is a hypothesis, based on observations and experiments.

→ According to Einstein’s theory, mass m is equivalent to energy E given by the relation E = mc², where c is speed of light in vacuum.

→ Optical fibres works on the principle of total internal reflection.

→ Rocket propulsion works on the principle of Newton’s laws of motion.

→ Aeroplane works on the principle of Bernoullis principle.

→ Electron microscope works on the principle of work nature of electrons.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 1st Lesson Atomic Structure Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 1st Lesson Atomic Structure

Very Short Answer Questions

Question 1.
What is the charge, mass and charge to mass ratio of an electron ?
Answer:

• Charge of an electron = – 1.602 × 10^-19 coloumbs (or)
  -4.8 × 10^-19 esu
• Mass of an electron = 9.1 × 10^-28 gms
• Charge to mass ratio of an electron \(\left(\frac{\mathrm{e}}{\mathrm{m}}\right)\) i.e., specific charge = 1.758 × 10¹¹ coloumbs/kg

Question 2.
Calculate the charge of one mole of electrons.
Answer:
One electron has charge – 1.602 × 10^-19 coloumbs.
One mole of electrons has charge -6.023 × 10²³ × 1.602 × 10^-19
= 9.648846 × 10⁴ = 96488.5 coloumbs.

Question 3.
Calculate the mass of one mole of electrons.
Answer:
Mass of electron = 9.1 × 10-31 kg (or) 9.1 × 10^-28 gms.
One mole of electrons has mass 6.023 × 10²³ × 9.1 × 10^-31 = 54.8 × 10^-8 = 5.48 × 10^-7 kg.

Question 4.
Calculate the mass of one mole of protons. ”
Answer:
One proton has mass 1.672 × 10^-27 kg
One mole protons has mass 6.023 × 10²³ × 1.672 × 10^-27
= 10.0704 × 10^-4
= 1.00704 × 10^-3 kg.

Question 5.
Calculate the mass of one mole of neutrons.
Answer:
One neutron has mass 1.675 × 10^-27 kg
One mole neutrons has mass 6.023 × 10²³ × 1.675 × 10^-27
= 10.088 × 10^-4
= 1.0088 × 10^-3 kg.

Question 6.
How many neutrons and electrons are present in the nuclei of ₆C¹³, ₈O¹⁶, ₁₂Mg²⁴, ₂₆Fe⁵⁶ and ₃₈Sr⁸⁸.
Answer:

Question 7.
What is a black body ?
Answer:
The body which is perfect absorber and emmiter of all type of radiations incident on it is called a black body.

Question 8.
Which part of electromagnetic spectrum does Balmer series belong?
Answer:
Balmer series (n = 2) belongs to visible region of electromagnetic spectrum.

Question 9.
What is an atomic orbital?
Answer:
In an atom, the region around the nucleus where the probability of finding the electron is maximum is known as atomic orbital.

• From the value of magnitude of square of wave function (|\(\psi^2\)|) this region can be predicted.

Question 10.
When an electron is transferred in hydrogen atom from n = 4 orbit to n=5 orbit to which spectral series does this belong?
Answer:

• By the absorption of energy electron jumps from n = 4 orbit to n = 5 orbit.
• The electron present in n = 5 orbit emitts energy and return to n = 4 orbit. Hence the spectral lines series obtained in Brackett series (IR region)

Question 11.
How many “p” electrons are present in sulphur atom?
Answer:
Sulphur has electronic configuration – 1s² 2s² 2p⁶ 3s² 3p⁴
∴ Sulphur has 10 ‘p’ electrons.

Question 12.
What are the values of principal quantum number (n) and azimuthal quantum number (l) for a 3d electron?
Answer:
For a 3d – electron principal quantum number (n) = 3 and
Azimuthal quantum number (l) = 2.

Question 13.
What is the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)?
I) Z = 4, A = 9 ;
II) Z = 17, A = 35 ;
III) Z = 92, A = 233.
Answer:
I) Z = 4, A = 9 Complete symbol is ₄Be⁹
II) Z = 17, A = 35 Complete symbol is ₁₇Cl³⁵
III) Z = 92, A = 233 complete symbol is ₉₂U²³³.

Question 14.
Draw the shape of \(\mathrm{d}_{\mathrm{z}^2}\) orbital.
Answer:

Question 15.
Draw the shape of d_{x² – y²} orbital.
Answer:

Question 16.
What is the frequency of radiation of wavelength 600nm?
Answer:
Formula:
\(v=\frac{c}{\lambda}\)
= \(\frac{3 \times 10^8}{6 \times 10^{-7}}\)
= \(\frac{1}{2} \times 10^{15}\)
= 0.5 × 10¹⁵ = 5 × 10¹⁴ sec^-1
λ = 600 nm
= 600 × 10^-9 m
= 6 × 10^-7 m
C = 3 × 10⁸ m/sec.

Question 17.
What is Zeeman effect?
Answer:
The splitting up of spectral lines in presence of strong external magnetic field is called as Zeeman effect.

Question 18.
What is Stark effect?
Answer:
The splitting of spectral lines in presence of strong electric field is called as Stark effect.

Question 19.
To which element does the following electronic configuration correspond?
I) 1s²2s² 2p⁶3s²3p¹
II) 1s²2s²2p⁶3s²3p⁶
III) 1s²2s²2p⁵
IV) 1s²2s²2p².
Answer:
I) 1s²2s² 2p⁶ 3s² 3p¹ (Atomic no. (Z) = 13) – Aluminium.
II) 1 s²2s²2p⁶3s²3p⁶(Atomic no. (Z) = 18) – Argon.
III) 1s²2s²2p⁵ (Atomic no. (Z) = 9) – Fluorine.
IV) 1s²2s²2p² (Atomic no. (Z) = 6) – Carbon.

Question 20.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 4000 A. What is the threshold frequency (\(v_0\))?
Answer:
Formula:
hv = hv₀ + \(\frac{1}{2} m v^2\)
hv = hv₀ + \(\frac{1}{2} m(0)^2\)
hv = hv₀
λ = 4000 A
= 4 × 10³ × 10^-10 = 4 × 10^-7 m.
V = 0
C = 3 × 10⁸ m/sec.
⇒ v = v₀
∴ v = \(\frac{\mathrm{C}}{\lambda}\) = \(\frac{3 \times 10^8}{4 \times 10^{-7}}\) = \(\frac{3}{4}\) × 10¹⁵
= 0.75 × 10¹⁵
= 7.5 × 10¹⁴ sec^-1

Question 21.
Explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle:
According to this principle
“No two electrons in an atom can have the same set of four quantum numbers”. This can also be stated as “only two electrons may exist in the same orbital and these electrons must have opposite spins”.

This means that the two electrons can have the same value of three quantum numbers n, l and m_l but have the opposite spin quantum number Ex : Consider ‘K’ shell of the atom having two electrons

Question 22.
What is Aufbaus principle ?
Answer:
Aufbau’s principle:
This principle states
“In the ground state of the atoms, the orbitals are filled in order of their increasing energies”. In other words electrons first occupy the lowest energy orbital available to them and enter into higher energy orbitals only after the lower energy orbitals are filled.
The order in which the orbitals are filled as follows :
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 4f < 5d < 6p < 7s

Question 23.
What is Hund’s rule ?
Answer:
Hund’s rule: This rule deals with the filling of electrons in degenerate orbitals. It states “Pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each (i.e.,) all the orbitals are singly occupied”.
Since there are three ’p’, five ‘d’ and seven ‘f’ orbitals, therefore the pairing of electrons will start in the p, d and f orbitals with the entry of 4^th, 6^th and 8^th electrons respectively.
Ex : ‘₈O’ electronic configuration is

Question 24.
Explain Heisenberg’s uncertainty principle.
Answer:
Heisenberg uncertainty principle : “Simultaneous and exact determination of the position and momentum of a sub-atomic particle, like electron moving with high speed is impossible.”
If Δx and Δp represents the uncertainties in the position and momentum respectively. Then according to Heisenberg
Δx. Δp ≥ \(\frac{\mathrm{h}}{4 \pi}\) ——- (1)

The product of uncertainties in position (Δx) and momentum (Δp) of an electron cannot be less than \(\frac{h}{4 \pi}\). It can be equal or greater than \(\frac{h}{4 \pi}\).
Since momentum = mass x velocity, the equation (1) can be written as
Δx × m (Δv) ≥ \(\frac{\mathrm{h}}{4 \pi}\) = Δx × Δv ≥ \(\frac{\mathrm{h}}{4 \pi \mathrm{m}}\)
If the position is determined accurately Δx = 0 and Δv = ∝. That means the inaccuracy in measuring the velocity is ∝. If velocity is determined accurately Δv = 0 and Δx = ∝.

Question 25.
What is the wavelength of an electron moving with a velocity of 2.0 × 10⁷m/s ?
Answer:
Formulae:
λ = \(\frac{h}{m v}\)
= \(\frac{6.625 \times 10^{-34}}{9.1 \times 10^{-31} \times 2 \times 10^7}\)
= 0.3640 × 10^-34 × 10⁺²⁴
= 0.3640 × 10¹⁰ m
= 0.3640 A
h = 6.625 × 10^-34 J.Sec
m = 9.1 × 10^-31 kg
V = 2.0 × 10⁷ m/sec.

Question 26.
An atomic orbital has n = 2, what are the possible values of l and m_l?
Answer:
For n = 2, l values are 0, 1
For l = 0 → m_l = 0
For l = 1 → m_l = -1, 0, +1.

Question 27.
Which of the following orbitals are possible? 2s, 1p, 3f, 2p.
Answer:
2s, 2p orbitals are possible among 2s, 1p, 3f, 2p and 1 p, 3f orbitals are not possible.

Question 28.
The static electric charge on the oil drop is – 3.2044 × 10^-19 C. How many electrons are present on it?
Answer:
Given static electric charge on oil drop = – 3.2044 × 10^-19 C
Charge of electron = – 1.602 × 10^-19 C.
Number of electrons present =

Question 29.
Arrange the following type of radiation in increasing order of frequency:
(a) X – rays
(b) visible radiation
(c) microwave radiation and
(d) radiation from radio waves.
Answer:
Increasing order of frequency of given radiations is
Radio waves < Micro waves < Visible radiation < X – rays.

Question 30.
How many electrons in an atom may have n = 4 and m_s = +1/2 ?
Answer:
For n = 4 → l values are 0, 1, 2, 3
l = 0 → s contains 1 electron with m_s = + 1/2
l = 1 → p contains 3 electron with m_s = + 1/2
l = 2 → d contains 5 electron with m_s = + 1/2
l = 3 → f contains 7 electron with m_s = + 1/2
∴ Total no.of electrons with ms = +1/2 for n = 4
= 1 + 3 + 5 + 7 = 16.

Question 31.
How many sub-shells are associated with n = 5 ?
Answer:
For n = 5
l values are 0, 1, 2, 3, 4
l = 0 → s – orbital
l = 1 → p — orbital
l = 2 → d – orbital
l = 3 → f – orbital
l = 4 → g — orbital
→ Five subshells are associated with n = 5.

Question 32.
Explain the particle nature of electromagnetic radiation.
Answer:

• According to earlier days concepts light was supposed to be made of particles. This assumption was made by Newton in his corpuscular theory. He called the particles as corpuscales.
• The particle nature of light explains the black body radiations and photo electric effect satisfactorily.
• The particle nature of light could not satisfactorily explains the phenomenon of diffraction and Interferance.

Question 33.
Explain the significance of Heisenberg’s Uncertainty principle.
Answer:
Significance of Uncertainty Principle:

1. This principle rules out the existence of definite paths or trajectories of electrons and other similar particles.
2. This principle is significant only for motion of microscopic objects, and is negligible for that of macroscopic objects.
3. In dealing with milligram size or heavier objects, the associated uncertainties are hardly of any real consequence.

Question 34.
What series of lines are observed in hydrogen spectra?
Answer:
The series of lines observed in hydrogen spectra are

Additional Answer Questions

Question 35.
How many newtrons and electrons are present in the nuclei of \({ }_6 C^{13}\), \({ }_8 \mathrm{O}^{16}\), \({ }_{12} \mathrm{Mg}^{24}\), \({ }_{26} \mathrm{Fe}^{56}\), \({ }_{38} \mathrm{Sr}^{88}\)
Answer:

Additional Problems

Question 36.
Calculate the wave no. and wave length of first line of lyman series.
Answer:

Question 37.
Calculate the wave no. of and wave length of first line of Balmer series.
Answer:

Short Answer Questions

Question 38.
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 5 to an energy level with n = 3 ?
Answer:
Formulae:

R = 1,09,677 cm^-1
n₁ = 3
n₂ = 5.
\(\bar{v}\) = 7799.25 cm^-1
λ = \(\frac{1}{\bar{v}}\) = \(\frac{1}{7799.25}\) = 1.2821 × 10^-4 cm

Question 39.
An atom of an element contains 29 electrons and 35 neutrons. Deduce

1. the number of protons and
2. the electronic configuration of the element.

Answer:
Given no.of electrons ’29’, ∴ Z = 29

1. So, no.of protons = 29
2. Electronic configuration of the element (Z = 29)
   = 1s²2s²2p²3s²3p⁶4s¹3d¹⁰ [anamalous electronic configuration]

Question 40.
Explain giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0, l = 0, m_l = 0, m_s = +\(\frac{1}{2}\)
b) n = 1, l = 0, m_l = 0, m_s = –\(\frac{1}{2}\)
c) n = 1, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
d) n = 2, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
e) n = 3, l = 3, m_l = – 3, m_s = +\(\frac{1}{2}\)
f) n = 3, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
Answer:
Following set of quantum numbers are not possible.
a) n = 0, l = 0, m_l = 0, m_s = +\(\frac{1}{2}\)
Reason:
‘n’ is principal quantum number, whose values are from 1 to n. The value of ‘n’ never equal to zero. But given n = 0.

c) n = 1, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
Reason:
Values of ‘l’ are from 0 to (n – 1).
If n = 1 then the value of ‘l’ is zero not equal to ‘1’.

e) n = 3, l = 3, m_l = – 3, m_s = +\(\frac{1}{2}\)
Reason:
If n = 3, possible values of ‘l’ are 0, 1, 2, but not equal to ‘3’.

Question 41.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
Consider the Bohr’s angular momentum equation.
mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
i.e The angular momentum of an electron is integral multiple of ‘\(\frac{\mathbf{h}}{2 \pi}\)‘
mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
2πr = \(\frac{\mathrm{nh}}{\mathrm{mv}}\)
According to de-Broglie’s wavelength λ = \(\frac{h}{m v}\)
2πr = \(n\left(\frac{h}{m v}\right)\)
2πr = nλ.
Thus the circumference of the Bohr orbit is integral multiple of de-Broglie’s wave length.

Question 42.
The longest wavelength doublet absorption transition is observed at 589.0 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
Given largest wave length doublet absorption transition is observed at 589.0 and 589.6 nm.
∴ \(v_1=\frac{c}{\lambda_1}\)
= \(\frac{3 \times 10^8}{589 \times 10^{-9}}\) = 0.005093 × 10⁺⁷
= 5.093 × 10¹⁴ sec^-1
λ₁ = 589 × 10^-9 m
∴ \(v_2=\frac{c}{\lambda_2}\)
= \(\frac{3 \times 10^8}{589.6 \times 10^{-9}}\)
= 0.005088 × 10¹⁷
= 5.088 × 10¹⁴ sec^-1
λ₁ = 5.089 × 10^-9 m
Energy difference between two states = h[\(v_1\) – h\(v_2\)]
= h[\(v_1\) – \(v_2\)]
= 6.625 × 10^-34[5.093 × 10^-14 – 5.088 × 10¹⁴]
= 6.625 × 10^-34 × 0.005 × 10¹⁴
= 0.0331 × 10^-20 J.

Question 43.
What are the main features of quantum mechanical model of an atom?
Answer:
Important features of quantum mechanical model of atom:

1. The energy of electrons in an atom is quantized (it can only have certain specific values).
2. The existence of quantized electronic energý levels is a direct result of the wave like properties at electrons and are allowed solution at Schrodinger wave equation.
3. All the information about the electron in an atom is contained in its orbital wave function ‘Ψ’ and quantum mechanics makes it possible to extract this information from “Ψ’.
4. The path of the electron can never be determined accurately. Therefore, we find only the probability of the electron at different points in space, around an atom.
5. The probability of finding an electroñat a point within an atom is proportional to the square of the orbital wave function i.e., \(|\Psi|^2\) at that point. \(|\Psi|^2\) is known as probability density and is always positive. From the value of \(|\Psi|^2\) at different points with in the atom, it is possible to predict the region around the nucleus where electron will most probably be found.

Question 44.
What is a nodal plane? How many nodal planes are possible for 2p – and 3d – orbitals?
Answer:
The plane at which the probability of finding the electron is zero is called as nodal plane.

• For 2p orbitaIs one nodal plane is possible for each ‘p’ orbital.
• For 3d orbitais two nodal planes are possible for each ‘d’ orbital.

Question 45.
The Lyman series occurs between 91.2 nm and 121.6 nm, the Balmer series occurs between 364.7 nm and 656.5 nm and the Paschen series occurs between 820.6 nm and 1876 nm. Identify the spectral regions to which these wavelengths correspond?
Answer:
In electromagnetic spectrum,
a) 91.2 – 121.6 nm (Lyman senes) corresponds to u.v. region.
b) 364.7 – 656.5 nm (Balmer series) corresponds to visible region.
c) 820.6 – 1876 nm (Paschen series) corresponds to I.R. region.

Question 46.
How are the quantum numbers n, l, m_l, for hydrogen atom obtained?
Answer:
Electronic configuration of hydrogen is 1s¹
For ns¹
Principal quantum no. (n) = 1
Azimuthal quantum no. (l) = 0
Magnetic quantum no. (m_l) = 0
and Spin quantum no. (m_s) = + 1/2

Question 47.
A line in Lyman series of hydrogen atom has a wavelength of 1.03 × 10^-7 m. What is the initial energy level of the electron?
Answer:
Given λ= 1.03 × 10^-7 m = 1.03 × 10^-5 cm
n₂ = 1 (for Lyman series)
We have, R = 109677 cm^-1

⇒ n₂ = 3 (i.e.,) original energy level of electron is 3.

Question 48.
If the position of the electron is measured within an accuracy of ±0.002 nm. Calculate the uncertainty in the momentum of the electron.
Answer:
Formulae:
Δx × Δp = \(\frac{h}{4 \pi}\)
Δp = \(\frac{h}{\Delta x \times 4 \pi}\)
= \(\frac{6.625 \times 10^{-34}}{4 \times 3.14 \times 2 \times 10^{-12}}\)
= 0.2637 × 10^-22
= 2.637 × 10^-23 J/m.
Δx = 0.002 nm
= 2 × 10^-3 × 10⁹ m
= 2 × 10^-12 m
h = 6.625 × 10^-34 J.sec.
∴ Uncertainity in momentum of electron = 2.637 × 10^-23 J/m.

Question 49.
If the velocity of the electron is 1.6 × 10⁶ m/s^-1. Calculate de Brogue wavelength associated with this electron.
Answer:
Formulae:
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\)
= \(\frac{6.625 \times 10^{-34}}{9.1 \times 10^{-31} \times 1.6 \times 10^6}\)
= 0.455 × 10^-9 m
= 0.455 nm.
v = 1.6 × 10⁶ m/sec
h = 6.625 × 10^-34 J.sec
m = 9.1 × 10^-31 Kg.

Question 50.
Explain the difference between emission and absorption spectra. (A.P. Mar. ‘15)
Answer:
Emission spectrum

1. It is produced by analysing the radiant energy emitted by an excited substance.
2. It consists of bright lines on dark back ground.
3. Produced due to the emission of energy by electrons.
4. Emission spectra contains bright ineson dark back ground.

Absorption spectrum

1. It is produced when white light is passed through a substance and the transmitted light is analysed by a spectrograph.
2. It consists of dark lines on bright background.
3. Produced due to the adsorption of energy by electrons.
4. Absorption spectra contains dark lines on bright back ground.

Question 51.
The quantum numbers of electrons are given below. Arrange them in order of increasing energies.
a) n = 4, l = 2, m_l = -2, m_s = +\(\frac{1}{2}\)
b) n = 3, l = 2, m_l = -1, m_s = –\(\frac{1}{2}\)
c) n = 4, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
d) n = 3, l = 1, m_l = -1, m_s = –\(\frac{1}{2}\)
Answer:
a) n = 4, l = 2, → 4d
b) n = 3, l = 2, → 3d
c) n = 4, l = 1, → 4p
d) n = 3, l = 1, → 3p
∴ 3p < 3d < 4p < 4d
According (n + l) values
Hence d < b < c < a is order of increasing energy.

Question 52.
The work function for Cesium atom is 1.9 eV. Calculate the threshold frequency of the radiation. If the Cesium element is irradiated with a wavelength of 500 nm, calculate the kinetic energy of the ejected photoelectron?
Answer:
Case-I
Photo electric effect equation is
hv = hv₀ + 1/2 mv²
w = hv₀

Case-II
Photo electric effect equation is
E = \(\frac{\mathrm{hc}}{\lambda}\)
= \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7}}\)
= \(\frac{19.878 \times 10^{-26}}{5 \times 10^{-7}}\)
= 3.9756 × 10^-19 J.

Given work function hv₀ = 1.9 ev
= 1.9 × 1.602 × 10^-19J.

Kinetic Energy (KE) = \(\frac{1}{2} m v^2\)
From Photo electric effect
\(\frac{1}{2} m v^2\) = hv – hv₀
K.E. = 3.9756 × 10^-19 × 1.602 × 10^-19
= 3.9756 × 10^-19 – 3.0438 × 10^-19 = 0.9318 × 10^-19 = 9.318 × 10^-20J.

Question 53.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
Given the radius of orbit from which it started = 1.35225 × 10^-9 m = 1 3.225Å
In general radius of orbit = 0.529 × n²Å
n² = \(\frac{13.225}{0.529}\) = 25
n² = 25 ⇒ n = 5
Given that the
Radius of orbit at which the transition ended = 211.6 pm
= 211.6 × 10^-12m
= 2.116A
Similarly as above
n² = \(\frac{2.116}{0.529}\) = 4
n² = 4 ⇒ n = 2
∴ transition takes place from n = 5 to n = 2 level
∴ spectral lines are obtained in Balmer series (visible region)

Question 54.
Explain the difference between orbit and orbital.
Answer:
Orbit

1. A circular path which is present around the nucleus in which electrons revolve is called as orbit.
2. Orbits are circular and are non directional paths.
3. The maximum no.of electrons in any orbit is given by the formula 2n² (n = orbit number).

Orbital

1. The 3 – dimension space where the probability of finding the electron is maximum around the nucleus is called as orbital.
2. These have definite shape and these are directional except’s orbital.
3. Each orbital can occupy a maximum of two electrons.

Question 55.
Explain photoelectric effect.
Answer:
The ejection of electrons from a metal surface, when the radiations of suitable frequency strikes the metal surface is called photoelectric effect.

Explanation using Einstein’s quantum theory:

1) To explain photoelectric effect, Einstein utilised Quantum theory.
2) When a photon strikes metal surface, it uses some part of its energy to eject the electron from the metal atom. The remaining part of the total energy is given to the ejected electrons in the form of kinetic energy.
Hence we can write hv = W + KE ⇒ hv = hv₀ + \(\frac{1}{2} m_e v^2\)
where hv = energy of photon,
v₀ = Threshold frequency,
m_e = mass of electron
W = energy required to overcome the attractive forces on the electron in the metal (work function)
KE = kinetic energy of ejected electron,
V = Velocity of ejected electron.

3) If a photon of sufficient energy struck the metal surface and could eject the electron. But if a photon has insufficient energy, it cannot eject the electron from the metal.
eg. : A photon of violet light [high frequency] can eject the electrons from the surface of potassium but a photon of red light [low frequency] cannot eject the electrons.

Question 56.
Explain Rutherford’s nuclear model of an atom. What are its drawbacks?
Answer:
Rutherford’s Planetary model: Rutherford drew some conclusions regarding the structure of atom.

1. Most of the space in the atom is empty (as most of the α – particles passed through the foil undeflected).
2. A few — positive charges were deflected. The deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson predicted. The positive charge is concentrated in a very small volume. Which is responsible for the deflection of α — particles.

On the basis of the above observations. Rutherford proposed the nuclear model. According to his model.

1. The positive charge in the atom is concentrated in the small dense portion, called the NUCLEUS.
2. The nucleus is surrounded by the electrons that move around it in circular paths called the ORBITS. Thus Rutherford’s model resembles the solar system.
3. Electrons and the nucleus are held together by electrostatic forces of attraction.

Drawbacks of Rutherford model:

1. Rutherford’s atomic model of an atom is like a small scale solar system. This similarity suggests that electrons should move around the nucleus in well defined orbits. However, when a body is moving, it undergoes acceleration. According to electromagnetic theory, charged particles, when accelerated, should emit radiation. Therefore, an electron in an orbit will emit radiation, thus the orbit will continue to shrink. But this does not happen. Thus Rutherford’s model cannot explain the stability of the atom.

2. If we assume that electrons as stationary around the nucleus, the electrostatic attraction between the nucleus and the electrons would pull the electrons towards the nucleus to form a miniature version of Thomson’s model.

3. Rutherford model does not explain the electronic structure of the atom i.e., how the electrons are distributed around the nucleus and what are the energies of these electrons.
Before studying further developments that lead to the formulation of various atomic models, it is necessary to study about light and its nature.

Question 57.
Explain briefly the Planck’s quantum theory.
Answer:
The postulates. of Planck’s quantum theory are
a) The emission of radiation is due to vibrations of charged particles (electrons) in the body.
b) The emission is not continuous but in discrete packets of energy called quanta. This emitted radiation propagates in the form of waves.
c) The energy (E) associated with each quantum for a particular radiation of frequency V is given by E = hv, Here ‘h’ is Planck’s constant.
d) A body can emit or absorb either one quantum (hv) of energy or some whole number multiple of it. Thus energy can be emitted or absorbed as hv, 2hv, 3hv etc., but not fractional values. This is called quantisation of energy.
e) The emitted radiant energy is propagated in the form of waves.
f) Values of Planck’s constant in various units:
h = 6.6256 × 10^-27 erg.sec (or) g cm²s^-1
= 6.6256 × 10^-34J.s (or) kg m²s^-1 = 1.58 × 10^-34 cal.s.
Success of Planck’s quantum theory: This theory successfully explains the black body radiations. A black body is a perfect absorber and also a perfect radiator of radiations.

Question 58.
What are the postulates of Bohr’s model of hydrogen atom? Discuss the importance of this model to explain various series of line spectra In hydrogen atom. (T.S. Mar. ‘16)(A.P. Mar.’15. ‘13)
Answer:
Niels Bohr quantitatively gave the general features of hydrogen atom structure and it’s spectrum. His theory is used to evaluate several points in the atomic structure and spectra.
The postulates of Bohr atomic model for hydrogen as follows

Postulates : –

• The electron in the hydrogen atom can revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits (or) stationary states. These circular orbits are concentric (having same center) around the nucleus.
• The energy of an electron in the orbit does not change with time.
• When an electron moves from lower stationary state to higher stationary state absorption of energy takes place.
• When an electron moves from higher stationary state to lower stationary state emission of energy takes place.
• When an electronic transition takes place between two stationary states that differ in energy by ΔE is given by
  ΔE = E₂ – E₁ = hv
  ∴ The frequency of radiation absorbed (or) emitted v = \(\frac{E_2-E_1}{h}\) E₁ and E₂ are energies of lower, higher energy states respectively.
• The angular momentum of an electron is given by mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
  An electron revolve only in the orbits for which it’s angular momentum is integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

Line spectra of hydrogen, Bohr’s Theory:

• In case of hydrogen atom line spectrum is observed and this can be explained by using Bohr’s Theory.
• According to Bohr’s postulate when an electronic transition takes place between two stationary states that differ in energy is given by
  
  
• In case of absorption spectrum n_f > n_i → energy is absorbed (+ve) energy is absorbed (+Ve)

• In case of emission spectrum n_i > n_f → energy is emitted (- Ve)
• Each spectral line in absorption (or) emission spectrum associated to the particular transition in hydrogen atom
• In case of large no.of hydrogen atoms large no.of transitions possible they rsults in large no.of spectral lines.
  The series of lines observed in hydrogen spectra are
  

Question 59.
Explain the success of Bohr’s theory for hydrogen atom.
Answer:
Succes of Bohrs Theory for hydrogen atom:

• Bohr’s theory gave the information about the principal quantum number. Principal quantum number represents the stationary states. (n = 1, 2, 3 integral numbers)
• Bohr’s theory gave the information about the radius of the stationary states (or) orbits.
  r = 0.529 × n² A (or)
  r = 52.9 × n² pm
  \(\left[r=\frac{n^2 h^2}{4 \pi^2 m e^2}\right]\) (for hydrogenation)
• This theory gave the information about the energy of the electron of particular stationary state.
  E_n = -R_H \(\left[\frac{1}{n^2}\right]\)n = 1, 2, 3,……
  R_H = Ryd berg constant
  = 1,09,677 cm^-1.
• This theory explained the line spectra of hydrogenation.
• This theory can also be applicable to the ions containing only one electron. Eg. : He⁺, Li⁺², Be⁺³….
• This theory can also gave information about velocity of electrons moving in the orbits.

Question 60.
What are the consequences that lead to the development of quantum mechanical model of an atom?
Answer:
Consequences that lead to development of quantum mechanical model of an atom are as follows.

• Clàssical mechanics successfully explained the motion of macro scopic objects.
  Eg: Falling stone, Planets etc.,.
• Classical mechanics failed to explain the motion of microscopic objects like electrons, atoms, molecules etc.
• Classical mechanics ignores the concept of dual behaviour of matter and especially for sub atomic particles
  Quantum mechanics:
  The Branch of science deals with the dual behaviour of matter is called quantum mechanics.
• This deals with the motions of microscopic objects like electron.

Important features of quantum mechanical model of atom:

1. The energy of electrons in an atom is quantized (it can only have certain specific values).
2. The existence of quantized electronic energy levels is a direct result of the wave like properties at electrons and are allowed solution at schrodinger wave equations.
3. All the information about the electron in an atom is contained in its orbital wave function ‘\(\Psi^{\prime}\) and quantum mechanics makes it possible to extract this information from ‘\(\Psi^{\prime}\).
4. The path of the electron can never be determined accurately. Therefore, we find only the probability of the electron at different points in space, around an atom.
5. The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function i.e., \(|\Psi|^2\) at that point. \(|\Psi|^2\) is known as probability density and is always positive. From the value of \(|\Psi|^2\) at different points with in the atom, it is possible to
   predict the region around the nucleus where electron will most probably be found.

Question 61.
Explain the salient features of quantum mechanical model of an atom.
Answer:
Important features of quantum mechanical model of atom:

1. The energy of electrons in an atom is quantized (it can only have certain specific values).
2. The existence of quantized electronic energy levels is a direct result of the wave like properties at electrons and are allowed solution at schrodinger wave equations.
3. All the information about the electron in an atom is contained in its orbital wave function ‘\(\Psi^{\prime \prime}\) and quantum mechanics makes it possible to extract this information from \(\Psi^{\prime \prime}\).
4. The path of the electron can never be determined accurately. Therefore, we find only the probability of the electron at different points in space, around an atom.
5. The probability of finding an electron at a point within an atom is proportional to the square of the
   orbital wave function i.e., \(|\Psi|^2\) at that point. \(|\Psi|^2\) is known as probability density and is always positive. From the value of \(|\Psi|^2\) at different points with in the atom, it is possible to predict the region around the nucleus where electron will most probably be found.

Question 62.
What are the limitations of Bohr’s model of an atom?
Answer:
Limitations:

1. Spectra of multielectron atoms: Bohr’s theory could explain the spectra of Hydrogen and single electron species like He⁺, Li²⁺, Be³⁺, but it fails to explain the spectra of multielectron atoms.
2. Fine structure: It fails to explain this fine structure of Hydrogen atom.
3. Splitting up of spectral lines : The theory fails to explain Zeeman effect and Stark effect.
   The splitting up of spectral lines when an atom is subjected to strong magnetic field is called Zeeman effect.
   The splitting up of spectral lines when an atom is subjected to strong electric field is called Stark effect.
4. Flat model: Bohr’s theory gives a flat model of the orbits. Bohr’s theory predicts definite orbits for electrons considering them as particles. But according to de Brogue electron has both wave nature and particle nature. Bohr’s theory cannot explain this dual role.
5. It fails to support the uncertainty principle proposed by Heisenberg.
6. It could not explain the ability of atoms to form molecules by chemical bonds.

Question 63.
What are the evidences in favour of dual behaviour of electron?
Answer:

• The particle nature of light explains the phenomenon of blackbody radiations and photo electric effect but it could not explain about wave nature of light.
• Wave nature of light explains the phenomenon of interference and diffraction.
• So, light has dual nature i.e it behaves as a wave (or) as a stream of particles.
• According to de-Broglie, light has dual behaviour i.e both particle and wave nature.
  de-Broglies gave the following relationship
  λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\)
  λ = wave length
  P = momentum
• Heisen bergs uncertainty principle also a consequence of dual behaviour of matter and radiation.
  Statement :— It is impossible to determine simultaneously, the exact momentum and exact position of a small particle like lectron.
  Δx × Δp ≥ \(\frac{h}{4 \pi}\)
  Δx = uncertainty in position
  Δp = uncertainty in momentum

Question 64.
How are the quantum numbers n, l and m_l, arrived at? Explain the significance of these quantum numbers. (A.P. Mar. ‘16)(T.S. Mar. ‘15, ‘14)
Answer:

• In general a large no.of orbitals are possible in an atom.
• These orbitaIs are distinguished by their size, shape and orientation.
• An orbital of smaller size means there is more chance to find electron near the nucleus.
• Atomic orbitals are precisely distinguished by quantum numbers. Each orbital is designated by three major quantum numbers.

1) Principal quantum number (n)
2) Azimuthal quantum number (l)
3) Magnetic quantum number (m)

1) Principal quantum number : The principal quantum number was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the distance between the nucleus and the orbit also increases.
It is denoted by the letter ‘n’. It can have any simple integer value 1, 2, 3, ……. but not zero. These are also termed as K, L, M, N etc.
The radius and energy of an orbit can be determined basing on ”n” value.
The radius of n^th orbit is r_n = \(\frac{n^2 h^2}{4 \pi^2 m e^2}\)
The energy of n^th orbet is E_n = \(\frac{-2 \pi^2 m e^4}{n^2 h^2}\)

2) Azimuthal quantum number: It was proposed by Sommerfeld. it is also known as angular momentum quantum number or subsidiary quantum number.
it indicates the shapes of orbitals. It is denoted by ‘l’. The values of ‘l’ depend on the values of ‘n’, ‘l’ has values ranging from ‘0 to (n – 1) i.e.. l = 0, 1, 2,….. (n – 1). The maximum number of electrons present in the subshells s, p, d, f are 2, 6, 10, 14 respectively.

3) Magnetic quantum number: It was proposed by Lande. It shows the orientation of the orbitals in space. ‘p’ — orbital has three orientations. The orbital oriented along the x-axis is called p_x orbital, along the y-axis is called p_y -orbital and along the z-axis is called p_z orbital. In a similar way d – orbital has five orientations. They are d_xy, d_yz, d_zx, d_{x² – y²} and d_z². It is denoted by ‘m’. Its values depends on azimuthal quantum number, ‘m’ can have all the integral values from -l to +l including zero. The total number of ‘m’ values are (2l + 1).

Question 65.
Explain the dual behaviour of matter. Discuss its significance to microscopic particles like electrons.
Answer:

• The particle nature of light explains the phenomenon of blackbody radiations and photo electric effect but it couldnot explain about wave nature of light.
• Wave nature of light explains the phenomenon of interference and diffraction.
• So, light has dual nature i.e it behaves as a wave (or) as a stream of particles.
• According to de-Broglie light has dual behaviour i.e both particle and wave nature.
  de-Broglies gave the following relationship
  λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\)
  λ = wavelength
  P = momentum
• Heisen bergs uncertainty principle also a consequence of dual behaviour of matter and radiation.
  Statement :— It is impossible to determine simultaneously, the exact momentum and exact position of a small particle like electron.
  Δx × Δp ≥ \(\frac{h}{4 \pi}\) = uncertainty in position
  Δx = uncertainty in position
  Δp = uncertainty in momentum

Significance of Uncertainty Principle:

1. This principle rules out the existence of definite paths or trajectories of electrons and other similar
   particles.
2. This principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.
3. In dealing with milligram size or heavier objects, the associated uncertainties are hardly of any real consequence.

Question 66.
What are various ranges of electromagnetic radiation ? Explain the characteristics of electromagnetic radiation.
(or)
Explain diagrammatically the boundary surfaces for three 2p orbitais and five 3d
Answer:
Electromagnetic radiation : When electrically charged particle is accelerated alternating electric and magnetic fields are produced and transmitted. These fields are transmitted in the form of waves called electromagnetic waves or electromagnetic radiation.

Important Characteristics of a wave:

1) These are produced by oscillating charged particles in a body.
2) These radiations can pass through vacuum also. So medium for transmission is not required.
3) Velocity (c) : It is defined as the linear distance travelled by the wave in one second.
Units : cm sec^-1 (or) m see^-1
All kinds of electromagnetic waves have the same velocity.
(3 × 10⁸ m sec^-1 or 3 × 10¹⁰ cm sec^-1)

4) Wavelength (λ) : It is defined as the distance between any two successive crests or troughs of wavez.
Units:A ; m ; cm ; nm or pm 1 A° = 10^-10 m
1 nm = 10^-9 m = 10^-7 cm
1 pm = 10^-12m.

5) Frequency (v): It is defined as the number of waves passing through a point in one second.
Units: Hertz (Hz); cycles sec^-1 or sec^-1.
v = \(\frac{c}{\lambda}\)

6) Wave number (\(\bar{v}\)): It is defined as the number of waves present in one unit length. It is equal
to the reciprocal of the wavelength.
\(\bar{v}=\frac{1}{\lambda}=\frac{v}{c}\)
Relation between wavelength and frequency :
c = v × λ ⇒ v = \(\frac{c}{\lambda}\)

7) Amplitude (A) is the height of the crest (or) depth of through of a wave, It determines intensity (or) brightness of the wave.

Question 67.
Define atomic orbital. Explain the shapes of s, p and d orbitals with the help of diagrams.
Answer:
Atomic orbital : A three dimensional space around the nucleus in an atom. Where the probability of finding an electron is maximum (i.e.,) \(\Psi^2\) is maximum is called an atomic orbital.

Shapes of orbitals:

a) s – orbital : Boundary surface diagram for ‘s’ orbital is spherical in shape, ‘s – orbitals are spherically symmetric (i.e.,) the probability of finding the electron at a given distance is equal in all directions.

b) p – orbitals : p – orbital consists of two sections called lobes that are either side of the plane that passes through the nucleus. The size, shape and energy of the three orbitals are identical. They differ only in the orientation. These are mutually perpendicular to each other and oriented along x, y and z axes. Each p-orbital is of dumb-bell shape.

c) d – orbitals: Five d-orbitals are designated as d_xy, d_yz, d_zx, d_{x² – y²} and d_z². The shapes of first four d – orbitals are similar to each other of double dumb-bell whereas that of the fifth one dd_z² is different from others, but all five d-orbitals are equivalent in energy.

Question 68.
Illustrate the reasons for the stability of completely filled and half filled subshells.
Answer:
Chromium and Copper shows anamalous electronic configurations
Cr – [Ar] 4s¹ 3d⁵
Cu – [Ar] 4¹ 3d¹⁰

• Cr — gets half filled 3d— shell electronic configuration.
• Cu — gets full filled 3d— shell electronic configuration.
• Half filled and full filled subshells are more stable than others.
  Causes of Stability of Completely filled and Half filled Sub-shells
  The completely filled and half filled sub-shells are stable due to the following reasons :

1. Symmetrical distribution of electrons : It is well known that symmetry leads to stability. The completely filled or half filled subshells have symmetrical distribution of electrons in them and are therefore more stable. Electrons in the same subshell (here 3d) have equal energy but different spatial distribution, Consequently, their shielding of one another is relatively small and the electrons are more strongly attracted by the nucleus.


2. Exchange Energy: The stabilizing effect arises whenever two or more electrons with the same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half filled or completely filled.
As a result the exchange energy is maximum and so is the stability.

The extra stability of half-filled and completely filled subshell is due to:

1. relatively small shielding,
2. smaller coulombic repulsion energy and
3. larger exchange energy.

Question 69.
Explain emission and absorption spectra. Discuss the general description of line spectra in hydrogen atom.
Answer:
Emission spectrum:

1. It is produced by analysing the radiant energy emitted by an excited substance.
2. It consists of bright lines on dark background.
3. Produced due to the emission of energy by electrons.
4. Emission spectra contains bright lines on dark background.

Absorption spectrum:

1. It is produced when white light is passed through a substance and the transmitted light is analysed by a spectrograph.
2. It consists of dark lines on bright background.
3. Produced due to the adsorption of energy by electrons.
4. Absorption spectra contains dark lines on bright background.
   Line spectra of Hydrogen — Bohrs Theory:
   • In case of hydrogen atom line spectrum is observed and this can be explained by using Bohr’s Theory.
   • According to Bohrs postulate when an electronic transition takes place between two stationary states that differ in energy is given by
     ΔE = E_f – E_i
     E_f = final orbit energy
     E_i = initial orbit energy

In terms of wave numbers

• In case of absorption spectrum n_f > n_i → energy is absorbed (+ Ve)
• In case of emission spectrum n_i > n_f → energy is emitted (- Ve)
• Each spectral line in absorption (or) emission spectrum associated to the particular transition in hydrogen atom
• In case of large no.of hydrogen atoms large no.of transitions possible they results in large no.of spectral lines.

The Spectral Lines for Atomic Hydrogen

Solved Problems

Question 1.
Calculate the num Notons, neutrons and electrons species?
Solution:
In this case, \({ }_{35}^{80} \mathrm{Br}\), Z = 35, 80, species is neutral
Number of protons = number of electrons = Z = 35
Number of neutrons = 80 – 35 = 45, Mass number (A) = number of protons (Z) + number of neutrons (n).

Question 2.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
Solution:
The atomic number is equal to number of protons =16. The element is sulphur (S).
Atomic mass number = number of protons + number of neutrons = 16 + 16 = 32
Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge equal to excess electrons = 18 – 16 = 2.
Symbol is \(\frac{32}{16} \mathrm{~s}^{2-}\)
Note: Before using the notation find out whether the speclés is a neutral atom, acation or an anion, If it is a neutral atom, Atomic number (Z) = number of protons in the nucleus of an atom = number of electrons in a neutral atom is valid, i.e., number of protons = number of electrons = atomic number. If the species is an ion, determine whether the number of protons are larger (cation, positive ion) or smaller (anion, negative ion) than the number of electrons. Number of neutrons is always given by A-Z, whether the species is neutral or ion.

Question 3.
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electro-magnetic radiation emitted by transmitter. Which part of the electromagnetic specturm does it belong to?
Solution:
The wavelength, λ, is equal to c/v. where c is the speed of electromagnetic radiation in vacuum and v is the frequency. Substituting the given values, we have

Question 4.
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wave
lengths in frequencies (Hz). (1 nm = 10^-9 m).
Solution:
Using c = v λ frequency of violet light.
V = \(\frac{\mathrm{c}}{\mathrm{v}}=\frac{3.00 \times 10^8 \mathrm{~ms}^{-1}}{400 \times 10^{-9} \mathrm{~m}}\)
= 7.50 × 10¹⁴ Hz
Frequency of red light
v = \(\frac{c}{v}=\frac{3.00 \times 10^8 \mathrm{~ms}^{-1}}{750 \times 10^{-9} \mathrm{~m}}\) = 4.00 × 10¹⁴ Hz
The range of visible spectrum is from
4.0 × 10¹⁴ to 7.5 × 10¹⁴Hz in terms of frequency units.

Question 5.
Calculate
(a) wave number and
(b) frequency of yellow radiation having wavelength 5000 A.
Solution:
(a) Calculation of wavenumber \((\bar{v})\)
λ = 5800 A = 5800 × 10^-8 cm
= 5800 × 10^-10m
\((\bar{v})\) = \(\frac{1}{\lambda}=\frac{1}{5800 \times 10^{-10} \mathrm{~m}}\)
= 1.724 × 10⁶m^-1
= 1.724 × 10⁴ cm^-1

Question 6.
Calculate energy of one mole of photons of radiation whose frequency is 5 × 10¹⁴ Hz.
Solution:
Energy (E) of one photon is given by the expression
E = hv .
h = 6.626 × 10^-34 J s
v = 5 × 10¹⁴ s^-1 (given)
E = (6.626 × 10^-34J s) × (5 × 10¹⁴ s^-1)
= 3.313 × 10^-19 J
Energy of one mole of photons
= (3.313 × 10^-19J) × (6.022 × 10²³ mol^-1)
= 199.51 kJ mol^-1.

Question 7.
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Solution:
Power of the bulb = 100 watt
= 100 J s^-1
Energy of one photon E = hv = hc/λ.
= \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{400 \times 10^{-19} \mathrm{~m}}\)
= 4.969 × 10^-19J
Number of photons emitted
\(\frac{100 \mathrm{~J} \mathrm{~s}^{-1}}{4.969 \times 10^{-19} \mathrm{~J}}\) = 2.012 × 10²⁰ s^-1

Question 8.
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 × 10⁵ J mol^-1. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
Solution:
The energy (E) of a 300 nm photon is given by
hv = hc/λ.
= \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{300 \times 10^{-9} \mathrm{~m}}\)
= 6.626 × 10^-19 J
The energy of one mole of photons
= 6.626 × 10^-19 J × 6.022 × 10²³ mol^-1
= 3.99 × 10⁵ mol^-1
The minimum energy needed to remove one mole of electrons from sodium
= (3.99 – 1.68) 10⁵ J mol^-1
= 2.31 × 10⁵ J mol^-1
The minimum energy for one electron
= \(\frac{2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}}{6.022 \times 10^{23} \text { elelctrons } \mathrm{mol}^{-1}}\)
= 3.84 × 10^-19J
This corresponds to the wavelength
∴ λ = \(\frac{\mathrm{hc}}{\mathrm{E}}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{3.84 \times 10^{-9} \mathrm{~J}}\)
= 517 nm
(This corresponds to green light)

Question 9.
The threshold frequency v₀ for a metal is 7.0 × 10¹⁴ s^-1. Calculate the kinetic energy of an electron emitted when radiation of frequency v = 1.0 × 10¹⁵ s^-1 hits the metal.
Solution:
According to Einstein’s equation
Kinetic energy = 1/2 m_ev² = h(v – v₀)
= (6.626 × 10^-34 J s)
(1.0 × 10¹⁵ s^-1 – 7.0 × 10¹⁴ s^-1)
= (6.626 × 10^-34 J s)
(10.0 × 10¹⁴s^-1 – 7.0 × 10¹⁴s^-1)
= (6.626 × 10^-34 J s)
(3.0 × 10¹⁴s^-1) = 1.988 × 10^-19 J

Question 10.
What are the frequency and wave-length of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?
Solution:
Since n₁ = 5 and n_f = 2, this transition gives rise to a spectral line in the visible region of the Balmer series. From ΔE = \(\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_i^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right)\)
= 2.18 × 10^-18J\(\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)\)
ΔE = 2.18 × 10^-18J\(\left(\frac{1}{5^2}-\frac{1}{2^2}\right)\)
= -4.58 × 10^-19 J.
It is an emission energy.
The frequency of the photon (taking energy in terms of magnitude) is given by

Question 11.
Calculate the energy associated with the first orbit of He⁺. What is the radius this orbit?
Solution:

The radius of the orbit is given by

Question 12.
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s^-1?
Solution:
According to de Brogue λ = \(\frac{h}{m V}=\frac{h}{p}\)
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)}{(0.1 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)}\)
= 6.626 × 10^-34m (J = kg m² s^-2)

Question 13.
The mass of an electron is 9.1 × 10^-31 kg. If its K.E. is 3.0 × 10^-25 J, calculate its wavelength.
Solution:

Question 14.
Calculate the mass of a photon with wavelength 3.6 A.
Solution:
λ = 3.6 A = 36 × 10^-10 m
Velocity of photon = velocity of light
m = \(\frac{\mathrm{h}}{\lambda v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(3.6 \times 10^{-10} \mathrm{~m}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}\)
= 6.135 × 10^-29kg

Question 15.
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 A. What
is the uncertainty involved in the measurement of its velocity?
Solution:
∆x ∆p = \(\frac{h}{4 \pi}\) or ∆x m∆v = \(\frac{h}{4 \pi}\)
∆v = \(\frac{h}{4 \pi \Delta x m}\)
∆v = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 0.1 \times 10^{-10} \mathrm{~m} \times 9.1 \times 10^{31} \mathrm{kc}}\)
= 0.579 × 10⁷ m s^-1 (1 J = 1 kg m² s^-2)
= 5.79 × 10⁶ ms^-1

Question 16.
A golf ball has a mass of 40g and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.
Solution:
The uncertainty in the speed is 2%, i.e.,
4 × \(\frac{2}{100}\) = 0.9 ms^-1
Using the λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}\)
Δx = \(\frac{h}{4 \pi m \Delta v}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 40 \mathrm{~g} \times 10^{-3} \mathrm{~kg} \mathrm{~g}^{-1}\left(0.9 \mathrm{~ms}^{-1}\right)}\)
= 1.46 × 10^-33 m
This is nearly \(\sim\) 10¹⁸ times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

Question 17.
What is the total number of orbitals associated with the principal quantum number n = 3?
Solution:
For n = 3, the possible values of 1 are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and m_l = 0); there are three 3p orbitals
(n = 3, 1 = 1 and m_l = -1, 0, +1); there are five 3d orbitais (n = 3, l = 2 and m_l = -2, -1, 0, +1, +2).
Therefore, the total number of orbitals is 1 + 3 + 5 = 9
The same value can also be obtained by using the relation; number of orbitals = n²,
i.e. 3² = 9

Question 18.
Using s, p, d, f notations, describe the orbital with the following quantum numbers
(a) n = 2, l = 1,
(b) n = 4, l = 0,
(c) n = 5, l = 3,
(d) n = 3, l = 2
Solution:

Question 19.
Calculate its wave length of 1^st line in Balmer series of hydrogen spectrum.
Solution:
Ryd berg’s equation \(\bar{v}\) = \(\frac{1}{\lambda}\) = \(\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\)
For the 1^st line of Balmer series
n₁ = 2, n₂ = 3
R = 1,09,677 cm^-1
\(\bar{v}\) = \(\frac{1}{\lambda}\) = 1,09,677\(\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\)
= 1,09,677 × \(\frac{5}{36}\)
Wave no. \((\bar{v})\) = 15232.9 cm^-1
Wave length λ = \(\frac{1}{\bar{v}}\) = \(\frac{1}{15232.9}\) = 6.5 × 10^-5 cm^-1

Question 20.
Calculate the shortest wave length in lyman series of hydrogen spectrum (R_H = 1,09,677 cm^-1).
Solution:
To calculate shortest wave length

Question 21.
What is the maximum no.of emission lines when the excited electron of a ‘H’ atom in n =6 drops to ground state.
Solution:
The no. of spectral lines found when an electron return from n^th orbit to ground state.
= \(\frac{n(n-1)}{2}\) = \(\frac{6(6-1))}{2}\) = \(\frac{30}{2}\) = 15

Question 22.
Calculate the longest wavelength transition in the paschen series of He⁺.
Solution:

Question 23.
The no. of waves in the forth Bohr’s orbit of hydrogen is
a) 3
b) 4
c) 9
d) 12
Solution:

Question 24.
It the speed of the electron in 1^st Bohr’s orbit of hydrogen is x, then the speed of the electron in the 3^rd orbit of hydrogen is
Solution:
Given
Velocity of electron in 1^st Bohr’s orbit of hydrogen = x
Velocity of electron in 3^rd Bohr’s orbit of hydrogen = \(\frac{x}{n}\) = \(\frac{x}{3}\)

Question 25.
The ratio of radii of the fifth orbits of He⁺ and Li⁺² will be
a) 2 : 3
b) 3 : 2
c) 4 : 1
d) 5 : 3
Solution:
Z₂(li)=3
Z₁ (He) = 2
\(\frac{r_1}{r_2}\) = \(\frac{Z_2}{Z_1}\) = \(\frac{3}{2}\) = 3 : 2

Question 26.
What is the lowest value of ‘n’ that allows ‘g’ orbitals to exist ?
Solution:
The lowest value of ‘n that allows ‘g’ orbitals to exist is ‘5’.

Question 27.
What is the orbital angular momentum of a d-electron
Solution:
Orbital angular momentum = \(\sqrt{l(l+1)} \frac{\mathrm{h}}{2 \pi}\)
For a d electron l = 2
= \(\sqrt{2(2+1)} \frac{h}{2 \lambda}\)
= \(\frac{\sqrt{6 h}}{2 \lambda}\)

Question 28.
What is the total spin and magnetic moment of an atom with atomic number 7’?
Solution:
Z = 7 (Nitrogen)
Electronic configuration is 1s² 2s² 2p³ (in ground state)

Question 29.
The quantum number of electrons are given below. Arrange in order of increasing energies.

Solution:
a) 4d
b) 3d
c) 4p
d) 3d
e) 3p
f) 4p
∴ Increasing order of energy
e < b = d < c = f < a

Question 30.
If the value of n + l = 7 then what should be the increasing order of energy of the possible subshells.
Solution:
Given
n + l = 7

∴ The increasing order of energy
4f < 5d < 6p < 7s
(According Aufbau principle)

Question 31.
Which of the following sets of quantum number is not permitted?
a) n = 3, l = 3, m = +1, s = +\(\frac{1}{2}\)
b) n = 3, l = 3, m = +2, s = –\(\frac{1}{2}\)
c) n = 3, l = 1, m = +2, s = –\(\frac{1}{2}\)
d) n = 3, l = 0, m = 0, s = +\(\frac{1}{2}\)
Solution:
Only d is permitted i.e., 3s¹
In a, b n = l but n always > l
in c m = + 2 is not permitted
Because l = 1,’m’ has -1, 0, +1 values only.

Question 32.
Ground state electronic configuration of nitrogenators can be represented as

Solution:
(a) and (d) are correct representations.

Question 33.
Which of the following is possible
a) 3f
b) 4d
c) 2d
d) 3p
Solution:
4d and 3p are possible.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 10th Lesson The p-Block Elements – Group 13 Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 10th Lesson The p-Block Elements – Group 13

Very Short Answer Questions

Question 1.
Discuss the pattern of variation in the oxidation states of Boron to Thallium.
Answer:

• Boron exhibits – 3 oxidation state due to its small size and non metalic nature.
• Aluminium exhibits +3 oxidation state.
• Gallium, Indium and Thallium exhibits both +1 and +3 oxidation states.
• In Thallium +1 oxidation state is more stable than +3 due to inert pair effect.

Question 2.
How do you explain higher stability of TlCl₃ ?
Answer:
[TlCl₃ is not stable because Tl doesnot exist in Tl³ stable state.] [TlCl is stable because of inertpair effect Tl⁺¹ is stable].

Question 3.
Why does BF₃ behave as a Lewis acid ?
Answer:
BF₃ is a electron deficient molecule. It has the tendency to accept an electron pair. Electron pair acceptors are Lewis acids. Hence BF₃ behave as a Lewis acid.

Question 4.
Is boric acid a protic acid ? Explain.
Answer:
Boric acid is a weak mono basic acid. In Boric acid plannar BO₃ units are joined by hydrogen bonds. It has layer structure (polymeric). Hence it is not a protic acid. It does not give up a proton.

Question 5.
What happens when boric acid is heated ?
Answer:
Boric acid when heated above 370 K forms meta boric acid. This on heating forms Boric oxide.

Question 6.
Describe the shapes of BF₃ and BH₄^–. Assign the hybridization of boron in these species.
Answer:

• Shape of BF₃ molecule is Trigonal planar
  Hybridisation of ‘B’ in BF₃ is sp²
• Shape of BH₄^– molecule is Tetrahedral
  Hybridisation of ‘B’ in BH₄^– is sp³

Question 7.
Explain why atomic radius of Ga is less than that of ‘Al’.
Answer:
In Gallium penultimate shell contains 10-d electrons. Due to this 10-d electrons shielding effect becomes poor on outer most electrons. So nuclear charge increased in Gallium. Hence atomic radius of Ga is lessthan that of ‘Al’.

Question 8.
Explain inert pair effect.
Answer:
The reluctance of ‘ns’ pair of electrons to take part in bond formation is called inert pair effect.
(or)
The occurrence of oxidation states two unit lessthan the group oxidation states is called inert pair effect.
Eg : In Group – 13 Tl exhibits +1 oxidation state instead of +3 oxidation state due to inert pair effect.

Question 9.
Write balanced equations for
a) BF₃ + LiH →
b) B₂H₆ + H₂O →
c) NaH + B₂H₆ →
d) H₃BO₃
e) B₂H₆ + NH₃
Answer:
a) 2BF₃ + 6LiH → B₂H₆ + 6 LiF
b) B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂
c) B₂H₆ + 2NaH → 2NaBH₄
d) H₃BO₃ HBO₂ B₂O₃
e) B₂H₆ + 6NH₃ → 3[BH₂(NH₃)₂]⁺ (BH₄)^– 2B₃N₃H₆ + 12H₂

Question 10.
Why is boric acid polymeric ?
Answer:
Boric acid has layer like lattice. In this structure planar BO₃ units are joined by hydrogen bonds and forms a polymeric structure.

Question 11.
What is the hybridization of B in diborane and borazine ?
Answer:

• In diborane ‘B’ hybridisation is sp³
• In Borazine ‘B’ hybridisation is sp²

Question 12.
Write the electronic configuration of group – 13 elements.
Answer:
General outer electronic configuration of Group – 13 elements is ns²np¹

• B – 1s²2s²2p¹
• Al – [Ne] 3s²3p¹
• Ga – [Ar] 3d¹⁰4s²4p¹
• In – [Kr] 4d¹⁰5s²5p¹
• Tl – [Xe] 5d¹⁰ 6s²6p¹

Question 13.
Give the formula of borazine. What is its common name ?
Answer:

• The formula of borazine is B₃N₃H₆.
• It’s common name is “In organic benzene” because it is – iso structural with benzene.

Question 14.
Give the formulae of
a) Borax
b) Colemanite.
Answer:
a) Formula of Borax is Na₂B₄O₇. 10H₂O
b) Formula of Colemanite is Ca₂B₆O₁₁.5H₂O

Question 15.
Give two uses of aluminium.
Answer:
Uses of Aluminium :

• Aluminium is used in packing.
• Aluminium is used in utensil making.
• Aluminium alloys are used in shaping of pipes, tubes, wires etc.
• Aluminium alloys are used in making air craft bodies.

Question 16.
What happens when
a) LiAlH₄ and BCl₃ mixture in dry ether is warmed and
b) Borax is heated with H₂SO₄ ?
Answer:
a) When LiAlH₄ and BCl₃ mixture is warmed in dry ether diborane is formed.
4BF₃ + 3 LiAlH₄ → 2B₂H₆ + 3 LiF + 3 AlF₃
b) Borax is heated with H2S04 then boric acid is formed
Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃

Question 17.
Sketch the structure of Orthoboric acid.
Answer:

Question 18.
Write the structure of AlCl₃ as a climer.
Answer:

Question 19.
Metal borides (having ¹⁰B) are used as protective shield – Why ?
Answer:
Boron – 10 (¹⁰B) has the capacity to absorb neutrons. Hence metal borides (having ¹⁰B) are used as protective shields in nuclear industry.

Short Answer Questions

Question 1.
Write reactions to justify amphoteric nature of aluminium.
Answer:

• Amphoteric nature means having acidic as well as basic nature.
• Aluminium reacts with both mineral acids as well as aqueous alkalis.

a) Reaction with mineral acid :
‘Al’ reacts with dil.HCl and liberates hydrogen gas.
2Al + 6HCl → 2AlCl₃ + 3H₂ ↑

b) Reaction with aqueous alkali :
‘Al’ reacts with aqueous alkali (NaOH) and liberates hydrogen gas.
2Al + 2NaOH + 6H₂O → Na₂[Al(OH)₄] + 3H₂ ↑

Question 2.
What are electron deficient compounds ? Is BCl₃ an electron deficient species ? Explain.
Answer:
These are the compounds in which the available no.of valence electrons is lessthan the number required for normal covalent bond formation (or) for writting the Lewis structure of the molecule.

• These compounds are electron pair acceptors and acts as Lewis acids.
• BCl₃ is an electron deficient compound.
• In BCl₃ ‘B’ contains only six electrons instead of eight electrons.
• BCl₃ has the tendency to accept an electron pair and acts as Lewis acid.
  Eg : Formation of BCl₃ . NH₃ :-
  BCl₃ accepts an electron pair from NH₃ and forms the compound BCl₃.NH₃ (Tetrahedral)
  

Question 3.
Suggest reasons why the B – F bond lengths in BF₃ (130 pm) and BF₄^– (143 pm) differ.
Answer:
About BF₃ :

• In BF₃ the central atom ‘B’ contains three bond pairs in the valency shell.
• ‘B’ under goes sp² hybridisation.
• Shape of the molecule is trigonal planar.

About BF₄^– :

• In BF₄^– the central atom ‘B’ contains four bond pairs in the valency shell.
• ‘B’ under goes sp³ hybridisation.
• Shape of the molecule is tetrahedral.
• The above reasons suggent that the difference in bond lengths of BF₃ (130 pm) and BF₄^– (143 pm).

Question 4.
B – Cl bond has a bond moment. Explain why BCl₃ molecule has zero dipolemoment.
Answer:

• B – Cl bond is a polar bond so it has bond moment.
• BCl₃ molecule is non-polar because of its symmetrical structure. (Trigonal planar structure)
• Symmetrical molecules has zero dipole moment.
  ∴ Dipole moment of BCl₃ (μ) = 0

Question 5.
Explain the structure of boric acid.
Answer:

• Boric acid has a layer lattice.
• In this layer lattice planar BO₃ units are joined by hydrogen bonds.
• The structure of Boric acid is polymeric as shown in following figure.
  
• In the above structure dotted lines represents the hydrogen bonds

Question 6.
What happens when
a) Borax is heated strongly
b) Boric acid is added to water
c) Aluminium is heated with dilute NaOH
d) BF₃ is treated with ammonia
e) Hydrated alumina is treated with aq.NaOH solution.
Answer:
a) Borax on heating first loses water molecules and forms sodium tetraborate. This on further heating forms a mixture of sodium metaborate and boric an hydride. This mixture is solidifies into glass like substance.

b) Boric acid is added to water, boric acid accepts a hydroxyl ion from water.
B(OH)₃ + 2H₂O → [B(OH)₄]^– + H₃O⁺

c) Aluminium is heated with dilute NaOH, sodium metaluminate is formed with the liberation of hydrogen gas.
2Al + 2NaOH → 2NaAlO₂ + H₂ ↑

d) BF₃ is treated with NH₃ an addition compound BF₃. NH₃ is formed. BF₃ accepts an electron pair from NH3 and forms a dative bond.
BF₃ + → [BF₃ ← NH₃] → [BF₃.NH₃]

e) Hydrated Alumina is treated with aq.NaOH to form sodium metaluminate.
Al₂O₃.2H₂O + 2NaOH_(aq) → 2NaAlO_(aq) + 3H₂O

Question 7.
Give reasons
a) Conc.HNO₃ can be transported in aluminium container.
b) A mixture of dil. NaOH and aluminium pieces is used to open drain.
c) Aluminium alloys are used to make aircraft body.
d) Aluminium utensils should not be kept in water overnight.
e) Aluminium wire is used to make transmission cables.
Answer:
a) Cone. HNO₃ can be transported in Aluminium containers because Al is passive towards Conc.HNO₃ due to the formation of thin layer of Al₂O₃ on the surface.

b) A mixture of dil.NaOH and aluminium pieces is used to open drain because it acts as cleaning agent.
2Al + 2NaOH → 2NaAlO₂ + H₂

c) Aluminium alloys are used to make air craft body because it is a light metal, soft, malleable, ductile and tenacious. It shows resistance to atmosphere corrosion.

d) Aluminium utensils should not be kept in water overnight because Aluminium reacts with water and liberates hydrogen gas and heat. It makes colour dissolving and sometimes Aluminium compounds are toxic in nature.

e) Aluminium wire used to make transmission cables because of it’s good conductivity (electrical) and resistance to atmospheric corrosion.

Question 8.
Explain why the electronegativity of Ga, In and Tl will not vary very much.
Answer:

1. Ga, In and Tl have the electro negativity values 1.6, 1.7 and 1.8 respectively.
2. In Ga, In and Tl the d-electrons (d¹⁰) in penultimate shell do not shield the outer most electrons from nuclear attraction effectively.
3. The reason for the above fact is.the shielding effect of various electrons in the orbitals follows the order
   s > p > d > f.
4. Hence the outer electrons are held more firmly by the nucleus. Because of this, atoms with d- electrons in the penultimate shell (d¹⁰) are smaller in size. Ga, In and Tl has same number of penultimate shell electrons.
5. So, Ga, In and Tl will not vary very much in their electronegativities.

Question 9.
Explain Borax bead test with a suitable example. [T.S. Mar. 16] [Mar. 13]
Answer:
Borax bend test: This test is useful in the identification of basic radicals in qualitative analysis. On heating borax swells into a white, opaque mass of anhydrous sodium tetra borate. When it is fused, borax glass is obtained. Borax glass is sodium meta borate and B203. The boric anhydride, B203, combined with metal oxides to form metal metaborates as coloured beads. The reactions are as follows :

Question 10.
Explain the structure of diborane. [A.P. Mar. 16] [A.P. & T.S. Mar. 15]
Answer:
Diborane is an electron deficient compound. It has ’12’ valency electrons for bonding purpose instead of ’14’ electrons.

In diborane each boron atom undergoes sp3 hybridization out of the four hybrid orbitals one is vacant.

Each boron forms two, σ – bonds (2 centred – 2 electron bonds) bonds with two hydrogen atoms by overlapping with their ‘1s’ orbital.

The remaining hybrid orbitals of boran used for the formation of B-H-B bridge bonds.

In the formation of B-H-B bridge, half filled sp3 hybrid ofbital of one boron atom and vacant sp³ hybrid orbital of second boron atom overlap with 1s orbital of H-atom.

These three centred two electron bonds are also called as banana bonds. These bonds are present above and below the plane of BH₂ units.

Diborane contains two coplanar BH₂ groups. The four hydrogen atoms are called terminal hydrogen atoms and the remaining two hydrogens are called bridge hydrogen atoms.

Bonding in diborane, Each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2- electron bonds but the two bridge bonds are 3-centre-2- electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 11.
Explain the reactions of aluminium with acids.
Answer:
Reactions of ‘Al’ with adds :
i) DiS. (or) cone. HCl dissolves Al and gives H₂.

ii) a) Dil.H₂SO₄ liberates H₂.

b) Cone. H₂SO₄ dissolves the metal ‘Al’ and gives SO₂.

iii) a) Very dil.HNO₃ is reduced to NH₄ NO₃ by Al.

b) Cone. HNO₃ makes ‘Al’ passive due to the formation of a thin film of oxide layer on the metal surface.

Question 12.
Write a short note on the anamalous behaviour of boron in group – 13.
Answer:

• Among Group – 13 elements ‘B’ is only the non metal.
• ‘B’ forms only covalents compounds.
• ‘B’ doesnot displaces hydrogen from acids.
• ‘B’ shows diagonal relationship with ‘Si’.
• ‘B’ forms acidic oxide where as other elements of group forms amphoteric oxides and basic oxides.
• ‘B’ has only two electrons in the penultimate shell.
• ‘B’ has covalency ‘4’ where as other elements has covalency of maximum ‘6’.

Question 13.
Aluminium reacts with dil.HNO₃ but not with conc.HNO₃ – explain.
Answer:

• Dilute HNO₃ reacts with Aluminium slowly and forms aluminium nitrate and ammonium nitrate.
  8Al + 30 HNO₃ → 8 Al(NO₃)₃ + 3NH₄NO₃ + 9H₂O
• Aluminium doesnot react with cone. HNO₃.

Reasons :

• Aluminium is passive towards cone. HNO₃ due to the formation of thin film of Al₂O₃ layer on the surface.
• Because of this passivity between Al and conc.HNO₃, conc.HNO₃ is transported in Aluminium containers.

Question 14.
Give two methods of preparation of diborane.
Answer:

1. In industries diborane is prepared by the reaction between boron tri fluoride and lithium hydride.
   
2. Boron trichloride and hydrogen mixture subjected to silent electric discharge at low pressure to from diborane.
   2BCl₃ + 6H₂ → B₂H₆ + 6HCl
3. Boron trichloride undergo reduction with LiAlH₄ to form diborane.
   

Question 15.
How does diborane react with
a) H₂O
b) CO
c) N(CH₃)₃ ?
Answer:
a) Diborane reacts with water to form boric acid and hydrogen.
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂↑

b) Diborane reacts with CO at 100° C and 20 atm. pressure to form borane carbonyl.

c) Diborane reacts with N(CH₃)₃ and form a adduct.
B₂H₆ + 2N(CH₃)₃ → 2BH₃.N(CH₃)₃ (adduct)
Reactions b, c are cleavage reactions.

Question 16.
Al₂O₃ is amphoteric – explain with suitable reactions.
Answer:

• Amphoteric oxides are the oxides which possess both acidic as well as basic nature.
  Al₂O₃ possess both acidic as well as basic behaviour.
• Al₂O₃ react with both acids as well as bases to produce salts and water.
  Supporting reactions for amphoteric nature of Al₂O₃
  a) With acids:
  
  b) With bases:
  

Question 17.
Na₂B₄O₇ + Cone. H₂SO₄ → A > B (Green edged flame) Identify A and B
Hint: A = H₃BO₃ B = (C₂H₅)₃ BO₃.
Answer:
Na₂B₄O₇ + Cone. H₂SO₄ + 5H₂O → 4H₃BO₃ A) B) 4(C₂H₅)₃BO₃

• ‘A’ is H₃BO₃
• ‘B’ is (C₂H₅)₃ BO₃.

Long Answer Questions

Question 1.
How are borax and boric acid prepared ? Explain the action of heat on them.
Answer:
Preparation of Borax:
Boric acid on heating first forms tetraboric acid. This on reaction with sodium hydroxide to form borax.

Preparation of Boric acid : Borax is treated with conc.H₂SO₄ boric acid is formed.
Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + Na₂SO₄.

Heating of Borax :
Borax on heating first loses the water molecules and forms sodium tetraborate. This on further heating forms a mixture of sodium meta borate and boric anhydride.
Na₂B₄O₇.10H₂O Na₂B₄O₇ imgg 2 2 NaBO₂ + B₂O₃.

Heating of Boric acid :
Boric acid on heating forms boric anhydride. The reaction depends on the temperature used.

Question 2.
How is diborane prepared ? Explain its structure.
Answer:
In industries diborane is prepared by the reaction between boran tri fluoride and lithium hydride.

Boron trichioride and hydrogen mixture subjected to silent electric discharge at low pressure to from diborane.
2BCl₃ + 6H₂ → B₂H₆ + 6HCl
Boron trichloride undergo reduction with LiAlH₄ to form diborane.

Diborane is an electron deficient compound. It has ’12’ valency electrons for bonding purpose instead of ’14’ electrons.

In diborane each boron atom undergoes sp3 hybridization out of the four hybrid orbitals one is vacant.

Each boron forms two, σ – bonds (2 centred – 2 electron bonds) bonds with two hydrogen atoms by overlapping with their ‘1s’ orbital.

The remaining hybrid orbitals of boran used for the formation of B-H-B bridge bonds.

In the formation of B-H-B bridge, half filled sp3 hybrid ofbital of one boron atom and vacant sp³ hybrid orbital of second boron atom overlap with 1s orbital of H-atom.

These three centred two electron bonds are also called as banana bonds. These bonds are present above and below the plane of BH₂ units.

Diborane contains two coplanar BH₂ groups. The four hydrogen atoms are called terminal hydrogen atoms and the remaining two hydrogens are called bridge hydrogen atoms.

Bonding in diborane, Each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2- electron bonds but the two bridge bonds are 3-centre-2- electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 3.
Write any two methods of preparation of diborane. How does it react with
a) Carbon monoxide and
b) Ammonia ?
Answer:
Preparation of diborane:
In industries diborane is prepared by the reaction between boroh tri fluoride and lithium hydride.

Boron trichloride and hydrogen mixture subjected to silent electric discharge at low pressure to from diborane.
2BCl₃ + 6H₂ → B₂H₆ + 6HCl
Boron trichloride undergo reduction with LiAlH₄ to form diborane.

a) Reaction with carbon monoxide :
Diborane reacts with CO at 100° C and 20 atm. pressure to form borane carbonyl.

b) Reaction with ammonia
Diborane reacts with ammonia at 120° C first forms B₂H₆.2NH₃ (or) [BH₂(NH₃)₂]⁺[BH₄]^– and on further heating forms borazole (or) borazine. Which is also called as “In organic benzene”. It has iso structural with benzene. Flence it is named as “Inorganic Benzene”.

Solved Problems

Question 1.
Standard electrode potential values, E^Θ for Al³⁺ / Al is – 1.66 V and that of Tl³⁺ / Tl is + 1.26 V. Predict about the formation of M³⁺ ion in solution and compare the electropositive character of the two metals.
Solution:
Standard electrode potential values for two half cell reactions suggest that aluminium has high tendency to make Al³⁺ (aq) ions, whereas Tl³⁺ is not only unstable in solution but is a powerful oxidising agent also. Thus Tl⁺ is more stable in solution than Tl³⁺. Aluminium being able to form +3 ions easily, is more electropositive than thallium.

Question 2.
White fumes appear around the bottle of anhydrous aluminium chloride. Give reason.
Solution:
Anhydrous aluminium chloride is partially hydrolysed with atmospheric moisture to liberate HCl gas. Moist HCl appears white in colour.

Question 3.
Boron is unable to form \(\mathrm{BF}_6^{3-}\) ion. Explain.
Solution:
Due to non – availability of d orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4.

Question 4.
Why is boric acid considered as a weak acid ?
Solution:
Because it is not able to release H⁺ ions on its own. It receives OH⁺ ions from water molecule to complete its octet and in turn releases H⁺ ions.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14 Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14

Very Short Answer Questions

Question 1.
Discuss the variation of oxidation states in the group -14 elements.
Answer:

• The common oxidation states exhibited by group – 14 elements are +4 and +2.
• Carbon exhibits negative oxidation states.
• Heavier elements exhibits +2 oxidation state.
• The tendency to show +2 oxidation state increases in the order Ge < Sn < pb.
• pb exhibits +2 oxidation state as stable state because of inert pair effect.

Question 2.
How the following compounds behave with water
a) BCl₃
b) CCl₄.
Answer:
a) BCl₃ reacts with water (hydrolysis) to form boric acid.

b) CCl₄ does not undergo hydrolysis due to lack of d-orbitals in the central atom ‘C1 and due to its highly non polar nature, CCl₄ does not acts as Lewis acid.

Question 3.
Are BCl₃ and SiCl₄ electron-deficient compounds ? Explain.
Answer:

• BCl₃ and SiCl₄ are electron-deficient compounds.
• These two compounds behave as Lewis acids.
• These compounds are electron pair acceptors.
• The following reacts support the electron deficiency of BCl₃ and SiCl₄.
  

Question 4.
Give the hybridization of carbon in
a) CO₃^-2
b) diamond
c) graphite
d) fullerene
Answer:
a) In C0₃^-2 carbon atom undergoes sp² hybridisation.
b) In Diamond carbon atom undergoes sp³ hybridisation.
c) In Graphite carbon atom undergoes sp² hybridisation.
d) In Fullerenes carbon atom undergoes sp² hybridisation.

Question 5.
Why is’CO’poisonous ? [T.S. Mar. 16]
Answer:
‘CO’ gas is highly poisonous because it has the ability to form a stable complex with haemoglobin.

Carboxy haemoglobin is 300 times more stable than oxyhaemoglobin.

Question 6.
What is allotropy ? Give the crystalline allotropes of carbon. [Mar. 13]
Answer:

• The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
• Crystalline allotropes of carbon are
  a) Diamond
  b) Graphite
  c) Fullerenes.

Question 7.
Classify the following oxides as neutral, acidic, basic or amphoteric,
a) CO
b) B₂O₃
c) SiO₂
d) CO₂
e) Al₂O₃
f) PbO₂
g) Tl₂O₃
Answer:
a) CO is neutral oxide.
b) B₂O₃ is acidic oxide.
c) SiO₂ is acidic oxide.
d) CO₂ is acidic oxide.
e) Al₂O₃ is amphoteric oxide.
f) PbO₂ is amphoteric oxide.
g) Tl₂O₃ is basic oxide.

Question 8.
Name any two manmade silicates.
Answer:
Glass and cement are man made silicates.

Question 9.
Write the outer electron configuration of group -14 elements.
Answer:
The general outer most electronic configuration of group – 14 elements is ns²np².
1) Carbon – [He] 2s²2p²
2) Silicon – [Ne] 3s²3p²
3) Germanium – [Ar] 3d¹⁰4s²4p²
4) Tin – [Kr] 4d¹⁰ 5s² 5p²
5) Lead – [Xe] 4f⁴ 5d¹⁰ 6s² 6p²

Question 10.
How does graphite function as a lubricant ?
Answer:
Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

Question 11.
Graphite is a good conductor – explain.
Answer:
In graphite carbon undergoes sp² hybridisation. Each carbon forms three a – bonds with three neighbouring carbon atoms. Fourth electron forms TC – bond and it is delocalised. Due to the presence of these moving (or) free electrons graphite acts as good conductor.

Question 12.
Explain the structure of silica.
Answer:

• Silica is a giant molecule with 3 – dimensional structure.
• In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
• Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
• Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
• In SiO₂ Silicon atom undergoes sp³ hybridisation.

Question 13.
What is ‘Synthesis gas’?
Answer:

• Water gas is also called as synthesis gas.
• It is a mixture of CO and H₂.
• It s prepared by passing steamover hot coke.
• It is used for the synthesis of methanol and a number of hydrocarbons. Hence it is called synthesis gas.

Question 14.
What is producer gas?
Answer:

• Producer gas is mixture of CO and N₂.
• It is prepared by passing air over hot coke.

Question 15.
Diamond has high melting point – Explain.
Answer:

• In Diamond each carbon undergoes sp³ hybridisation and it is surrounded by four other carbon atoms with strong a – bonds tetrahedrally.
• The C – C bond energy in diamond is very high and it has 3 – dimensional structure.
• Due to these reasons diamond has high melting point.
• It has melting point 4200 K.

Question 16.
Give the use of CO₂ in photosynthesis.
Answer:
The process of converting the atmospheric CO₂ into Carbohydrates by green plants is known as ‘photosynthesis’.
In Photosynthesis CO₂ changes to carbohydrates such as glucose.

Question 17.
How does CO₂ increase the green house effect ?
Answer:

• Green plants absorbs CO₂ gas for photosynthesis and releases O₂ gas.
• Due to deforestation, decomposition of lime stone and burning of fossil fuels CO₂ concentration is increased in atmosphere.
• The increase of CO₂ level disturbs the O₂ – CO₂ balance in the atmosphere and it is responsible for green house effect (or) global warming.

Question 18.
What are silicones ?
Answer:

• Silicones are the organo Silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – O – Si.
• Linkage preparation : These are formed by the hydrolysis of chlorosilanes.

Question 19.
Give the uses of silicones.
Answer:
Uses of silicones :

• These are used in surgical and cosmetic plants.
• These are used in preparation of silicone rubbers.
• These are used as sealoant, greases etc.
• These are used in preparing water proof clothes and papers.
• These are used as insulators.
• These are used in paints and enamels.

Question 20.
What is the effect of water on tin ?
Answer:

• Tin metal reacts with steam to form tin dioxide and dihydrogen gas.
• In this reaction steam is decomposed.
  Sn + 2H₂O SnO₂ + 2H₂

Question 21.
Write an account of SiCl₄.
Answer:

• Silicon tetrachloride (SiCl₄) is also called as tetra chloro silico methane.
• SiCl₄ can acts as Lewis acid due to availability of 3d orbital in ‘Si’.
• SiCl₄ undergoes hydrolysis due to presence of vacant 3d-orbital. Here water molecules forms dative bonds with empty 3d-orbitals of Siratom.

Uses :

• SiCl₄ and NH₃ mixture used to produce smoke screens.
• Ultra pure Silicon is used to make transistors.
• SiO₂ prepared from SiCl₄ used in epoxypaints, resis etc..

Question 22.
SiO₂ is a solid while CO₂ is a gas – explain.
Answer:

• Silica (SiO₂) has giant molecular structure.
• In SiO₂ ‘Si’ undergoes sp³ hybridisation.
•  It is a 3 – dimension structure in which each ‘Si’ atom is tetrahedrally surrounded by four oxygen atoms.
• Hence it exists as solid compound.
• CO₂ has linear structure.
• In CO₂ ‘C’ undergoes sp hybridisation.
• In between CO₂ molecule weak Vander Waal’s forces are present.
• In CO₂ molecule two double bonds are present.
•  Hence CO₂ exists as a gas.

Question 23.
Write the use of ZSM – 5.
Answer:

1. ZSM – 5 is a zeolite.
2. It is used to convert alcohols directly into gasoline.

Question 24.
What is the use of dry ice ?
Answer:

1. Solid CO₂ is called as dry ice.
2. It is used as refrigirent for frozen food and ice – creams.

Question 25.
How is water gas prepared ?
Answer:
Water gas is prepared by passing superheated steam over hot coke.

Question 26.
How is producer gas prepared ?
Answer:
Producer gas is prepared by passing air over white not coke.

Question 27.
C-C bond length in graphite is shorter than C-C bond length in diamond – explain.
Answer:

1. In graphite each carbon undergoes sp² hybridisation and hence bond length is 1.42 A° (or) 141.5 pm.
2. Graphite has hexagonal layer like lattice. It is a 2-dimensional structure.
3. In diamond each carbon undergoes sp3 – hybridisation and hence bond length is 1.54 A° (or) 154 pm.
4. Diamond has regular tetrahedral giant polymeric structure. It is a 3-dimensional structure.

Question 28.
Diamond is used as precious stone – explain.
Answer:

• Diamonds are used as precious stones.
• Diamonds are clear, colourless form of pure carbon.
• These are hardest substances occurring naturally.
• The weight of diamond expressed in carats.
  1 carat = 200 mg.

Question 29.
Carbon never shows co-ordination number greater than four while other members of carbon family show co-ordination number as high as six – explain.
Answer:
Carbon never shows co-ordination number greater than four because of absence of d-orbitals in carbon atom.

The other members of carbon family show co-ordination number as high as six because of availability of d – orbitals.

Question 30.
Producer gas is less efficient fuel than water gas – explain.
Answer:

1. Producer gas has calorific value 5439.2 KJ/m³
2. Water gas has calorific value 13000 KJ/m³.
3. Due to high calorific value of watergas, it is more efficient fuel than producer gas (or) producer gas is less efficient than watergas.

Question 31.
SiF₆^-2 is known while SiCl₆^-2 is not. Explain. [A.P. Mar. 16]
Answer:
SiF₆^-2 is known while SiCl₆^-2 is not because

• Si⁺⁴ has small size so it cannot be accomodate six large chloride ions.
• The interaction between lone pairs of Cl^– ion and Si⁺⁴ is not very strong.

Short Answer Questions

Question 1.
Explain the difference in properties of diamond and graphite on the basis of their structure.
Answer:
Diamond
a) Each carbon is sp³ hybridised.
b) Each carbon is bonded to 4 other carbons tetrahedrally.
c) It has a 3 dimensional structure.
d) C – C bond length is 1.54 Å and bond angle is 109° 28′.
e) Carbon atoms are firmly held with strong covalent bonds.
f) Diamond is very hard.
g) Density = 3.5 g/cc.
h) Graphite is a conductor due to the presence of free electrons.
i) It is transparent to light and X-rays. It has high refractive index (2.45).

Graphite
a) Each carbon is sp² hybridised.
b) Each carbon is bonded to 3 other carbon atoms to form hexagonal rings. It has sheet like structure.
c) It has a 2 dimensional structure.
d) C – C bond length in hexagonal rings is 1.42 A° and bond angle is 120°.
e) The distance between two adjacent layers is 3.35 A°. These layers are held by weak Vander Waal’s forces.
f) Graphite is soft.
g) Density – 2.2g/cc.
h) Diamond is an insulator due to the absence of free electrons.
i) It has layer, lattice. The layers are slippery. Hence it is greasy.

Question 2.
Explain the following.
a) PbCl₂ reacts with Cl₂ to give PbCl₄
b) PbCl₄ is unstable to heat,
c) Lead is not known to form PbI₄.
Answer:
a) PbCl₂ + Cl₂ → PbCl₂
But PbCl₄ is unstable than PbCl₂. Because compounds of lead in +2 oxidation state are stable than +4 oxidation state.

b) PbCl₄ is unstable to heat:

• In PbCl₄ lead exhibits +4 oxidation state.
• he compounds of lead in +2 oxidation state are stable than +4 oxidation state. Hence PbCl₄ is unstable to heat.

c) Lead is not known to form PbI₄:

• Pb – I bond formed initially during the reaction does not release enough energy to unpair the 6s electrons.
• Lead compounds in +2 state are stable than +4 state.
  Due to inert pair effect Pb exhibits stable +2 oxidation state.

Question 3.
Explain the following :
a) Silicon is heated with methyl chloride at high temperature in presence of copper.
b) SiO₂ is treated with HF.
c) Graphite is a Lubricant
d) Diamond is an abrasive.
Answer:
a)

• Methyl chloride reacts with silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si.
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (silicone) (Me = CH₃ – group)
  

b) SiO₂ is treated with HF to form SiF₄. Which on hydrolysis to form H₄SiO₄ and H₂SiF₆.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 4H₂O → H₄SiO₄ + 2H₂SiF₆

c) Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

d) The covalent bonds in diamond are very strong and difficult to break, Hence diamond is used as an abrasive for sharpening hard tpols, in making dyes and in the manufacturing of tungsten filaments etc.

Question 4.
What do you understand by
a) Allotropy
b) Inert pair effect
c) Catenation.
Answer:
a) Allotropy :
The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
Crystalline allotropes of carbon are
a) Diamond
b) Graphite.

b) Inert pair effect : The reluctance of ‘ns’ pair of electrons to take part in bond formation is known as inert pair effect.
(or)
The occurrence of oxidation states two units less than the group oxidation states is known as inert pair effect.
Eg : Lead exhibits +2 oxidation state as stable oxidation state due to inert pair effect. (Instead of +4 state).

c) Catenation : The phenomenon of self linkage of atoms among themselves to form long chains (or) rings is called as catenation.
Carbon has highest catenation tendency due to its high bond energy (348 KJ/mole).

Question 5.
If the starting material for the manufacturing of Silicons is RSiCi3. Write the structure of the product formed.
Answer:
When RSiCl₃ type of compound is used for the manufacturing of silicones a cross – linked silicon is formed.
Eg : When Methyl trichloro silane (CH₃SiCl₃) undergoes hydrolysis to give monomethyl silane triol. This undergoes polymerisation to form a very complex cross-linked polymer (Silicone).

Question 6.
Write a short note on Zeolites.
Answer:
Zeolites : The three dimensional structure contain no metal ions. If some of the Si⁺⁴ in them is replaced by Al⁺³ and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si₂O₈]^-2n form zeolites. Zeolites act as ion exchanges and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH₃, CO₂ and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca⁺² ions from hard water and replace them by Na⁺² ions.
Uses of Zeolites :
Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

Question 7.
Write a short note on Silicates.
Answer:
Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.
The silicates can be divided into six types. They are mentioned here.

1. Orthosilicate or Nesosilicates : Their general formula may be M₂¹¹ (SiO₄).
   Ex : Willemite Zn₂ (SiO₄).
2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO₇^-6 units. Pyrosilicates are rare.
   Ex: Thorveitite Ln₂ (Si₂O₇).
3. Chain silicates : They have the units (SiO₃)_n^2n-
   Ex : Spodumene LiA/ (SiO₃)₂.
   Amphiboles are one type chain silicates. Generally double chains are formed in them.
4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
   Ex : Beryl Be₃Al₂ (Si₆O₁₈)
5. Sheet silicates : When SiO₄ units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si₂O₅)_n^2n-. These compounds appear in layer struc-tures. They can be cleaved.
   Ex : Kaolin Al₂ (OH)₄ Si₂O₅.
6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO₄ tetrahedron results in three dimensional lattice of formula SiO₂.
   Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na₈ [Al₆ Si₆ O₂₄] S₂), zeolites.

Question 8.
What are Silicones ? How are they obtained ?
Answer:

• Silicones are the organo silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – 0 – Si.
  Preparation : These are formed by the hydrolysis of chlorosilanes.
• Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH₃ – group).
  

Question 9.
Write a short note on Fullerene.
Answer:
Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – membered rings.
  
•  In C₆₀ 6 – membered rings can combine with 5 (or) 6 – mem-bered rings while 5 – membered rings only combine with 6 – membered rings.
• In Fullerens each carbon undergoes sp2 hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1,43A° and 1.38A° respectively.

Question 10.
Why SiO₂ does not dissolve in water ?
Answer:
Silica (SiO₂) is a non reactive compound in it’s normal state.

1. This non reactivity is due to very high Si – O bond enthalpy.
2. Silica is a giant molecule with 3-dimensional structure.
3. In Silica each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms
4. Hence SiO₂ is insoluble in water.
5. But slightly dissolves at high pressures when heated.

Question 11.
Why is diamond hard ?
Answer:
In diamond, each carbon undergoes sp3 hybridization. A carbon atom is bound to four carbon atoms, arranged in a tetrahedral symmetry, with single bonds. A three dimensional arrangement of the tetrahedral structures result in giant molecule. The bond energy is very high (348 kJ mol^-1). It is very difficult to break the bonds. So, diamond is hard.

Question 12.
What happens when the following are heated
a) CaCO₃
b) CaCO₃ and SiO₂
c) CaCO₃ and excess of coke.
Answer:
CaCO₃ up on heating gives Quick lime.


Quick lime (CO) with silica gives Calcium silicate.


Coke reacts with Quick lime and form Carbides.

Question 13.
Why does Na₂CO₃ solution turn into a suspension, when saturated with CO₂ gas.
Answer:
An aq. solution of Na₂CO₃ when saturated with CO₂, gives Sodium bicarbonate (NaHCO₃).
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃
NaHCO₃ is less soluble compared to Sodium carbonate, hence suspension is formed.

Question 14.
What happens when
a) CO₂ is passed through slaked lime
b) CaC₂ is heated with N₂.
Answer:
a) Slaked lime, Ca(OH)₂ is turned milky on passing CO₂ with the formation of insoluble calcium
carbonate Ca(OH)₂ + CO₂ → CaCO₃ + H₂O on passing more ‘CO₂‘, CaCO₃ is converted into Calcium bicarbonate.
CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂
b) CaC₂ on heating with N2 gives calcium cyanamide.

Question 15.
Write a note on the anomalous behaviour of carbon in the group -14.
Answer:
Carbon shows ariamalous behaviour in group – 14 elements. The following facts support that anamalous behaviour.
Except carbon all other elements of group -14 has available d-orbitals and can expand octet in valency shell.
Carbon occurs in free state but not the other elements of this group.

• Maximum covalency of carbon is four but for silicon is six.
• C – C bond energy is very high (348 kJ/Mde).
• Carbon can form multiple bonds with C, O, S, etc.
• Hydrocarbons are more stable thermally than silanes.

Long Answer Questions

Question 1.
What are Silicones ? How are they prepared ? Give one example. What are their uses?
Answer:

• Silicones are the organo silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – 0 – Si.
  Preparation : These are formed by the hydrolysis of chlorosilanes.
• Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH₃ – group).
  

Uses of Silicones:

• These are used in surgical and cosmetic plants.
• These are used in preparation of Silicons rubbers.
• These are used as sealoant, greases etc.
• These are used in preparing water proof clothes and papers.
• These are used as insulators.
• These are used in paints and enamels.

Question 2.
Explain the structure of Silica. How does it react with
a) NaOH and
b) HF.
Answer:

• Silica is a giant molecule with 3 – dimensional structure.
• In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
• Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
• Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
• In SiO₂ Silicon atom under goes sp³ hybridisation.
  

a) Silica reacts with NaOH and forms Sodium Silicate (Na₂SiO₃)
SiO₂ + 2 NaOH → Na₂SiO₃ + H₂O

b) SiO₂ is treated with HF to form SiF₄. Which on hydrolysis to form H₄SiO₄ and H₂SiF₆.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 4H₂O → H₄SiO₄ + 2H₂SiF₆

Question 3.
Write a note on the allotropy of carbon.
Answer:
The property of an element to exist in two or more physical forms due to difference in the arrangement of atoms is called Allotropy. Allotropes have more or less similar chemical properties but different physical properties.

Diamond, graphite and fullerenes are the crystalline allotropes of carbon.
Structure of Diamond : In diamond, each carbon atom bonded to four carbon atoms situated tetrahydrally around it.
In diamond, each carbon atom is in sp³ hybridisation and is linked to four carbon atoms by single covalent bonds.
C – C bond distance in diamond is 1.54 A°.
bond angle in diamond is 109° 28″.
Uses of Diamond :

1. Diamonds are used as precious stones for jewellery because of their ability to reflect light.
2. Diamonds are used for cutting glass and drilling rocks due to their remarkable hardness.

Structure of Graphite : Graphite consists of a series of layers in which hexagonal rings made up of carbon atoms.
In Graphite, each carbon atom undergo sp² hybridisation and forms three covalent bonds with three other carbon atoms.

The fourth electron present in the pure ‘p orbitaI which is unhybridised. The electron become Free Electron.
The C — C bond length in graphite is 1.42 A°.

The distance between the two layers in graphite is 3,4 A°.
These layers are held together by Vander Waal’s forces which are weak.
Graphite is a layer lattice structure.
Uses of Graphite:

1. Graphite is used as a lubricant.
2. It is used in the manufacturing of lead pencils.
3. It is used in the manufacturing of Electrodes and Refractory crucibles.

Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – membered rings.
• In C₆₀ 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
•  In Fullerens each carbon undergoes sp² hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1.43A° and 1.38A° respectively.
• Amorphous allotropes of carbon are coal, coke, animal charcoal wood charcoal, lamp black, carbon black, gas carbon, petroleum coke and sugar charcoal.

Question 4.
Write a note on
a) Silicates
b) Zeolites
c) Fullerenes.
Answer:
a) Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.

The silicates can be divided into six types. They are mentioned here.

1. Orthosilicate or Nesosilicates : Their general formula may be M₂¹¹ (SiO₄).
   Ex : Willemite Zn₂ (SiO₄).
2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO₇^-6 units. Pyrosilicates are rare.
   Ex : Thorveitite Ln₂ (Si₂O₇).
3. Chain silicates : They have the units (SiO₃)_n^2n-
   Ex : Spodumene LiAl (SiO₃)₂.
   Amphiboles are one type chain silicates. Generally double chains are formed in them.
4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
   Ex : Beryl Be₃Al₂ (Si₆O₁₈)
5. Sheet silicates : When SiO₄ units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si₂O₅)_n^2n-. These compounds appear in layer struc-tures. They can be cleaved.
   Ex : Kaolin Al₂ (OH)₄ Si₂O₅.
6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO₄ tetrahedron results in three dimensional lattice of formula SiO₂.
   Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na₈ [Al₆ Si₆ O₂₄] S₂), zeolites.

b) Zeolites : The three dimensional structure contain no metal ions. If some of the Si⁺⁴ in them is replaced by Al⁺³ and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si₂O₈]^-2n form zeolites. Zeolites act as ion ex-changes and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH₃, CO₂ and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca⁺² ions from hard water and replace them by Na⁺² ions.

Uses of Zeolites :

• Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
• Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

c) Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon. .
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – mem- bered rings.
  
• In C₆₀. 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
• In Fullerens each carbon undergoes sp² hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1.43A° and 1.38A° respectively.

Solved Problems

Question 1.
Select the member(s) of group 14 that

1. forms the most acidic dioxide
2. is commonly found in +2 oxidation state
3. used as semiconductor.

Solution:

1. Carbon
2. lead
3. Silicon and germanium

Question 2.
[SiF₆]^2- is known whereas [SiCl₆]^2- not. Give possible reasons.
Solution:
The main reasons are :

1. Six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.
2. Interaction between lone pair of chloride ion and Si⁴⁺ is not very strong.

Question 3.
Diamond is covalent, yet it has high melting point. Why ?
Solution:
Diamond has a three – dimensional network involving strong C – C bonds, which are very difficult to break and in turn has high melting point.

Question 4.
What are Silicones ?
Solution:
Simple Silicones consist of chains in which alkyl or phenyl groups occupy the remaining bonding positions on each silicon. They are hydrophobic (water repellant) in nature.