AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure

Students can go through AP Inter 1st Year Chemistry Notes 3rd Lesson Chemical Bonding and Molecular Structure will help students in revising the entire concepts quickly.

AP Inter 1st Year Chemistry Notes 3rd Lesson Chemical Bonding and Molecular Structure

→ Atoms combine among themselves to get stability in terms of energy and electronic configuration.

→ Electrons of the outermost shell (valence shell) only take part in bonding.

→ Eight electrons ns2 np6 is known as octet. According to octet rule atoms or ions with octets are most stable than others.

→ The attractive force between two atoms in a molecule is known as chemical bond.

→ There are three important types of bonds. Ionic bond or electrovalent bond proposed by Kossel and Lewis, covalent bond proposed by Lewis; dative or co-ordinate covalent bond proposed by Sidgwick.

AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure

→ The electrostatic force of attraction that exists between oppositely charged ions is known as ionic bond. It is formed between a highly electropositive element and highly electronegative element. The best ionic compound is CsF.

→ Shapes of simple covalent molecules can be known with the help of valence shell electron pair repulsion theory by counting the number of electron pairs at the central atom in its valence shell.

→ The presence of lone pairs cause deviation in bond angles and distortion in shapes of molecules. Lone pair – lone pair repulsions is greater than lone pair – bond pair is greater than bond pair – bond pair.

→ Linear overlap of orbitals gives a sigma bond while sidewise overlap of orbitals gives a ‘pi’ bond.

→ Bond angle is influenced by the type of hybridisation at the central atom and the number of ion pairs on it.

→ Hybridisation is a hypothetical concept and is a modification of valence bond theory. Concept of hybridisation was proposed by Pauling.

→ Mixing of atomic orbitals of nearly same energy to give the same number of new set of orbitals of equal energy and shape is called hybridisation.

→ sp hybridisation : BeCl2, CO2, C2H2 etc., linear shape with a bond angle 180°.

→ sp2 hybridisation : BCl3, BF3, C2H4, SO3 etc., trigonal shape with 120° bond angle.

→ sp3 hybridisation: CH4, C2H6, tetrahedral, bond angle 109°28′. But NH3 has pyramidal shape with 107° bond angle. In case of water the shape is angular with a bond angle 104°36′.

AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure

→ sp3d hybridisation : PCl5, TeCl4, ClF3, XeF2 etc. Bond angles 120° and 90°.

→ sp3d2 hybridisation : SF6, IF5, XeF4 etc. Bond angle is 90°.

→ Molecular orbital is a region around the nuclei of the bonded atoms in a molecule where the probability of finding electrons is maximum (or) the wave function of a molecule.

→ Order of energies. Bonding orbitals < non bonding orbitals < anti bonding orbitals.

→ Sequence in the order of increasing energy is given below.
For lighter elements:
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 1
For higher elements:
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 2
Dipole Moment: The product of magnitude of the charge and the distance between two poles is called dipole Moment.

→ The electrostatic force of attraction between partially positive charged Hydrogen atom of a polar molecule and highly electronegative atom of same (or) different molecule is called Hydrogen bond.

→ Substance having intramolecular H – bonds can be steam distilled.
ex.: o – hydroxy benzaldehyde is purified by steam distillation.
P – hydroxy benzaldehyde cannot be purified by steam distillation.

AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure

→ Inter molecular H – bonding is responsible for association of molecules in water. Boiling point of water is higher than that of HF. NH3 forms hydrogen bonds while l does not.

→ Writing the structures of molecules:
Total no. of electron pairs in a molecule can be calculated by the formula 1
P = \(\frac{1}{2}\)(V + M – c + a)
V = No. ofvalency electrons in central atom
M = No. of monovalent atoms attached to central atom
c = No. of positive charges (if cation)
a = No. of negative charges (if anion)
P = No. of b.p + No. of l.p
b.p = bond pairs
l.p = lone pairs
Bond pairs = No. of atoms around the central atom
l.p = P- b.p
Eg. :
1) SF6 (Contains no multiple bonds)
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 3
P = \(\frac{1}{2}\)(V + M – c + a)
= P = \(\frac{1}{2}\)(6 + 6 – 0 – 0)
= 6
in this molecule B.P = 6
∴ l.p = 6 – 6 = 0
∴ SF6 have regular octahedral shape
Other example for practice:
BeF2, Bcl3, CCl4, PCl3, PCl5, NH3, H2O, OF2, SF4, ClF3, IF7, XeF2 etc ….

→ CoCl2 (Contains multiple bonds)
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 4
P = \(\frac{1}{2}\)(V + M – c + a)
= \(\frac{1}{2}\) (4 + 2 – 0 + 0)
= 3
In this molecule B.P = 3
l.p = 3 – 3 = 0
CoCl2 have regular trigonal / planar structure
H3o+ (Cation)
P = \(\frac{1}{2}\) (V + M – c + a)
= \(\frac{1}{2}\) (6 + 3 – 1 + 0)
= \(\frac{8}{2}\) =4
B.P = 3
l.p = 4 – 3 = 1
H3o+ have a pyramidals shape with one l.p and 3 b.p.s with one l.p. interaction tetrahydral shape other E.g. NH4 +.

→ No3 (Anion)
P = \(\frac{1}{2}\) (V + M – c + a)
= \(\frac{1}{2}\) (5 + 0 – 0 + 1)
= \(\frac{6}{2}\) = 3
B.P = 3
l.p = 3 – 3 = 0
It has regular trigonal plasal shape
Other E.g.: No2 , Po4 -3, Co3 -2, So2 -2, So3 -2 etc….

AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure

→ ICl4
P = \(\frac{1}{2}\) (V + M – c + a)
= \(\frac{1}{2}\) (7 + 4 – 0 + 1)
= 6
B.P = 4
l.p = 6 – 4 = 2
So, ICl4 have square planar structure.
Two l.ps occupying opposite positions of the octahedrum
Other E.g.: I3 , ClF2 etc….

→ Molecular Orbital Theory (MOT)
(Homonuclear diatomic molecule)
1) For carbon molecule Molecular orbital (MO)
Electronic configuration, Bond order and.Magnetic property mentioned in the following table
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 5

→ For O2 molecule (Home nuclear diatomic molecule)
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 6

AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure

→ Hetero nuclear diatomic molecules
E.g.: 1) Co – 14 Electrons
(6e from ‘C’ + 8e from ‘o’)
Bond order = 3 (Diamagnetic)
2) No – 15 Electrons
(7e from ‘N’ + 8e from ‘o’)
Bond order = 2.5 (Paramagnetic)
3) CN – 14 Electrons
(6e from ‘C’ + 7e from ‘N’ + One -ve charge)
Bond order = 3.0 (Diamagnetic)

→ Dipolemoment : The product of magnitude of charge and the distance between the two poles is called dipole moment (μ)
Applications of Dipolemoment:
μ = q × d
q = charge , d = distance , Units:Debye

→ Dipole moment is used to calculate the percentage of ionic character of ionic character
% of ionic character = \(\frac{\mu \text { observed(experimated })}{\mu \text { calculated }} 100\)

→ Dipole moment is used to calculate the bond moment
μobs =2 × Bond moment × Cos(\(\frac{\theta}{2}\))
μobs = \(\sqrt{\mu_1^2+\mu_2^2+2 \mu_1 \mu_2 {Cos} \theta}\)
Bond moment = The contribution of individual bonds in the dipole moment of a poly atomic molecule.

→ In Cis Isomer μ ≠ 0
In Trans isomer μ = 0

→ In ortho, meta,para isomers ortho, meta, isomers have non zero dipole moment. Para isomer has zero dipole moment.
Eg. :
AP Inter 1st Year Chemistry Notes Chapter 3 Chemical Bonding and Molecular Structure 7

→ Symmetry of the molecules can be known by dipole moment.

→ Dipole moment can be useful in predicting shape of molecule.

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