Chemical Bonding & Molecular Structure

The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound. Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state. The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.

1. Charge on the ions: The lattice energy is directly proportional to the product of the charges of the cation and the anion. Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy. This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.

Mathematically:

$U \propto \frac{q_1 \cdot q_2}{r}$

where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.

2. Size of the ions: The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii. Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy. Larger ions, being further apart, result in a lower lattice energy.

Considering these two factors, the correct choice is:

Option D: Charge on the ion and size of the ion

TIPS/Formulae :

A diamagnetic substance contains no unpaired electron.

${H_2}$ is diamagnetic as it contains all paired electrons

$\mathop {{H_2} = \sigma _b^2}\limits_{\left( {diamagnetic} \right)} \,\,,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {paramagnetic} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {paramagnetic} \right)} $

$\mathop {\sigma _a^{ * 1};He_2^ + }\limits_{\left( {paramagnetic} \right)} = \mathop {\sigma _b^2,\sigma _a^{ \ne 1}}\limits_{\left( {paramagnetic} \right)} $

Here eight 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4.

Here twelve 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4, $\angle $6M1, $\angle $6M2, $\angle $6M3, $\angle $6M4.

Here four 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1.

Here six 90^o angles between bond pair and bond pair. Those angles are $\angle $1M3, $\angle $1M4, $\angle $1M5, $\angle $2M3, $\angle $2M4, $\angle $2M5.

Structure of  H₃BO₃  is



Here Boron has 3 sigma bond. So it is sp² hybridised.

And oxygen has two sigma bond and two lone pair. So it sp³ hybridised.

1. In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

2. In NH₃, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107^o.



3. $\,\,\,\,$ BF₃ has sp² hybridization. So bond angle is 120^o.



4.
Here Si is sp³ hybridised and shape is regular tetrahedral and bond angle 109^o 28'

Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Note :
(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So, bond length in NO > NO⁺

Regular Tetrahedral structure is possible in sp³ hybridization where central atom has 4 bond pair and no lone pair.

(a)   XeF₄  is sp³d² hybridised and structure is square planar.



(b)   [Ni(CN)₄]^$-$2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni⁺² is $=$ [Ar]3d⁸



But because of CN^$-$ ion which is a strong field ligand , it can perform pairing of electron.



And the structure of dsp² hybridization is square planar.



(C) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral.



(d)   

SF₄ is sp³d hybridised and structure is see-saw.

Here $\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $_total = $\mu $₁ $-$ $\mu $₂ = 0

$\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $₁ + $\mu $₂ + $\mu $₃ = $\mu $₄

   $\mu $_total = 0

In NO₂ and O₃ have permanent dipole moment as both of them are not symmetric molecule.

In CO₂ and SiF₄ each individual bond c = O and S- $-$ F have dipole moment but because of symmetric structure individual dipole moment get's cancelled.

(a)   H₂O is sp³ hybridized, and oxygen atom has 2 bond pair and 2 lone pair. So the angle between two O $-$ H bond is 104.5^o

(b)   In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(s) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

(c)   NH₃ has sp³ hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107^o

(d)   SO₂ is sp² hybridized and bond angle is 117^o.

(a) $\,\,\,\,$

(b) $\,\,\,\,$

(c) $\,\,\,\,$

(d) $\,\,\,\,$

So, option (B) is correct

Note :
If in carbon(C) all 4 bonds are sigma bond then it is sp³ hybridization and if there is 3 sigma bond then sp² and if there is only 2 sigma bond there sp hybridization.

(A) Electronegativity of F is 4 and Electronegativity of H is 2.1.

$ \therefore $ Electronegativity difference between F and H is = 4 - 2.1 = 1.9

Now Electronegativity of Br is 2.8 and Electronegativity of H is 2.1.

$ \therefore $ Electronegativity difference between Br and H is = 2.8 - 2.1 = 0.7

As in HF Electronegativity difference is more then it's polar nature is also more.

(B) At equilibrium pure H₂O produce very less H⁺ and OH^- ions.

(C) Chemical bond formation takes place when forces of attraction overcome the forces of repulsion.

(D) In covalenrt bond, sharing of electons take place. Transfer of electrons take place in ionic bond.

Hybridization of square planar complex is dsp² .



In square planar complex all 4 surrounding atoms and central atom are in the same plane and let this plane is x $-$ y plane. As it is dsp² hybridized, so it has 1 d orbital, 1 s orbital and 2 p orbital.

s orbital is non-directional, so we just write it as s.



In p_x the two lobes are along x-axis and in p_y two lobes are along y-axis

One d orbital is d_{x² $-$ y²}, it means out of four lobes of d orbital two along x-axis and two along y-axis.

(a) $\,\,\,$ AlH₃ + H^$-$ $ \to $ AlH$_4^ - $

Steric number of AlH₃ is = ${1 \over 2}$ [3 + 3] = 3

$\therefore\,\,\,$ AlH₃ is sp² hybridized.

Steric number of AlH$_4^ - $ = ${1 \over 2}$ [3 + 4 +1] = 4

$\therefore\,\,\,$ AlH$_4^ - $ is sp³ hybridized.

(b) $\,\,\,$ H₂O + H⁺ $ \to $ H₃O⁺

Steric no of H₂O = ${1 \over 2}$ (6+ 2) = 4

$\therefore\,\,\,\,$ H₂O s sp³ hybridized.

Steric no of H₃O⁺ = ${1 \over 2}$ [ 8 + 3 $-$1] = 4

$\therefore\,\,\,$ H₃O⁺ s also sp³ hybridized.

(c) $\,\,\,$ NH₃ + H⁺ $ \to $ NH₄⁺

Steric no of NH₃ = ${1 \over 2}$ [5 + 3] = 4

$\therefore\,\,\,$ hybridization of NH₃ is sp³

Steric number of NH₄⁺ = ${1 \over 2}$ [5 + 4 $-$ ] = 4

$\therefore\,\,\,$ Hybridization of NH₄⁺ is sp³

Note :

(1) $\,\,\,\,$ Bond strength $ \propto $ Bond order

(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond strength $ \propto $ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

Bond angle 109^o 28' means Regular Tetrahedral geometry and hybridization is sp³.

Steric Number (SN) for sp³ hybridization is 4.

In sp³ molecules can have different shapes which is decided by SN number

(1) $\,\,\,\,$ 4 bond pair in a molecule then angle between bonds 109^o 28^'

(2) $\,\,\,\,$ 3 bond pair and 1 lone pair in a molecule then angle between bonds 107^o.

(3) $\,\,\,\,$ 2 bond pair and 2 lone pair in a molecule then angle between bonds 104.5^o

(4) $\,\,\,\,$ 1 bond pair and 3 lone pair in a molecule then bond angle is undefined as there is only one bond.

(1) $\,\,\,\,$ In NH₃, 3 Bond pair(BP) +1 lone pair (LP) present so angle between bond 107^o.



(2) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^'.



(3) In NH$_4^ + $, 4 bond pair present so angle between bond is 109^o 28^'



(4) $\,\,\,\,$ BF₃ has sp² hybridization. So bond angle is 120^o.



(5) $\,\,\,\,$ In NH$_2^ - $, 2 bond pair and 2 lone pair present, so bond angle is 104.5^o



So, $NH{}^ + $ and BF$_4^ - $ has bond angle 109^o 28^'.