Chemical Bonding & Molecular Structure

(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^'.