Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry

Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus
In $$N{F_3}:H = \frac{1}{2}\left( {5 + 3 - 0 + 0} \right) = 4$$
Thus $$N$$ in $$N{F_3}$$  is $$s{p^3}$$  hybridized as 4 orbitals are involved in bonding.
In $$NO_3^ - :H = \frac{1}{2}\left( {5 + 0 - 0 + 1} \right) = 3$$
Thus $$N$$ in $$NO_3^ - $$  is $$s{p^2}$$  hybridized as 3 orbitals are involved in bonding
$$B{F_3}:H = \frac{1}{2}\left( {3 + 3 - 0 + 0} \right) = 3$$
Thus $$B$$ in $$B{F_3}$$  is $$s{p^2}$$  hybridized and 3 orbitals are involved in bonding
In $${H_2}{O^ + }:H = \frac{1}{2}\left( {6 + 3 - 1 + 0} \right) = 4$$
Thus $$O$$ in $${H_2}{O^ + }$$  is $$s{p^3}$$  hybridized as 4 orbitals are involved in bonding.
Thus, isostructural pairs are $$\left[ {N{F_3},{H_3}{O^ + }} \right]$$   and $$\left[ {NO_3^ - ,B{F_3}} \right].$$

Due to polar nature, water molecules show intermolecular hydrogen bonding as

whereas the ice has open structure with large number of vacant spaces. So, density of ice is lower than water.

$${{p_z}}$$ orbital by convention undergoes end to end overlapping and $${{p_x},{p_y}}$$  undergo sidewise overlapping.

$${p_x}$$  and $${p_y}$$  orbitals do not have proper orientation to overlap and hence no bond is formed.

In $$N{H_3}$$  and $$BF_4^ - $$  the hybridisation is $$s{p^3}$$ and the bond angle is almost $${190^0}\,\,{28'}.$$

NOTE THIS STEP: Write the electronic configuration of each species according to molecular orbital theory.
$$NO\left( {15{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^0} \right.\,\,1\,{\text{unpaired}}\,{\text{electron}}.} \right.$$
$$CO\left( {14{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\sigma 2p_x^2} \right.\,\,\,\,\,{\text{no}}\,{\text{unpaired}}\,{\text{electron}}$$
$$C{N^ - }\left( {14{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\sigma 2p_x^2} \right.$$
$${O_2}\left( {16{e^ - }} \right) - \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1;} \right.} \right.\,\,{\text{Two}}\,\,\,{\text{unpaired}}\,{\text{electrons}}{\text{.}}$$

$$B{F_3} - $$  Trigonal planar with bond angle $${120^ \circ }.$$
$$PC{l_5} - $$  Trigonal bipyramidal with bond angles $${120^ \circ }$$ and $${90^ \circ }.$$

$$N_2^ + :\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {\sigma 2p_z^1} \right)$$
$$\eqalign{ & B.O. = \frac{1}{2} \times \left( {9 - 4} \right) = 2.5 \cr & {N_2}:B.O. = \frac{1}{2} \times \left( {10 - 4} \right) = 3 \cr} $$
$$O_2^ + :\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$      $$\left( {{\pi ^ * }2p_x^1} \right)$$
$$B.O. = \frac{1}{2} \times \left( {10 - 5} \right) = 2.5;$$      $${O_2}:B.O. = \frac{1}{2} \times \left( {10 - 6} \right) = 2$$
Since $$N_2^ + $$ has lower bond order than $${N_2},$$  bond length of $$N-N$$  in $$N_2^ + $$ increases. In $$O_2^ + ,$$  bond order increases from 2 to 2.5 hence, bond length decreases.

Diamagnetic species have no unpaired electrons
$$\,O_2^{2 - }\, \Rightarrow \sigma 1{s^{ + 2}},{\sigma ^*}1{s^2},\sigma 2{s^{2,}}{\sigma ^*}2{s^2},\sigma 2p_x^2,$$          $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^2} \right.} \right.$$
Whereas paramagnetic species has one or more unpaired electrons as in
$$\,{O_2} \to \,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$         $$\,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1 - 2} \right.\,\,{\text{unpaired}}\,{\text{electrons}}} \right.$$
$$O_2^ + \to \sigma 1{s^2}\,,\,{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$       $$\,\left\{ {\pi 2p_y^2 = \pi 2p_z^2\,\,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^0 - 1} \right.\,\,{\text{unpaired}}\,{\text{electrons}}} \right.$$
$$NO \to \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},{\sigma ^ * }2{s^2},\sigma 2p_x^2,$$       $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2,} \right.\left\{ {{\pi ^ * }2p_y^1 = {\pi ^ * }2p_z^0 - 1} \right.$$       $${\text{unpaired electron}}$$