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

In $$NO_3^ - $$  ion

Nitrogen has four bond pair and zero lone pair of electrons, due to the presence of one coordination bond.

Paramagnetic character is based upon presence of unpaired electron.
$${}_{17}C{l^ - } = 1{s^2},2{s^2}2{p^6},3{s^2}3p_x^23p_y^23p_z^2$$
In $$C{l^ - }$$ no unpaired electron, so it is in nature diamagnetic.
$${}_4Be = 1{s^2},2{s^1}2p_x^1$$
$${}_{10}N{e^{2 + }} = 1{s^2},2{s^2}2p_x^22p_y^12p_z^1$$
$${}_{33}A{s^ + } = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},$$       $$4{s^2}4p_x^14p_y^14p_z^0$$
While all others have unpaired electron, so they are paramagnetic in nature.

According to molecular orbital theory,
$$O_2^{2 - }\left( {8 + 8 + 2 = 18} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$      $$\sigma 2p_z^2,\pi 2p_x^2 \approx \pi 2p_y^2,$$     $$\mathop \pi \limits^* 2p_x^2 \approx \mathop \pi \limits^* 2p_y^2$$
$$\eqalign{ & {\text{Bond}}\,\,{\text{order}}\left( {BO} \right) \cr & = \frac{{{N_b} - {N_a}}}{2} \cr & = \frac{{10 - 8}}{2} \cr & = 1 \cr & {B_2}\left( {5 + 5 = 10} \right) \cr} $$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$     $$\pi 2p_x^1 \approx \pi 2p_y^1$$
$$\eqalign{ & BO = \frac{{6 - 4}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 1 \cr} $$
Thus, $$O_2^{2 - }$$  and $${B_2}$$  have the same bond order.
NOTE
$$BO\,\,{\text{of}}\,\,O_2^ + = 2.5,\,N{O^ + } = 3,$$      $$NO = 2.5,\,CO = 3,\,{N_2} = 3$$       $${\text{and}}\,\,{O_2} = 2$$

In ethyne, $$H - C \equiv C - H;C$$     is $$sp$$  hybridised.
In graphite, one $$C$$ atom is attached to 3 other $$C$$ atoms which is $$s{p^2}$$ hybridised.
In diamond, one $$C$$ atom is attached to 4 other $$C$$ atoms which is $$s{p^3}$$  hybridised.

Thus, option (C) is correct.

NOTE : Compounds having $$F\,or\,\,O\,\,or\,\,N$$   attached to $$H$$  form hydrogen bond.

∴ Ethanol having $$H$$  attached to $$O$$  atom will form hydrogen bond. Rest of the compounds do not hydrogen bonds.

TIPS/Formulae :

$$\eqalign{ & \left( {\text{A}} \right)\,\,{\text{For}}\,\,Al{H_3}, \cr & {\text{Hybridisation of }}Al\,\,{\text{atom}} = \frac{1}{2}\left[ {3 + 3 - 0 + 0} \right] = 3 = s{p^2} \cr & {\text{For}}\,\,AlH_4^ - , \cr & {\text{Hybridisation of }}Al\,\,{\text{atom}} = \frac{1}{2}\left[ {3 + 4 - 0 + 1} \right] = 4 = s{p^3} \cr & \left( {\text{B}} \right)\,\,{\text{For}}\,{H_2}O, \cr & {\text{Hybridisation of}}\,\,O\,\,{\text{atom}}\, = \frac{1}{2}\left[ {6 + 2 - 0 + 0} \right] = 4 = s{p^3} \cr & {\text{For}}\,{H_3}{O^ + },\,\,{\text{Hybridisation of }}\,O\,\,\,{\text{atom}}\, = \frac{1}{2}\left[ {6 + 3 - 1 + 0} \right] = 4 = s{p^3} \cr & \left( {\text{C}} \right)\,\,{\text{For}}\,\,N{H_3} \cr & {\text{Hybridisation of }}N\,{\text{atom}}\, = \frac{1}{2}\left[ {5 + 3 - 0 + 0} \right] = 4 = s{p^3} \cr & {\text{For}}\,\,NH_4^ + ,\,\,{\text{Hybridisation of}}\,N\,{\text{atom}}\, = \frac{1}{2}\left[ {5 + 4 - 1 + 0} \right] = 4 = s{p^3} \cr} $$
$${\text{Thus hybridisation changes only in option (A)}}{\text{. }}$$

In $$P - O$$  bond, $$\pi - {\text{bond}}$$   is formed by the sidewise overlapping of $$d - {\text{orbital}}$$   of $$P$$  and $$p - {\text{orbital}}$$   of oxygen.
Hence, it is formed by $$p\pi $$  and $$d\pi - $$ overlapping.

In nitrogen and carbon, no vacant $$d - {\text{orbital}}$$   is present. So, they do not form $$p\pi - d\pi $$   bond.

$$O_2^ - :\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$      $$\left( {{\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^1} \right)$$
$$O_2^ - \,{\text{has 1 unpaired electron}}{\text{.}}$$
$$\mu = \sqrt {n\left( {n + 2} \right)} \,B.M. = $$       $$\sqrt {1\left( {1 + 2} \right)} = \sqrt 3 = 1.732\,B.M.$$

The geometry of $$BH_4^ - $$  is tetrahedral. $$NH_2^ - $$  has $$V$$ - shape, $$CO_3^{2 - }$$  has trigonal planar and $${H_3}{O^ + }$$  has pyramidal shape.