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

On the basis of Fajans' rule, lower the size of cation, higher will be its polarising power and higher will be covalent character.
$$\therefore {\text{Polarising}}\,{\text{power}} \propto \frac{1}{{{\text{size}}\,{\text{of}}\,{\text{cation}}}}$$
Covalent character $$ \propto $$ polarising power
So, the correct order is $$NaCl < LiCl < BeC{l_2}$$
( $$\because $$  The order of size of cation $${N{a^ + } > L{i^ + } > B{e^{2 + }}}$$     )

$$\left( {\text{A}} \right){H_2}O \Rightarrow $$  
$$\left[ {bp = {\text{bond}}\,{\text{pair}}\,{\text{and}}\,lp = {\text{lone}}\,{\text{pair}}} \right]$$
$$\left( {\text{B}} \right)B{F_3} \Rightarrow $$   
$$\left( {\text{C}} \right)NH_2^ - \Rightarrow $$   
$$\left( {\text{D}} \right)PC{l_3} \Rightarrow $$   
Thus, in $$PC{l_3},$$  the central $$P - {\text{atom}}$$   is surrounded by three bond pairs and one lone pair.

Electrons from outermost shells $$ns$$  and $$\left( {n - 1} \right)d$$   take part in bond formation for transition elements.

$$CC{l_4}$$  does not show dipole moment because it has tetrahedral symmetrical structure.

TIPS/Formulae : Dipole moment of compound having regular geometry and same type of atoms is zero. It is vector quantity.
The zero dipole moment of $$B{F_3}$$  is due to its symmetrical (triangular planar) structure. The three fluorine atoms lie at the corners of an equilateral triangle with boron at the centre.
NOTE: The vectorial addition of the dipole moments of the three bonds gives a net sum of zero because the resultant of any two dipole moments is equal and opposite to the third. The dipole moment of $$N{H_3}$$  is $$1.46$$ $$D$$  indicating its unsymmetrical structure. The dipole moment of$$C{H_2}C{l_2}$$   ( the molecule uses $$s{p^3}$$  hybridisation but is not symmetric ) is $$1.57\,{\text{D}}.$$

In $$SO_3^{2 - },$$  $$S$$  is $$s{p^3}$$  hybridised, so

\[\underset{\begin{smallmatrix} \text{(Sulphur atom in} \\ \text{excited state)} \end{smallmatrix}}{\mathop{_{16}S=1{{s}^{2}},2\,{{s}^{2}}2{{p}^{6}},}}\,\]     \[\underbrace{3{{s}^{2}}3p_{x}^{1}\,\,3p_{y}^{1}3p_{z}^{1}}_{s{{p}^{3}}\,\,\text{hybridisation}}\,\,\underset{\text{Unhybridised}}{\mathop{3d_{xy}^{1}}}\,\]
In $$'S'$$   the three $$p - {\text{orbitals}}$$   forms $$\sigma - {\text{bonds}}$$   with three oxygen atoms and unhybridised $$d - {\text{orbitals}}$$   is involved in $$\pi - {\text{bond}}$$   formation.
$${{\text{O}}_8} = 1{s^2},2{s^2}2p_x^22p_y^12p_z^1$$
In oxygen two unpaired $$p - {\text{orbitals}}$$   are present, one is involved in $$\sigma - {\text{bonds}}$$   formation while other is used in $$\pi - {\text{bond}}$$   formation.
Thus in $$SO_3^{2 - },$$  $$p$$  and $$d - {\text{orbitals}}$$   are involved for $$p\pi - d\pi $$   bonding.

KEYCONCEPT
(i) Bond length $$ \propto \frac{1}{{{\text{Bond}}\,{\text{order}}}}$$
(ii) Bond order is calculated by either the help of molecular orbital theory or by resonance.
(i) Bond order of $$CO$$  as calculated by molecular orbital
theory = $$3\left\{ {b.o. = \frac{1}{2}\left[ {{N_b} - {N_a}} \right]} \right\}$$
(ii) Bond order of $$C{O_2}$$  (by resonance method)
$$\eqalign{ & = \frac{{{\text{No}}{\text{. of bonds in all possible sides}}}}{{{\text{No}}{\text{. of resonating structure }}}} \cr & = \frac{4}{2} \cr & = 2 \cr} $$
(iii) Bond order in $$CO_3^{2 - }$$  (by resonance method)
$$ = \frac{4}{3} = 1.33$$
∴ Order of bond length of $$C - O$$  is $$CO < C{O_2} < CO_3^{2 - }$$

$${B_2}$$ molecule has the following configuration : $$\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {{\sigma ^ * }2{s^2}} \right)$$      $$\left( {\pi 2p_x^1 = \pi 2p_y^1} \right)$$

Higher the bond order, higher is the bond enthalpy.

According to the molecular orbital theory $$\left( {MOT} \right).$$
$${N_2}\left( {7 + 7 = 14} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$     $$\pi 2p_x^2 \approx 2p_y^2,\sigma 2p_z^2$$
$${\text{Bond}}\,{\text{order}} = \frac{{10 - 4}}{2} = 3$$
$$N_2^ - \left( {7 + 7 + 1 = 15} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$     $$\sigma 2p_z^2,\pi 2p_x^2 \approx 2p_y^2,\mathop \pi \limits^* 2p_x^1$$
$${\text{BO}} = \frac{{10 - 5}}{2} = 2.5$$
$$N_2^{2 - }\left( {7 + 7 + 2 = 16} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_z^2,$$      $$\pi 2p_x^2 \approx \pi 2p_y^2,\mathop \pi \limits^* 2p_x^1 \approx \mathop \pi \limits^* 2p_y^1$$
$${\text{BO}} = \frac{{10 - 6}}{2} = 2$$
Hence, the increasing order of bond order is, $$N_2^{2 - } < N_2^ - < {N_2}$$