NEET Practice Questions (MCQs) with Answers & Solutions

(b) BeF₂ - sp hybridisation

       BF₃ - sp₂ hybridisation

     C₂H₂ - sp hybridisation

      NH₃ - sp₃ hybridisation

The correct order of C-O bond length is CO < CO₂ < CO₃^2-. The bond order decreases from CO to CO₂ to CO₃^2-, resulting in an increase in bond length due to weaker bond strength.

(b) According to the molecular orbital theory (MOT),

$$\begin{gathered} {\mathrm{N}}_2(7+7=14) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \\ \approx 2{\mathrm{p}}_{\mathrm{y}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \\ \mathrm{Bond} \mathrm{Order} = \frac{10-4}{2} = 3 \\ {\mathrm{N}}_2^{-}(7+7+1=15) =\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx 2{\mathrm{p}}_{\mathrm{y}}^2, \\ \mathrm{BO}= \frac{10-5}{2}= 2.5 \\ {\mathrm{N}}_2^{2-}(7+7+2=16)=\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx 2{\mathrm{p}}_{\mathrm{y}}^2,\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1\approx \stackrel{*}{\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1} \\ \mathrm{BO} = \frac{10-6}{2}=2 \end{gathered}$$

Hence, the increasing order of bond order is,

                     ${\mathrm{N}}_2^{2-} < {\mathrm{N}}_2^{-} < {\mathrm{N}}_2$

(b) Water is a polar colvent and have dielectric constant 80. As NaCl is a polar compound and like dissolves like so, forces of attraction between Na⁺ and Cl^- ion will reduce to $\frac{1}{80}$in water.

(c) CN^- and CO are isoelectric because they have equal number of electrons.

     In CN^- the number of electrons = 6 + 7 + 1 =14

     In CO the number of electrons = 6 + 8 =14

(a) In the formation of d²sp³ hybrid orbitals, two (n-1) d-orbitals of eg set, i.e. (n-1)d_z2 and (n-1)d_x2-_y2 orbitals, one n s and three np (np_x, np_y and np_z) orbitals combine together.

The correct answer is o2: sp^2 hybridization. Equilateral or trigonal planar shapes have sp^2 hybridization. In this hybridization, one s orbital and two p orbitals mix to form three sp^2 hybrid orbitals arranged in a trigonal planar geometry.