Chemistry MCQs for NEET — Practice Questions with Answers

(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.

Bond strength depends upon two factors

(i) Size of orbitals (ii) Extent of overlapping

Thereby, σ bond is always stronger than π bonds. Smaller the size of orbitals, stronger is
the bond. Therefore, bond formed by 1s will be stronger. However, p-orbitals are more
directional than s-orbital.

σ _1s-2p  is the strongest amongst four.

the stronger s-p overlap is due to the "sharpness" of the p orbitals which allow the electrons of e.g. the carbon atom's orbits to penetrate deeper toward the hydrogen nucleus

$$\begin{gathered} {\mathrm{NO}}_2^{+} \\ \mathrm{n}= \mathrm{No}. \mathrm{of} \mathrm{valance} \mathrm{electrons} \pm \mathrm{charge}. \\ \mathrm{n}= 5\times 1+6\times 2-1=16 \\ 82160 \\ \mathrm{Hybridisation} \mathrm{of} \mathrm{N} \mathrm{in} {\mathrm{NO}}_2^{+} \mathrm{is} \mathrm{sp} \end{gathered}$$

$$\begin{gathered} {\mathrm{NO}}_3^{-} \\ \mathrm{n}= 5\times 1+6\times 3+1=24 \\ 8324\frac{24}{0} \\ \mathrm{Hybridisation} \mathrm{of} \mathrm{N} \mathrm{in} {\mathrm{NO}}_3^{-} \mathrm{is} {\mathrm{sp}}^2 \\ {\mathrm{NH}}_4^{+} \\ \mathrm{n}=5\times 1+1\times 4-1=8 \\ 248\frac{8}{0} \\ \mathrm{Hybridisation} \mathrm{of} \mathrm{N} \mathrm{in} {\mathrm{NH}}_4^{+} \mathrm{is} {\mathrm{sp}}^3. \end{gathered}$$

Polarizing power of cation $\propto \frac{\mathrm{charge} \mathrm{on} \mathrm{ion}}{\mathrm{size} \mathrm{of} \mathrm{ion}}$

$\therefore {\mathrm{Li}}^{+} < {\mathrm{Be}}^{2+} < {\mathrm{B}}^{3+} < {\mathrm{C}}^{4+} (\mathrm{Polarising} \mathrm{power})$

Covalent character depends upon extent of polarizing. Higher the polarisation, higher is the covalent character.

LiCl < BeCl₂ < BCl₃ < CCl₄ (covalent character)

 $$\begin{gathered} \mathrm{For} \left(2\right), {\mathrm{He}}_2 \\ {\mathrm{\sigma }}_{1\mathrm{s}}^2{\mathrm{\sigma }}_{1\mathrm{s}}^{*2} \\ \mathrm{BOnd} \mathrm{order} = \frac{1}{2}(2-2) = 0 \\ \mathrm{Therefore}, {\mathrm{He}}_2 \mathrm{is} \mathrm{not} \mathrm{stable} \mathrm{and} \mathrm{hence} \mathrm{cannot} \mathrm{exist}. \\ {\mathrm{He}}_2^{+} \\ {\mathrm{\sigma }}_{1\mathrm{s}}^2{\mathrm{\sigma }}_{1\mathrm{s}}^{*1} \\ \mathrm{Bond} \mathrm{order} = \frac{1}{2}(2-1) = 0.5 \\ \mathrm{therefore}, {\mathrm{He}}_2^{+} \mathrm{can} \mathrm{exist}. \\ \mathrm{For} \left(3\right), \mathrm{Bond} \mathrm{order} \mathrm{of} {\mathrm{N}}_2 \mathrm{is} \mathrm{highest} \mathrm{amongst} \mathrm{homonuclear} \mathrm{diatomic} \mathrm{molecules} \mathrm{of} \mathrm{second} \mathrm{period} \mathrm{elements} ({\mathrm{Li}}_2, {\mathrm{Be}}_2, {\mathrm{B}}_2, {\mathrm{C}}_2, {\mathrm{N}}_2, {\mathrm{O}}_2, {\mathrm{F}}_2, {\mathrm{Ne}}_2) \\ \mathrm{For} \left(4\right), \mathrm{Order} \mathrm{of} \mathrm{energies} \mathrm{of} \mathrm{molecular} \mathrm{orbitals} \mathrm{for} {\mathrm{N}}_2 \mathrm{is} \\ {\mathrm{\sigma }}_{1\mathrm{s}} < {\mathrm{\sigma }}_{1\mathrm{s}}^{*} <{\mathrm{\sigma }}_{2\mathrm{s}} < {\mathrm{\sigma }}_{2\mathrm{s}}^{*} <{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}} = {\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}} <{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}} < {\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{*} = {\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{*} < {\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^{*} \end{gathered}$$