Chemistry MCQs for NEET — Practice Questions with Answers

(A) The magnitude of and element’s electron affinity depends on the element’s valence shell electrons
configuration.

${}_{25}Mn ={\left[Ar\right]}^{18}3d^{5}4s^2$ configuration,

${}_{25}Cr ={\left[Ar\right]}^{18}3d^{5}4s^1$ configuration

${}_{25}Fe ={\left[Ar\right]}^{18}3d^{6}4s^2$ configuration

(A) High electronegativity and small size, has only – oxidation state.

$$\begin{gathered} △H =△H_{sub}+I{E}_1+I{E}_2+△H_{hydration} \\ =1000+750+12000+(-1800) \\ △H=1150 KJ mo{l}^{-1} \end{gathered}$$

${\mathrm{BaSO}}_4$ is insoluble.

(d) According to Fajan's rule,

Covalent character $$\begin{gathered} \propto \frac{1}{\mathrm{size} \mathrm{of} \mathrm{cation}} \\ \propto \mathrm{size} \mathrm{of} \mathrm{anion} \end{gathered}$$

In the given options, cation is same but anions are different. Among halogens the order of size is

                      F < Cl < Br < I

$\therefore$Order of covalent character is

                   MI > MBr > MCl > MF

(a) Isostructural Compounds having same structure and same hybridisation are known as isostructural species. e.g. XeF₂ and IF₂^-   are sp³d hybridised and both have linear shape.

F-I-F    F-Xe-F

(b) Molecules with zero bond order, do not exist.

According to molecular orbital theory,

$$\begin{gathered} \left(\mathrm{a}\right) {\mathrm{Be}}_2^{+} (4+4-1=7) \\ = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^1 \\ \mathrm{Bond} \mathrm{order} \left(\mathrm{BO}\right) = \frac{4-3}{2} = 0.5 \\ \left(\mathrm{b}\right) {\mathrm{Be}}_2(4+4=8)= \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2 \\ \mathrm{BO} = \frac{4-4}{2}=0 \\ \left(\mathrm{c}\right) {\mathrm{B}}_2(5+5=10) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1 \approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1 \\ \left(\mathrm{d}\right) {\mathrm{Li}}_2 (3+3=6) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2 \\ \mathrm{BO} = \frac{4-2}{2}=1 \\ \mathrm{Thus}, {\mathrm{Be}}_2 \mathrm{does} \mathrm{not} \mathrm{exist} \mathrm{under} \mathrm{normal} \mathrm{conditions}. \end{gathered}$$

(c) Paramagnetic character is shown by those atoms or molecules which have unpaired electrons.

In the given compounds CO is not paramagnetic since, it does not have unpaired electrons. The configuration of CO molecule is 

CO(14) = $\sigma 1s^2, \stackrel{*}{\sigma }1s^2, \sigma 2s^2, \stackrel{*}{\sigma }2s^2, \sigma 2p_x^2, \pi 2p_y^2 \approx \pi 2p_z^2$