NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} {\mathrm{C}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1}^{+}\left(\mathrm{g}\right)\to {\mathrm{C}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^2}^{2+}\left(\mathrm{g}\right)+ {\mathrm{e}}^{-} \\ {\mathrm{N}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^2}^{+}\left(\mathrm{g}\right)\to {\mathrm{N}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1}^{2+} \left(\mathrm{g}\right)+{\mathrm{e}}^{-} \\ {\mathrm{O}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3}^{+}\left(\mathrm{g}\right)\to {\mathrm{O}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^2}^{2+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-} \\ {\mathrm{F}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^4}^{+}\left(\mathrm{g}\right)\to {\mathrm{F}}_{1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3}^{2+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-} \end{gathered}$$

According to ionic size and effective nuclear charge, their order of decreasing second ionization potential should be

F > O > N > C

However, O⁺(g) has half-filled electronic configuration which is more stable than partially-filled electronic configuration in F⁺(g). Therefore, second ionization potential of O is more than that of F. Hence, the correct order is 

O > F > N > C

Element above the element with atomic number 43 is Manganese (Mn).

Mn$\to$25

Electronic configuration of M $\to$ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

IE₁ of B is lower than that of Be because in B, a 2p-electron is to be removed while in Be it is 2s-electron. Similarly, IE₁ of Al is lower than that of Mg because in Al, a 3p-electron is to be removed while in Mg it is the 3s-electron. IE₁ of N is higher than that of O because os extra stability of exactly half-filled 2p subshell in N. IE₁ of Mg is higher than that of Na because of higher nuclear charge and completely filled 3s-orbital in Mg. There is no exception in Na & Mg.

IE₁<<<IE₂ ( for Na as Na⁺ (g) has noble gas configuration) second ionization enthalpy of Na is very high and that is why it does not form Na²⁺

All the elements in group l have one electron in their valence shells. Transfer of one electron from the valence shell of elements in group l to the valence shall of elements in group VII results in :

(i) Octet formation for both.

(ii) Formation of ions having charge of +1 and -1 respectively.

The combining ratio 1 : 1. This ratio of ions results in a neutral aggregate with the formula AB, where A stands for the cations of the group l elements , and B stands for the anions of group VII.

The ionisation potential of Mg is higher than Na as well as Al due to presence of stable configuration i.e ns²

decreasing order of size (Na+ smallest size)

N^3- > O^2- > F^- > Na⁺

The electron affinity of F is less than Cl. Hence electron affinity of F is greater than I is the correct option.

Due to presence of stable configuration,

[Ne]3s²3p³ has highest energy.