Classification of elements and Periodicity in properties MCQs for NEET — Chemistry Questions with Answers

(d) All are the characteristics of transition metals.

(c) It is a fact.

MCl_n does not undergo hydrolysis. This implies than Mn⁺ on hydrolysis form a strong base i.e. M(OH)_n.

Elements of first grioup forms strong bases and that of third group, fifth group and transition metals form weak bases.

Group 14 elements exhibit oxidation states of +4 and +2. In +4 state the compounds are ionic. The +4 state becomes less important in Ge, Sn and Pb because they prefer to exhibit +2 state for ionic bond formation due to inert pair effect. The inert pair effect decreases down the group from Ge to Pb and stability of +4 oxidation state decreases and that of +2 state increases. Thus stability of M²⁺ ions is in the order Ge²⁺<Sn²⁺<Pb²⁺.

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

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.