Question

Which of the following order is /are incorrect?

• A. $O<Se<S\left[E{A}_1\text{order}\right]$
• B. $O^{2-}<N{a}^{+}<F^{-}<M{g}^{2+}$ [Ionic radius order]
• C. $Li<Be<B<C\left[I{E}_1\text{order}\right]$
• D. $Cl<F<Br<I\left[E{A}_1\text{order}\right]$

Answer 1

Answer: (B), (C), (D)

$Cl<F<Br<I\left[E{A}_1\text{order}\right]$

a) As we move down the group, electron affinity decreases due to the decrease in effective nuclear charge. However, due to very small size, oxygen has the lowest electron affinity in its group. Hence the statement is correct.

b) Size of the cation is smaller than the size of the parent atom and size of the anion is greater than the size of the parent atom. For a isoelectric species greater the postive charge, smaller the size because of increase in charge of the effective nuclear charge compared to the parent atom. Hence, the correct order is $O^{2-}>F^{-}>N{a}^{+}>M{g}^{2+}$

c) Ionization energy decreases, as we move from left to right along the periodice table. However, due to more penetration of 2s orbital as compared to 2p orbital, ionization energy of Be is greater than that of B.
Hence the correct order is $Li<B<Be<C$

d) Electron affinity decreases down the group due to decrease in effective nuclear charge. However, due to small size of F, electron affinity of F is smaller than that of Cl.
Hence the correct order of EA is $I<Br<F<Cl$