Question

Correct order of ionisation enthalpy among the following is:

• A. $Pb>Sn$
• B. $N{a}^{+}>M{g}^{+}$
• C. $L{i}^{+}<O^{+}$
• D. $B{e}^{+}<C^{+}$

Answer 1

Answer: (A), (B), (D)

$B{e}^{+}<C^{+}$

(a) We know that, the ionization enthalpy decreases regularly down the group; $Pb$, however, shows a higher value than $Sn$ due to poor shielding of inner $f-$orbitals as a result of which effective nuclear charge experienced by outer shell electrons becomes more in $Pb$.
(b) $N{a}^{+}:1s^{2}2s^{2}2p^{6}M{g}^{+}:1s^{2}2s^{2}2p^{6}3s^1$
As can be seen from the electronic configuration of both the ions, $N{a}^{+}$ has fully filled orbital hence removal of electron is tough and requires higher energy as compared to $M{g}^{+}$.
(c) $L{i}^{+}$ will have higher ionization enthalpy than $O^{+}$ as $L{i}^{+}:1s^2$ has noble gas configuration and its size is small as compared to $O^{+}$.
(d) Electronic configuration of $B{e}^{+}:1s^{2}2s^1\text{and}{C}^{+}:1s^{2}2s^{2}2p^1$. Since, effective nuclear charge of $B{e}^{+}$ is low, removal of an electron from $B{e}^{+}$ is easier as compared to $C^{+}$.