In how many of following pairs, second species has higher ionisation energy than first species :
(i) $Na^{+}, Mg^{2 +}$ (ii) $S, Cl$ (iii) $Cu, Zn$ (iv) $Xe, Kr$
(v) $B, Be$ (vi) $O, O^{+}$ (vii) $Al, Si$ (viii) $Cl^{-}, Cl$

Question

In how many of following pairs, second species has higher ionisation energy than first species :

(i) $Na^{+}, Mg^{2 +}$ (ii) $S, Cl$ (iii) $Cu, Zn$ (iv) $Xe, Kr$

(v) $B, Be$ (vi) $O, O^{+}$ (vii) $Al, Si$ (viii) $Cl^{-}, Cl$

TestHub · Accepted Answer

Ionization energy (IE) is the energy required to remove an electron.

(i) $Na^{+}$ ( $[N e]$ ) vs $Mg^{2 +}$ ( $[N e]$ ): $Mg^{2 +}$ has higher nuclear charge, so higher IE.

(ii) $S$ ( $[N e] 3 s^{2} 3 p^{4}$ ) vs $Cl$ ( $[N e] 3 s^{2} 3 p^{5}$ ): $Cl$ has higher nuclear charge, so higher IE.

(iii) $Cu$ ( $[A r] 3 d^{10} 4 s^{1}$ ) vs $Zn$ ( $[A r] 3 d^{10} 4 s^{2}$ ): $Zn$ has a stable full $4 s$ orbital, so higher IE.

(iv) $Xe$ vs $Kr$ : $Kr$ is smaller, so higher IE. (IE decreases down a group)

(v) $B$ ( $[He] 2 s^{2} 2 p^{1}$ ) vs $Be$ ( $[He] 2 s^{2}$ ): $Be$ has a stable full $2 s$ orbital, so higher IE. (Exception to trend)

(vi) $O$ vs $O^{+}$ : $O^{+}$ has higher effective nuclear charge for the remaining electrons, so higher IE. (Successive IE)

(vii) $Al$ ( $[N e] 3 s^{2} 3 p^{1}$ ) vs $Si$ ( $[N e] 3 s^{2} 3 p^{2}$ ): $Si$ has higher nuclear charge, so higher IE.

(viii) $Cl^{-}$ vs $Cl$ : $Cl$ is neutral and smaller than the anion $Cl^{-}$ , so higher IE.

All 8 pairs have the second species with higher ionization energy.

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