Question

Which of the following relations is correct wrt first ($I_1$) and second ($I_2$) ionisation potentials of $Na$ and $Mg$?

• A. $I_{1\left(Na\right)}>I_{1\left(Mg\right)}$
• B. $I_{1\left(Mg\right)}=I_{2\left(Na\right)}$
• C. $I_{2\left(Mg\right)}>I_{2\left(Na\right)}$
• D. $I_{2\left(Na\right)}>I_{2\left(Mg\right)}$

Answer 1

Answer: (D)

$I_{2\left(Na\right)}>I_{2\left(Mg\right)}$

The correct relation is $I_{2\left(Na\right)}>I_{2\left(Mg\right)}$
The electronic configurations of neutral atoms are given below.
$Na:\left[Ne\right]3s^1$
$Mg:\left[N3\right]3s^2$
$Na$ loses an electron to form $N{a}^{+}:\left[Ne\right]3s^0$ with stable electronic configuration of Noble gas with completed octet.
The removal of second electron will break the octet and will require higher energy. On the other hand, when $Mg$ loses two electrons, it forms $M{g}^{2+}:\left[Ne\right]3s^0$ with stable electronic configuration of Noble gas with completed octet. This requires less amount of energy.