Question

Based on the following, which comparison(s) is/are correct?

• A. $I_2<I_2^{{}^{\prime }}$
• B. $I_1^{{}^{\prime }}>I_1^{{}^{\prime \prime }}$
• C. $I_1<I_1^{{}^{\prime }}<I_2^{\prime \prime }$
• D. $I_1<I_1^{\prime \prime }<I_1^{{}^{\prime }}$

Answer 1

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

The correct options are
B $I_1^{{}^{\prime }}>I_1^{{}^{\prime \prime }}$
C $I_1<I_1^{{}^{\prime }}<I_2^{\prime \prime }$
D $I_1<I_1^{\prime \prime }<I_1^{{}^{\prime }}$

The electronic configuration of
$Na=1s^{2}2s^{2}2p^{6}3s^1$
$Mg=1s^{2}2s^{2}2p^{6}3s^2$
$Al=1s^{2}2s^{2}2p^{6}3s^{2}3p^1$
Since removal of one electron gives a stable electronic configuration to sodium, thus the first ionizatine energy of sodium is lowest among the three.

But electronic configuration of
$N{a}^{+}=1s^{2}2s^{2}2p^6$
$M{g}^{+}=1s^{2}2s^{2}2p^{6}3s^1$
Since $N{a}^{+}$ has stable electronic configuration of $Ne$, thus the second ionization energy of of $Na$ is more than $Mg$.

Since $Mg$ has stable electronic configuration, filled outermost subshell and $Al$ has only 1 electron in the $p-orbital$, thus $I_1^{{}^{\prime }}>I_1^{{}^{\prime \prime }}$
Again the second I.E of $Al$ is more than the first I.E of $Mg$ followed by the first IE of $Na$

So first ionization energy of $Mg$ is maximum among the three and first ionization energy of $Na$ is minimum.