Question

The decreasing order of the first ionisation energy (in $kJmo{l}^{-1}$ ) of $He,Mg$ and $Na$ is $He>Mg>Na$. The increasing order of the $2nd$ ionisation energy $($ in $kJmo{l}^{-1})$ of these elements will be

• A. $Na<Mg<He$
• B. $Mg<Na<He$
• C. $Mg<He<Na$
• D. $Na<He<Mg$

Answer 1

Answer: (B)

In case of sodium metal removal of $2^{\text{nd}}$ electron takes place from $2p$ subshell which is more closer to nucleus arid thus more energy is required to remove the electron, whereas removal of $2^{\text{nd}}$ electron from magnesium takes place from $3s$ subshell.
$He\underset{\text{removal of}1s^2{e}^{-}}{\overset{I{E}_1}{\to }}\underset{(1s^1)}{H{e}^{+}}\underset{\text{removal}1s^1{e}^{-}}{\overset{I{E}_2}{\to }}\underset{1s^0}{H{e}^{2+}}$
Noble gases have the highest ionisation energies in their respective periods. In helium, removal of $2^{\text{nd}}$ electron takes place from $1s^1$ subshell which is nearest to nucleus.
Therefore, the increasing order of $I{E}_2$ (in $kJmo{l}^{-1}$ ) of these elements is as follows $Mg<Na<He$