Question

Which one of the following is correct order of second ionization potential of $Na,Ne,Mg$ and $Al$?

• A. $Al<Na<Mg<Ne$
• B. $Ne<Al<Na<Mg$
• C. $Mg<Al<Ne<Na$
• D. $Na<Mg<Ne<Al$

Answer 1

Answer: (C)

$Mg<Al<Ne<Na$

The correct order of second ionisation potential of $Na,Ne,Mg$ and $Al$, is $Mg<Al<Ne<Na$.

When $Mg$ loses a second electron, a stable electronic configuration of $Ne$ is obtained. Hence, $Mg$ has the lowest value of second ionisation potential.

When sodium loses a second electron, a stable electronic configuration of $Ne$ is broken. Hence, $Na$ has a maximum second ionisation potential.

Hence, the correct option is $\text{C}$