Question

From the electronic configuration of the given elements $K,L,M$ and $N$, which one has the highest ionization potential?

• A. $M=\left[Ne\right]3s^{2}3p^2$
• B. $L=\left[Ne\right]3s^{2}3p^3$
• C. $K=\left[Ne\right]3s^{2}3p^1$
• D. $N=\left[Ar\right]3d^{10},4s^{2}4p^3$

Answer 1

Answer: (B)

$L=\left[Ne\right]3s^{2}3p^3$

Let us consider all the elements

element $M$ has two elements in the outermost shell which makes it easy to remove as compared to $L$ which has half filled $porbital$ and is therefore stabler. Even $K$ has only one electron in $porbital$ which makes it easy to remove.

The element $N$ is large since it goes to $3d$ orbital also. Large size makes it easier to remove an electron since the energy holding it is lesser. Therefore since $L$ has half filled $porbital$ and is quite small in size it will have highest ionization potential.