Question

Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is

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

Answer 1

The correct option is A $\left[Ne\right]3s^{2}3p^3$

Solution:

In groups: On moving down the group,number of shells increases, so attraction of nucleus on outer most electrons decreases.

Thus, the ionization enthalpy decreases.

$IE=As,p$

In periods: On moving left to right, effective nuclear charge $Z_{eff}$ increases, ionization enthalpy also increases.

${}_{13}AI-\left[Ne\right]3s^{2}3p^1$

${}_{14}Si-\left[Ne\right]3s^{2}3p^2$

${}_{15}P-\left[Ne\right]3s^{2}3p^3$

${}_{33}As-\left[Ar\right]3d^{10}4s^{2}4p^3$

Order of effective nuclear charge $\left(Z_{eff}\right)=Al<Si<p$

Ionisation enthalpy order $=Al<Si<p$

$p$ has maximum ionisation enthalpy. Ionisation enthalpy of "As" is low since it has 1 shell more than AI, Si and p.

Hence, option (B) is correct.