The following is the radius of Group $15$ elements:
$$\begin{array}{c|c} \hline \text{Element} & \text{Covalent Radius }(\pu{pm}) \\ \hline \ce{N} &75 \\ \ce{P} &110 \\ \ce{As} &121 \\ \ce{Sb} &140 \\ \ce{Bi} &155 \\ \hline \end{array}$$ Source: General Properties and Reactions. (2020, August 15). Retrieved April 6, 2021, from https://chem.libretexts.org/@go/page/31730
From the above table it is clear that the increase in radius decreases as we go down the group.
My Attempt to Reason it
The electronic configurations of $\ce{N, P}$ and $\ce{As}$ are:
- $\ce{N: [He]} \ \mathrm{2s^2 2p^3}$
- $\ce{P: [Ne]} \ \mathrm{3s^2 3p^3}$
- $\ce{As: [Ar]} \ \mathrm{4s^2 3d^10 4p^3}$
First let us compare $\ce{N}$ and $\ce{P}$
$\ce{P}$ has $\ce{8 e-}$s and $\ce{8 p+}$s more than $\ce{N}$. The $\ce{8 p+}$s increase the nuclear charge. But, due to the addition of $\ce{e-}$s, the shielding effect also increases i.e., there is more repulsions caused by the core $\ce{e-}$s.
Out of the $\ce{8 e-}$s, $3$ goes into $\mathrm{2p}$, $2$ goes into $\mathrm{3s}$ and $3$ goes into $\mathrm{3p}$. The shielding effect due to $\mathrm{s}$ and $\mathrm{p}$ is appreciable. As a result, the effective nuclear charge is not that high. But due to the addition of extra shell i.e., $3$, the size increases by almost $\pu{35 pm}$.
Now Let us compare $\ce{P}$ and $\ce{As}$
Now in $\ce{As}$, there are $18 \ce{e-}$s and $18 \ce{p+}$s more than that in $\ce{P}$. Out of the $18 \ce{e-}$s, $3$ goes into $\mathrm{3p}$, $2$ goes into $\mathrm{4s}$, $10$ goes into $\mathrm{3d}$ and $3$ goes into $\mathrm{4p}$.
In this case, majority of the $\ce{e-}$s are in the $\mathrm{3d}$ orbital. The shielding effect due to $\mathrm{3d}$ is not much. In fact, the order of the ability of orbitals to shield is: $$\mathrm{s} \gt \mathrm{p} \gt \mathrm{d} \gt \mathrm{f}$$
Due to this, the effective nuclear charge is more and therefore the size doesn't increase as much as it increase on going from $\ce{N}$ to $\ce{P}$.
Is my reasoning correct? If not, what is the reason behind this?