The number of f-electrons in the ground state electronic configuration of Np (Z = 93) is ……… . (Nearest integer)

The electronic configuration of Np = 93 is

       $\begin{array}{l}1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}\\ 5{\mathrm{d}}^{10}6{\mathrm{s}}^{2}6{\mathrm{p}}^{6}5{\mathrm{f}}^{4}6{\mathrm{d}}^{1}7{\mathrm{s}}^2\end{array}$

In ground state, total number of f electrons = 14 + 4 = 18