General configuration of outermost and penultimate shell is $\left(\mathrm{n}-1\right){\mathrm{s}}^2\left(\mathrm{n}-1\right){\mathrm{p}}^6\left(\mathrm{n}-1\right){\mathrm{d}}^{\mathrm{x}}{\mathrm{ns}}^2$. If n = 4 and x = 5 then no. of protons in the nucleus will be

Given n = 4 x = 5

So $\underset{3{\mathrm{s}}^2}{\left(4-1\right){\mathrm{s}}^2}\underset{3{\mathrm{p}}^6}{\left(4-1\right){\mathrm{p}}^6}\underset{3{\mathrm{d}}^5}{\left(4-1\right){\mathrm{d}}^5}\underset{4{\mathrm{s}}^2}{4{\mathrm{s}}^2}$

Total electron = 2 + 6 + 5 + 2 = 15
Electron in 1 + 2 orbit = 2 + 8 = 10
Total electron = 10 + 15 = 25
No. of electron = No. of proton
So total proton = 25