General configuration of outermost and penultimate shell is n−1s2n−1p6n−1dxns2. If n = 4 and x = 5 then no. of protons in the nucleus will be

Given n = 4 x = 5So 4−1s23s24−1p63p64−1d53d54s24s2Total electron = 2 + 6 + 5 + 2 = 15Electron in 1 + 2 orbit = 2 + 8 = 10Total electron = 10 + 15 = 25No. of electron = No. of protonSo total proton = 25