Chemistry - GENERAL CHEMISTRY Question with Solution | TestHub

The electronic configuration of valence shell of element below the given element is$4{\mathrm{s}}^{2}4{\mathrm{p}}^3$. But before the filling of 4p-orbitals start, the electrons enters the 3d-orbitals, which can accommodate 10 electrons. Hence, the electronic configuration of the element, which is just below the given element must be$1{\mathrm{s}}^2,2{\mathrm{s}}^{2}2{\mathrm{p}}^6,3{\mathrm{s}}^{2}3{\mathrm{p}}^6,4{\mathrm{s}}^{2}3{\mathrm{d}}^{10},4{\mathrm{p}}^3$. So its atomic number would be 33.