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Q. The electronic configuration of an element is $\text{1s}^{\text{2}} \text{,} \, \text{2s}^{\text{2}} \text{,} \, \text{2p}^{\text{6}} \text{,} \, \text{3s}^{\text{2}} \text{,} \, \text{3p}^{\text{3}}$ . What is the atomic number of the element, which is just below the above element in the periodic table?

NTA AbhyasNTA Abhyas 2020Structure of Atom

Solution:

The electronic configuration of valence shell of element below the given element is $4s^{2}4p^{3}$ . 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 $1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2}3d^{10},4p^{3}$ . So its atomic number would be 33.