Question

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?

• A. 33
• B. 34
• C. 31
• D. 49

Answer 1

Answer: (A)

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.