Chapter 7, Problem 9STP

Interpretation Introduction

Interpretation:

The correct orbital diagram for the third and fourth principal energy levels of vanadium needs to be identified.

Concept introduction:

Electron configuration shows that the orbitals that are occupied by electrons in an atom. Electron occupies orbitals to minimize the energy of the atom. So, lower energy level fills before higher energy orbitals. Each orbital can hold maximum of two electrons.

Answer to Problem 9STP

Option B is correct.

Explanation of Solution

Electron configuration of Vanadium is as follows:

Atomic number of V is 23.

${\text{V=1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{3}}$

According to the Hund’s rule of multiplicity the electrons first occupy the orbital singly with parallel spins rather than in pairs. Therefore, the correct orbital diagram for the third and fourth principal energy levels of vanadium is B.

Conclusion

In option A, the diagram shows that the 3d orbitals has four electrons but V has three electrons in 3d orbitals. Therefore, option A is incorrect.

In option C, 3d orbitals started to pair but according to Hund’s rule, electrons must singly occupy all the subshells before they started pairing up in any subshell. Therefore, option C is also incorrect.

In option D, 3d orbitals started to pair but according to Hund’s rule, electrons must singly occupy all the subshells before they started pairing up in any subshell. Therefore, option D is also incorrect.