Question

The total spin resulting from a ${d^7}$ configuration is:
A. $3/2$
B. $1/2$
C. $2$
D. $1$

Answer 1

In order to answer this question you must be aware of the concepts of atomic structure or coordination compounds. Firstly, do the electronic configuration of d orbital and fill 7 electrons according to the rules. And then evaluate the number of unpaired electrons and then use the concept that spin of one unpaired electron is ½. Then use the unitary method to find the spin of the resultant no. of unpaired electrons.

Complete step-by-step answer: Step 1: When you will do the electronic configuration of ${d^7}$ orbital, you will find that there are 3 (three) unpaired electrons in that d orbital. The paired set of electrons will not give any contribution in the total spin because the spins of paired electrons are in the opposite direction and hence they will simply cancel each other. Whereas, the spin of unpaired electrons will be preserved and will contribute to the total spin.
Step 2: Since we know that, the spin of one unpaired electron is $1/2$ . Then the electron spin of 3 unpaired electrons will be:
$\frac{1}{2}\, + \,\frac{1}{2}\, + \,\frac{1}{2}$ $= \,3 \times \frac{1}{2}\, =$ $\frac{3}{2}$ .
So, the total spin from a ${d^7}$ configuration which is having 3 unpaired electrons will be $\frac{3}{2}$ .

Hence, clearly option A is the correct answer.

Note: In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles, and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to the classical angular momentum of orbital revolution and appears when there is periodic structure to its wave function as the angle varies.