Question

The calculated spin-only magnetic moment of ${\mathrm{Cr}}^{2+}$ ion is ?

• A. $2.84\mathrm{BM}$
• B. $3.87\mathrm{BM}$
• C. $4.90\mathrm{BM}$
• D. $5.92\mathrm{BM}$

Answer 1

The correct option is C

$4.90\mathrm{BM}$

Explanation for the correct option:

Option(C):$4.90\mathrm{BM}$

Step 1:Formula for calculating spin magnetic moment :

The spin magnetic moment is a magnetic moment caused due to the spin of particles.

The formula to find the spin magnetic moment is $\sqrt{\text{n(n}+\text{2)}}$,

(where n is denoted as the number of unpaired electrons)

Step 2:Calculation of unpaired electrons from Electronic configuration of ${\mathrm{Cr}}^{2+}$:

The ground state electronic configuration of $\mathrm{Cr}$ (the atomic number is 26):$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{1}3{\text{d}}^5$
Now, after removing 2 electrons it forms ${\mathrm{Cr}}^{2+}$ , for which the electronic configuration can be given as:

${\mathrm{Cr}}^{2+}$:$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{0}3{\mathrm{d}}^4$
Here, the number of unpaired electrons available in the given ion is $=4$

Step 2:Calculation of the spin magnetic moment of ${\mathrm{Cr}}^{2+}$:

From the given data the spin-only magnetic moment of ${\mathrm{Cr}}^{2+}$can be calculated as:

$$\begin{gathered} =\sqrt{\mathrm{n}(\mathrm{n}+2)} \\ =\sqrt{4\times (4+2)} \\ =4.9\mathrm{BM} \end{gathered}$$

Hence, The calculated spin-only magnetic moment of ${\mathrm{Cr}}^{2+}$ ion is $4.90\mathrm{BM}$.