MCQ
$\mu = \sqrt{15}$ is true for the pair :-
- ✓$Co^{+2}, Cr^{+3}$
- B$Fe^{+2}, Cr^{+3}$
- C$Fe^{+3}, Fe^{+2}$
- D$Mn^{+2} , Fe^{+2}$
✓
Answer
Correct option: A.
$Co^{+2}, Cr^{+3}$
a
The electronic configuration of $\mathrm{Co}^{2+}$ is; $[\mathrm{Ar}] 4 \mathrm{s}^{0} 3 \mathrm{d}^{7}$
The electronic configuration of $\mathrm{Co}^{2+}$ is; $[\mathrm{Ar}] 4 \mathrm{s}^{0} 3 \mathrm{d}^{7}$
Hence there are three unpaired electrons in $\mathrm{Co}^{2+}$ Now the elctronic configuration of $\mathrm{Cr}^{3+}$ is; $[\mathrm{Ar}] 3 \mathrm{d}^{3}$
Hence there are 3 unpaired electrons in $\mathrm{Cr}^{3+}$ We know magnetic moment $\mu=\sqrt{(\mathrm{n}(\mathrm{n}+2)} \$ \$$
where n is number of unpaired electrons.
So when $n=3$ $\mu=\sqrt{(3(3+2)}=\sqrt{15}$
Hence option A is correct answer.
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