mu = sqrt{15} is true for the pair :- | vedclass

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Question

The electronic configuration of $\mathrm{Co}^{2+}$ is; $[\mathrm{Ar}] 4 \mathrm{s}^{0} 3 \mathrm{d}^{7}$

Hence there are three unpaired electrons i

Vedclass · Accepted Answer

$Co^{+2}, Cr^{+3}$

Vedclass · Answer

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.

$\mu = \sqrt{15}$ is true for the pair :-

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