The spin-only magnetic moment value of B_{2}^ | vedclass

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Question

(C) The total number of electrons in $B_{2}^{+}$ is $5 + 5 - 1 = 9$.The molecular orbital configuration of $B_{2}^{+}$ is $\sigma_{1s}^{2} \sigma_{1s}

Vedclass · Accepted Answer

$173$

Vedclass · Answer

(C) The total number of electrons in $B_{2}^{+}$ is $5 + 5 - 1 = 9$.The molecular orbital configuration of $B_{2}^{+}$ is $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2py}^{1}$.There is $1$ unpaired electron $(n=1)$.The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM$.Given $\sqrt{3} = 1.73$,so $\mu = 1.73 \ BM = 173 	imes 10^{-2} \ BM$.

The spin-only magnetic moment value of $B_{2}^{+}$ species is $...... \times 10^{-2} \ BM$. (Nearest integer)
[Given : $\sqrt{3}=1.73$ ]

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The spin-only magnetic moment value of $B_{2}^{+}$ species is $...... \times 10^{-2} \ BM$. (Nearest integer)[Given : $\sqrt{3}=1.73$ ]