JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 21)
The spin-only magnetic moment value of $$B_2^ + $$ species is _____________ $$\times$$ 10$$-$$2 BM. (Nearest integer) [Given : $$\sqrt 3 $$ = 1.73]
Answer
173
Explanation
$$B_2^ + \Rightarrow \sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\pi _{2py}^1 \simeq \pi _{2pz}^0$$
It has one unpaired electron.
Spin - only magnetic moment = $$\mu $$
= $$\sqrt {n\left( {n + 1} \right)} $$
n = Number of unpaired electrons
$$= \sqrt {1(1 + 2)} = \sqrt 3 $$ BM
= 1.73 BM
= 1.73 $$\times$$ 10$$-$$2 BM
It has one unpaired electron.
Spin - only magnetic moment = $$\mu $$
= $$\sqrt {n\left( {n + 1} \right)} $$
n = Number of unpaired electrons
$$= \sqrt {1(1 + 2)} = \sqrt 3 $$ BM
= 1.73 BM
= 1.73 $$\times$$ 10$$-$$2 BM
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