Unit 8 The D And F Block Elements (Intext Questions-6)
Intext Question
8.8 Calculate the ‘spin only’ magnetic moment of $\mathrm{M}^{2+}{ _(a q)}$ ion $(Z=27)$.
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Answer
Number of unpaired electrons
Given:
Atomic number $(Z)=27$
The valence electronic configuration of cobalt ( $\mathrm{Co}$ ) is $3 d^7, 4 s^2$.
$\mathrm{M}^{2+}(\mathrm{aq})$ ion means, it loses two electrons.
Hence, valence electronic configuration becomes $d^7$.
Now, Number of unpaired electrons $=3$
Spin only magnetic moment
We know that $\mu=\sqrt{n(n+2)}$
where $n=$ number of unpaired electrons
Putting $\mathrm{n}=3$
we get, $\mu=\sqrt{3(3+2)}=3.87 \mathrm{BM}$
