Question

In $Cu$ (Atomic No. 29)

• A. 13 electrons have spin in one direction and 16 electrons in other direction
• B. 14 electrons have spin in one direction and 15 electrons in other direction
• C. Total spin , $S=\frac{1}{2}$
• D. Spin magnetic moment is $\sqrt{3}B.M.$

Answer 1

Answer: (B), (C), (D)

Spin magnetic moment is $\sqrt{3}B.M.$

The electronic configuration of $Cu$ is
${}_{29}Cu=[Ar{]}^{18}3d^{10}4s^1$
the electron jumps from $4s$ to $3d$ orbital as in this $3d$ is completely filled and stable due to high exchange energy.

B) All electrons are paired except $4s^1$. Hence, $14e^{-}$ have spin in one direction and $15e^{-}$ in the other.

C) Total spin is given as
$S=n\times \frac{1}{2}$
where n is the number of unpaired electrons
Here, $n=1$ So,
$S=\frac{1}{2}$
D) Spin magnetic moment is given as
$\mu =\sqrt{n(n+2)}B.M$
n = no. of unpaired electrons
${}_{29}Cu$ has only one unpaired electron in $4s$ i.e. n=1
Therefore , $\mu =\sqrt{1(1+2)}$
$\mu =\sqrt{3}B.M.$
Hence, (b) ,(c) and (d) are correct.