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Question

In Cu (Atomic No. 29)

A
13 electrons have spin in one direction and 16 electrons in other direction
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B
14 electrons have spin in one direction and 15 electrons in other direction
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C
Total spin , S=12
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D
Spin magnetic moment is 3 B.M.
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Solution

The correct option is D Spin magnetic moment is 3 B.M.
The electronic configuration of Cu is
29Cu=[Ar]183d104s1
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 4s1. Hence, 14e have spin in one direction and 15e in the other.

C) Total spin is given as
S=n×12
where n is the number of unpaired electrons
Here, n=1 So,
S=12
D) Spin magnetic moment is given as
μ=n(n+2)B.M
n = no. of unpaired electrons
29Cu has only one unpaired electron in 4s i.e. n=1
Therefore , μ=1(1+2)
μ=3 B.M.
Hence, (b) ,(c) and (d) are correct.

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