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The orbital configuration of .(24)Cr is ...

The orbital configuration of `._(24)Cr` is `3d^(5) 4s^(1)`. The number of unpaired electrons in `Cr^(3+) (g)` is

A

1

B

2

C

3

D

4

Text Solution

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The correct Answer is:
C
`Cr-4s^(2) 3d^(5), Cr^(+)-4s^(0) 3d^(3)`
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