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Question

Chemistry Question on Electronic Configurations Of Elements And The Periodic Table

The orbital configuration oi 24Cr_{24}Cr is 3d54s13d^5 4s^1. The number of unpaired electrons in Cr3+(g)Cr^{3+} (g) is

A

3

B

2

C

1

D

4

Answer

3

Explanation

Solution

Cr3+Cr^{3+} has configuration 3d34s03d^3\, 4s^0