A) +4
B) +3
C) +2
D) +1
Correct Answer: A
Solution :
Magnetic moment \[(\mu )=\sqrt{n(n+2)}\] \[1.73=\sqrt{n(n+2)}\] \[3=n(n+2)\] \[n=1\] Hence, one unpaired electron in the vanadium ion. \[V(Z=23):[Ar]3{{d}^{3}}4{{s}^{2}}\] \[{{V}^{4+}}(Z=23):[Ar]3{{d}^{1}},4{{s}^{0}}\] Hence, oxidation state of vanadium is +4.You need to login to perform this action.
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