Question

The magnetic moments of complex given below are in order

I. $\left[Ni(CO{)}_4\right]$

II. ${\left[Mn(CN{)}_6\right]}^{4-}$

III. ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}$

IV. ${\left[Co{F}_6\right]}^{3-}$

• A. I > II > III > IV
• B. I < II < III < IV
• C. IV > II > I > III
• D. IV < II < I < III

Answer 1

Answer: (B)

Magnetic moment $\mu =\sqrt{n(n+2)}B.M.$

n = unpaired electrons.

$n=$ unpaired electrons.

I. $\left[Ni(CO{)}_4\right]n=0$, so $\vec{\mu }=0BM$

II. ${\left[Mn(CN{)}_6\right]}^{4-}n=1$, so $\vec{\mu }=\sqrt{3}BM$

III. ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}n=3$, so $\vec{\mu }=\sqrt{15}BM$

IV. ${\left[Co{F}_6\right]}^{3-}n=4$, so $\vec{\mu }=\sqrt{24}BM$

Order of magnetic moment is $I<II<III<IV$.