Question

The magnetic moments of complexes given below are in the order:

(I) $\left[Ni\right(CO{)}_4]$ (II) $\left[Mn\right(CN{)}_6{]}^{4-}$ (III) $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ (IV) $[Co{F}_6{]}^{3-}$.

• A. $\left(I\right)>\left(II\right)>\left(III\right)>\left(IV\right)$
• B. $\left(I\right)<\left(II\right)<\left(III\right)<\left(IV\right)$
• C. $\left(II\right)>\left(III\right)>\left(IV\right)>\left(I\right)$
• D. $\left(IV\right)<\left(II\right)<\left(I\right)<\left(III\right)$

Answer 1

The correct option is B $\left(I\right)<\left(II\right)<\left(III\right)<\left(IV\right)$

As magnetic moment $\mu =\sqrt{n(n+2)}$ , where n is no. of unpaired electron.

In $\left[Ni\right(CO{)}_4],d^8$ configuration, low spin complex, 0 unpaired electron.

In $\left[Mn\right(CN{)}_6{]}^{4-},d^5$ configuration, low spin complex, 1 unpaired electron.

In $\left[Cr\right(N{H}_3{)}_6{]}^{3+},d^3$ configuration, 3 unpaired electron.

In $[Co{F}_6{]}^{3-},d^6$ configuration, high spin complex, 4 unpaired electron.