Question

The species that has a spin only magnetic moment of $5.9\mathrm{BM}$ is

• A. $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{-2}(\mathrm{square}\mathrm{planar})$
• B. $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\left(\mathrm{Td}\right)$
• C. ${\left[{\mathrm{MnBr}}_4\right]}^{-2}\left(\mathrm{Td}\right)$
• D. ${\left[{\mathrm{NiCl}}_4\right]}^{-2}\left(\mathrm{Td}\right)$

Answer 1

The correct option is C

${\left[{\mathrm{MnBr}}_4\right]}^{-2}\left(\mathrm{Td}\right)$

The explanation for the correct option:

C: ${\left[{\mathrm{MnBr}}_4\right]}^{-2}\left(\mathrm{Td}\right)$

The expression used for the calculation of magnetic moment:

$\mathrm{\mu }=\surd \mathrm{n}(\mathrm{n}+2)$ where, $\mathrm{\mu }$ is the magnetic moment and $\mathrm{n}$ is the number of the unpaired electrons.

putting the required values in the above formula:

$\begin{array}{rcl}\mathrm{\mu }& =& \surd \mathrm{n}(\mathrm{n}+2)\\ & =& \surd 5(5+2)\\ & =& \surd 35\\ & =& 5.9\mathrm{BM}\end{array}$

The explanation for the incorrect options:

A: $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{-2}(\mathrm{square}\mathrm{planar})$

In the above electronic configuration there is no unpaired electron is present. Therefore, the magnetic moment is zero.

B: $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\left(\mathrm{Td}\right)$

In this too, the magnetic moment of the complex is zero.

D: ${\left[{\mathrm{NiCl}}_4\right]}^{-2}\left(\mathrm{Td}\right)$

In this complex, the magnetic moment is also zero because of zero unpaired electron.

Final Answer: Therefore, the correct option is C, i.e. ${\left[{\mathrm{MnBr}}_4\right]}^{-2}\left(\mathrm{Td}\right)$ has a spin-only magnetic moment $5.9\mathrm{BM}$.