Question

The pair in which both species have same magnetic moment (spin only value) is :

• A. $\left[Cr\right(H_{2}O{)}_6{]}^{2+},[CoC{l}_4{]}^{2-}$
• B. $\left[Cr\right(H_{2}O{)}_6{]}^{2+},\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• C. $\left[Mn\right(H_{2}O{)}_6{]}^{2+},\left[Cr\right(H_{2}O{)}_6{]}^{2+}$
• D. $[CoC{l}_4{]}^{2-},[Fe(H_{2}O{)}_6{]}^{2+}$

Answer 1

Answer: (B)

$\left[Cr\right(H_{2}O{)}_6{]}^{2+},\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

$H_{2}O$ is a weak field\quad ligand. Hence, it will not pair the electrons in the metal's orbital.

${Cr}^{2+}\left[{3d}^4{4s}^0\right]⟶\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\end{array}\right]$

${Fe}^{2+}\left[{3d}^6{4s}^0\right]⟶\left[\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]\left[\begin{array}{c}\upharpoonleft \end{array}\right]$

Both have 4 unpaired electrons in their d-orbital.

Magnetic moment $\mu =\sqrt{n(n+2)}BM$ where n is the number of unpaired electrons.

In this case,

$\mu =\sqrt{4(4+2)}=\sqrt{24}BM$

Hence, the value of magnetic moment is same for both.

Therefore option B is correct.