Question

Predict the order of ${\Delta }_0$ for the following compounds.
I. $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$
II. $\left[Mn\right(CN{)}_2\left(H_{2}O{)}_4\right]$
III. $\left[Mn\right(CN{)}_4(H_{2}O{)}_2{]}^{2-}$

• A. ${\Delta }_0\left(I\right)<{\Delta }_0\left(II\right)<{\Delta }_0\left(III\right)$
• B. ${\Delta }_0\left(II\right)<{\Delta }_0\left(I\right)<{\Delta }_0\left(III\right)$
• C. ${\Delta }_0\left(III\right)<{\Delta }_0\left(I\right)<{\Delta }_0\left(I\right)$
• D. ${\Delta }_0\left(I\right)<{\Delta }_0\left(III\right)<{\Delta }_0\left(II\right)$

Answer 1

The correct option is A ${\Delta }_0\left(I\right)<{\Delta }_0\left(II\right)<{\Delta }_0\left(III\right)$

More is the number of strong field ligand higher is the crystal field stabilisation energy.

So, $\left[Mn\right(Cn{)}_4(H_{2}O{)}_2{]}^{2-}$ having $4$ $C{N}^{-}$ ligands will have the highest value of ${△}_0$ then followed by $\left[Mn\right(CN{)}_2\left(H_{2}O{)}_4\right]$ then $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$.