Question

For $M{n}^{3+}$ ion, the electron pairing energy P is about 28,000 $c{m}^{-1},{\Delta }_{\circ }$ values for the complexes $\left[Mn\right(H_{2}O{)}_6{]}^{3+}$ and $\left[Mn\right(CN{)}_6{]}^{3-}$ are 15,000 $c{m}^{-1}$ and 38,500 $c{m}^{-1}$, respectively, which of the following complex is high spin:

• A. $\left[Mn\right(CN{)}_6{]}^{3-}$
• B. $\left[Mn\right(H_{2}O{)}_6{]}^{3+}$
• C. Both are high spin
• D. None of these

Answer 1

The correct option is B $\left[Mn\right(H_{2}O{)}_6{]}^{3+}$

As $\left[Mn\right(H_{2}O{)}_6{]}^{3+}$ have lower pairing energy than normal so it have less tendency to pair and it will form high spin complex while $\left[Mn\right(CN{)}_6{]}^{3-}$ form low spin complex.

$\left[Mn\right(CN{)}_6{]}^{3-}$ - strong ligand pairing happens low spin

$\left[Mn\right(H_{2}O{)}_6{]}^{3+}$ - weak ligand no pairing high spin

Hence option B is correct.