Question

One of the most important properties of d-block metal cations is the formation of coordinate compounds. Widely used theories that explain bonding in these compounds are valence bond theory and crystal field theory.
The pair of complex ions which have the same value of magnetic moments is:

• A. ${\left[Co{\left(H_{2}O\right)}_6\right]}^{3+},{\left[Ni{F}_6\right]}^{2-}$
• B. ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+},{\left[Fe{\left(H_{2}O\right)}_6\right]}^{3+}$
• C. ${\left[Mn{\left(N{H}_3\right)}_6\right]}^{2+},{\left[Fe{\left(CN\right)}_6\right]}^{3-}$
• D. ${\left[CoC{l}_4\right]}^{2-},{\left[NiC{l}_4\right]}^{2-}$

Answer 1

The correct option is A ${\left[Co{\left(H_{2}O\right)}_6\right]}^{3+},{\left[Ni{F}_6\right]}^{2-}$

To compare the magnetic moment, we need to be able to find out the number of unpaired electrons on the central atom.
1. ${\left[Co{\left(H_{2}O\right)}_6\right]}^{3+},{\left[Ni{F}_6\right]}^{2-}$
$C{o}^{3+}:\left[Ar\right]3d^{6}4s^0$
$H_{2}O$ behaves as a strong field ligand which means that it will force pairing. Hence, there will be 0 unpaired electrons.
$N{i}^{4+}:\left[Ar\right]3d^{6}4s^0$
$F^{-}$ behaves as a strong field ligand which will force pairing, hence there will be 0 unpaired electrons.

2. ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+},{\left[Fe{\left(H_{2}O\right)}_6\right]}^{3+}$
$C{r}^{3+}:\left[Ar\right]3d^{3}4s^0$
$H_{2}O$ behaves as a weak field ligand which means that it will not force pairing. Hence, there will be 3 unpaired electrons.
$F{e}^{3+}:\left[Ar\right]3d^{5}4s^0$
$H_{2}O$ behaves as a weak field ligand which means that it will not force pairing. Hence, there will be 5 unpaired electrons.

3. ${\left[Mn{\left(N{H}_3\right)}_6\right]}^{2+},{\left[Fe{\left(CN\right)}_6\right]}^{3-}$
$M{n}^{2+}:\left[Ar\right]3d^{5}4s^0$
$N{H}_3$ behaves as a weak field ligand which means that it will not force pairing. Hence, there will be 5 unpaired electrons.
$F{e}^{3+}:\left[Ar\right]3d^{5}4s^0$
$C{N}^{-}$ behaves as a strong field ligand which means that it will force pairing. Hence, there will be 1 unpaired electron.

4. ${\left[CoC{l}_4\right]}^{2-},{\left[NiC{l}_4\right]}^{2-}$
$C{o}^{2+}:\left[Ar\right]3d^{7}4s^0$
$C{l}^{-}$ behaves as a weak field ligand which means that it will not force pairing. Hence, there will be 3 unpaired electrons.
$N{i}^{2+}:\left[Ar\right]3d^{8}4s^0$
$C{l}^{-}$ behaves as a weak field ligand which means that it will not force pairing. Hence, there will be 2 unpaired electrons.

Hence, the required pair is ${\left[Co{\left(H_{2}O\right)}_6\right]}^{3+},{\left[Ni{F}_6\right]}^{2-}$