In the complex $\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, 6 \mathrm{NH}_{3}$ ligands are monodentate as well as neutral, so the oxidation state corresponding to the Fe atom can be calculated as follows:

$x+6(0)=+2 \\ x=+2$

The atomic number relative to Fe is 26 and its electronic configuration is $[\mathrm{Ar}] 3 \mathrm{d}^{6} 4 \mathrm{s}^{2}$ .

The complex constitutes $\mathrm{NH}_{3}$ ligand due to which no pairing will occur. The electronic configuration of $\mathrm{Fe}^{+2}$ will be $[\mathrm{Ar}] 3 \mathrm{d}^{6}$ . The 6 $\mathrm{NH}_{3}$ ligands will occupy the empty s, p and d orbitals giving rise to an outer orbital complex and $\mathrm{sp}^{3} \mathrm{d}^{2}$ hybridisation as follows:

The given complex is $\left[\mathrm{MnF}_{6}\right]^{3-}$ . It constitutes $\mathrm{F}^{-}$ ligand which is monodentate and its charge is . The oxidation state corresponding to the central Mn atom can be calculated as follows:

$x+6(-1)=-3 \\ x-6=-3 \\ x=+3$

The atomic number relative to Mn is 25 and its electronic configuration is $[\mathrm{Ar}] 3 \mathrm{d}^{5} 4 \mathrm{s}^{2}$ . The complex constitutes $\mathrm{F}^{-}$ ligands due to which no pairing will occur. The electronic configuration of $\mathrm{Mn}^{+3}$ will be $[\mathrm{Ar}] 3 \mathrm{d}^{4}$ . The 6 $\mathrm{F}^{-}$ ligands will occupy the empty s, p and d orbitals giving rise to an outer orbital complex and $\mathrm{sp}^{3} \mathrm{d}^{2}$ hybridisation as follows:

In complex $\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}, 6 \mathrm{H}_{2} \mathrm{O}$ ligands are monodentate as well as neutral so the oxidation state corresponding to Zn atom can be calculated as follows:

$x+4(0)=+3 \\ x=+3$

The atomic number relative to V is 23 and its electronic configuration is $[\mathrm{Ar}] 3 \mathrm{d}^{3} 4 \mathrm{s}^{2}$ . The complex constitutes $\mathrm{H}_{2} \mathrm{O}$ ligand due to which no pairing will occur. The electronic configuration of $\mathrm{V}^{+3}$ will be $[\mathrm{Ar}] 3 \mathrm{d}^{2}$ . The 6 $\mathrm{H}_{2} \mathrm{O}$ ligands will occupy the empty d, s as well as p orbitals giving rise to an inner orbital complex and $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation as follows:

In the complex $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}, 6 \mathrm{CN}^{-}$ ligands are monodentate and have charge. The oxidation state corresponding to the Fe atom can be calculated as follows:

$x+6(-1)=-4 \\ x-6=-4 \\ x=+2$

The atomic number relative to Fe is 26 and its electronic configuration is $[\mathrm{Ar}] 3 \mathrm{d}^{6} 4 \mathrm{s}^{2}$ . The complex constitutes $\mathrm{CN}^{-}$ ligand due to which pairing will occur. The electronic configuration of $\mathrm{Fe}^{+2}$ will be $[\mathrm{Ar}] 3 \mathrm{d}^{6}$ which will be rearranged. So, the 6 $\mathrm{CN}^{-}$ ligands will occupy the empty d, s as well as p orbitals giving rise to an inner orbital complex and $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation as follows:

So, complexes $\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ and $\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ are inner and outer orbital complexes,respectively.

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Introduction To Compounds

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