Question

Which complex compound possesses \[s{p^3}{d^2}\] hybridisation?
A. \[{\left[ {Fe{{(N{H_3})}_6}} \right]^{3 + }}\]
B. \[{\left[ {Fe{{(CN)}_6}} \right]^{4 - }}\]
C. \[{\left[ {Fe{{(CN)}_6}} \right]^{3 - }}\]
D. \[{\left[ {Fe{{(Cl)}_6}} \right]^{3 - }}\]

Answer 1

Hint: Hybridisation is the process of mixing of orbitals of different shapes and energies. When two atomic orbital combines, it forms a hybrid orbital. When one ‘s’ orbital and ‘3’ 3p orbitals and two ‘d’ orbital, are mixed, then the formed hybridization is\[s{p^3}{d^2}\] and its geometry is octahedral.

Complete step by step answer:
The outermost electronic configuration for iron can be written as
\[Fe = 3{d^6}4{s^2}\].
A.The oxidation state of iron in \[{\left[ {Fe{{(N{H_3})}_6}} \right]^{3 + }} = + 3\]
Thus, the outermost electronic configuration for \[F{e^{3 + }} = 3{d^5}4{s^0}\]
Every six ligands donate its two electrons to the central metal ion, \[F{e^{3 + }}\] And also \[N{H_3}\]is a strong field ligand and it makes the unpaired electron in the central metal ion for pairing. So the hybridization obtained as \[{d^2}s{p^3}\]
B.The oxidation state of iron in \[{\left[ {Fe{{(CN)}_6}} \right]^{4 - }} = + 2\]
Thus, the outermost electronic configuration for \[F{e^{2 + }} = 3{d^6}4{s^0}\]
Every six ligands donate its two electrons to the central metal ion, \[F{e^{2 + }}\] And also \[C{N^ - }\]is a strong field ligand and it makes the unpaired electron in the central metal ion for pairing. So the hybridization obtained as \[{d^2}s{p^3}\]
C.The oxidation state of iron in \[{\left[ {Fe{{(CN)}_6}} \right]^{3 - }} = + 3\]
The outermost electronic configuration for \[F{e^{3 + }}\]=\[3{d^5}4{s^0}\]
Every six ligands donate its two electrons to the central metal ion, \[F{e^{3 + }}\] And also \[C{N^ - }\] is a strong field ligand and it makes the unpaired electron in the central metal ion for pairing. So the hybridization obtained as \[{d^2}s{p^3}\]
D.The oxidation state of iron in \[{\left[ {Fe{{(Cl)}_6}} \right]^{3 - }} = + 3\]
The outermost electronic configuration for \[F{e^{3 + }} = 3{d^5}4{s^0}\]
Every six ligands donate its two electrons to the central metal ion, \[F{e^{3 + }}\] And also \[C{l^ - }\] is a weak field ligand and it does not help in the pairing. So the hybridization obtained as \[s{p^3}{d^2}\].

Thus, the correct answer is option D.

Note: The shapes of hybridization are linear, bent, trigonal planar, trigonal bipyramidal, square planar, tetrahedral, and octahedral respectively. In octahedral hybridization, six ligands are symmetrically occupied around the central metal atom. Both \[s{p^3}{d^2}\]and \[{d^2}s{p^3}\] have the geometry octahedral.