Inner orbital complex among the following is:

$\begin{array}{cc}\left(A\right)& {\left[\mathrm{Co}\right({\mathrm{NH}}_3\left)6\right]}^{3+}\\ \left(B\right)& {\left[\mathrm{COF}6\right]}^{3-}\\ \left(D\right)& {\left[\mathrm{MnCl}6\right]}^{3-}\\ \left(E\right)& {\left[\mathrm{FeF}6\right]}^{3-}\end{array}$

Choose the correct answer from the options given below:

The correct option is (A) and (C) only.

1. Understanding Inner and Outer Orbital Complexes:
According to Crystal Field Theory and Valence Bond Theory, octahedral complexes can be classified based on the d-orbitals used during hybridization:
• Inner Orbital Complexes: These complexes use inner _(n-1)d orbitals (specifically 3d orbitals for first-row transition metals) for hybridization, resulting in $d^{2}s{p}^3$ hybridization. This usually occurs in the presence of strong-field ligands that cause pairing of d-electrons.
• Outer Orbital Complexes: These complexes use outer _nd orbitals (specifically 4d orbitals) for hybridization, resulting in $s{p}^3{d}^2$ hybridization. This occurs with weak-field ligands that cannot pair the d-electrons.

Note on Option (C): Although complex (C) is missing from the list of formulas in the question description, in standard chemistry curriculum questions of this type, (C) represents the hexacyanomanganate(III) ion:
${\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

2. Step-by-Step Analysis of the Complexes:

(A) ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$:
• Cobalt is in the +3 oxidation state.
• The electronic configuration of ${\mathrm{Co}}^{3+}$ is $\left[\mathrm{Ar}\right]3d^6$.
• Ammonia (${\mathrm{NH}}_3$) acts as a strong-field ligand in this system.
• Under its influence, the 6 d-electrons pair up, occupying the lower energy $t_{2g}$ orbitals ($t_{2g}^6{e}_g^0$ configuration).
• This leaves two 3d orbitals vacant, which participate in $d^{2}s{p}^3$ hybridization.
• Thus, it is an inner orbital complex.

(B) ${\left[{\mathrm{CoF}}_6\right]}^{3-}$:
• Cobalt is in the +3 oxidation state ($3d^6$).
• Fluoride ($F^{-}$) is a weak-field ligand and cannot cause electron pairing.
• It undergoes $s{p}^3{d}^2$ hybridization using the outer 4d orbitals.
• Thus, it is an outer orbital complex.

(C) ${\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$:
• Manganese is in the +3 oxidation state.
• The electronic configuration of ${\mathrm{Mn}}^{3+}$ is $\left[\mathrm{Ar}\right]3d^4$.
• Cyanide (${\mathrm{CN}}^{-}$) is a strong-field ligand.
• It causes the d-electrons to pair up to form a low-spin configuration ($t_{2g}^4{e}_g^0$), vacating two 3d orbitals.
• These vacant inner orbitals are used for $d^{2}s{p}^3$ hybridization.
• Thus, it is an inner orbital complex.

(D) ${\left[{\mathrm{MnCl}}_6\right]}^{3-}$:
• Manganese is in the +3 oxidation state ($3d^4$).
• Chloride (${\mathrm{Cl}}^{-}$) is a weak-field ligand and does not force pairing.
• It undergoes $s{p}^3{d}^2$ hybridization (outer orbital complex).

(E) ${\left[{\mathrm{FeF}}_6\right]}^{3-}$:
• Iron is in the +3 oxidation state ($3d^5$).
• Fluoride ($F^{-}$) is a weak-field ligand, resulting in a high-spin $s{p}^3{d}^2$ hybrid configuration.
• Thus, it is an outer orbital complex.

Therefore, only (A) and (C) are inner orbital complexes.