Meaning of outer orbital complex

In octahedral complexes, an outer orbital complex uses the orbitals: $sp^3d^2$ involving the outer $d$-orbitals of the metal.

An inner orbital complex uses: $d^2sp^3$ involving the inner $(n-1)d$-orbitals.

Check each complex by oxidation state and $d$-electron count

Option A: $[Fe(CN)_6]^{4-}$

Let oxidation state of Fe be $x$: $x + 6(-1) = -4 \,\Rightarrow\, x=+2$ So, $Fe^{2+}$ has configuration: $3d^6$ Since $CN^-$ is a strong field ligand, pairing occurs.

Thus it forms a low-spin octahedral complex using inner $3d$ orbitals: $d^2sp^3$ Hence, this is an inner orbital complex.

Option B: $[Ni(NH_3)_6]^{2+}$

Oxidation state of Ni: $x + 6(0) = +2 \Rightarrow x=+2$ So, $Ni^{2+}$ is: $3d^8$ For an octahedral complex, to form an inner orbital complex, two vacant inner $3d$ orbitals are needed. But in $3d^8$, this is not possible.

Therefore, it uses outer $4d$ orbitals: $sp^3d^2$ So this is an outer orbital complex.

Option C: $[Co(NH_3)_6]^{3+}$

Oxidation state of Co: $x=+3$ So, $Co^{3+}$ is: $3d^6$ In $Co^{3+}$, even $NH_3$ generally causes pairing because of high oxidation state of cobalt.

Hence it forms low-spin complex with inner orbitals: $d^2sp^3$ So this is an inner orbital complex.

Option D: $[Mn(CN)_6]^{4-}$

Oxidation state of Mn: $x + 6(-1) = -4 \Rightarrow x=+2$ So, $Mn^{2+}$ is: $3d^5$ Since $CN^-$ is a strong field ligand, pairing occurs, giving a low-spin complex.

Therefore it uses inner orbitals: $d^2sp^3$ So this is an inner orbital complex.

Conclusion

Among the given options, only $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

Comparison with stored answer

Stored correct answer: B

Derived answer: B

Hence, the derived answer agrees with the stored answer.