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In the formation of coordination complexes, if the inner d orbital (n-1d) is used in hybridisation, the complex, is called an inner orbital or low spin or spin paired complex. And if it uses outer d orbital (nd) in hybridisation (like $sp^3d^2$), it is called outer orbital or high spin or spin free complex. The degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the complex. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. For coordination complexes, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula, $μ = \sqrt{n (n+2)}$ where $n$ is the number of unpaired electrons and $μ$ is the magnetic moment in units of Bohr magneton (BM). The coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine.

$[Co(NH_3)_6]^{3+}$ and $[Ni(NH_3)_6]^{2+}$ respectively are

Inner orbital complex, outer orbital complex

The correct answer is Option (4) → Inner orbital complex, outer orbital complex

These are octahedral complexes, and the type of complex (inner or outer orbital) depends on whether (n−1)d orbitals or nd orbitals are used for hybridization. Inner orbital complexes use d²sp³ hybridization, while outer orbital complexes use sp³d² hybridization.

(1) [Co(NH₃)₆]³⁺ 

Oxidation state of Co: Co + 6(0) = +3

Co³⁺

Electronic configuration: Co = [Ar] 3d⁷4s²

Co³⁺ = 3d⁶

NH₃ is a strong field ligand for Co³⁺, causing pairing of electrons. 

Thus the configuration becomes: t₂g⁶ e_g⁰

Two 3d orbitals become vacant, allowing hybridization: d²sp³

Thus it forms an inner orbital complex.

(2) [Ni(NH₃)₆]²⁺ 

Oxidation state of Ni: Ni + 6(0) = +2

Ni²⁺

Electronic configuration: Ni = [Ar] 3d⁸4s²

Ni²⁺ = 3d⁸

Here pairing does not create two empty 3d orbitals.

Therefore hybridization occurs using 4d orbitals: sp³d²

Thus it forms an outer orbital complex.