Read the passage carefully and answer the Questions.
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
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Inner orbital complex, outer orbital complex
In valence bond theory an octahedral complex uses six hybrid orbitals of the metal ion. If two inner 3d orbitals are free (after any pairing the ligand causes), the hybridisation is \( \mathrm{d^{2}sp^{3}} \) and the complex is an inner orbital (low spin) complex. If two 3d orbitals cannot be emptied, the metal uses two outer 4d orbitals: \( \mathrm{sp^{3}d^{2}} \), an outer orbital (high spin) complex.
\( \ce{[Co(NH3)6]^3+} \): \( \ce{NH3} \) is neutral, so cobalt is +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): \( \mathrm{3d^{6}} \). Ammonia makes the six 3d electrons pair up in three orbitals, leaving two 3d orbitals empty. The hybridisation is \( \mathrm{d^{2}sp^{3}} \): an inner orbital complex, diamagnetic.
\( \ce{[Ni(NH3)6]^2+} \): nickel is +2. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \). Eight electrons need at least four of the five 3d orbitals even when paired, so two 3d orbitals can never be emptied. The 4s, three 4p and two 4d orbitals are used: \( \mathrm{sp^{3}d^{2}} \), an outer orbital complex with two unpaired electrons.
So the two complexes are an inner orbital complex and an outer orbital complex, respectively.
NCERT (2026-27 edition):
Valence Bond Theory – “… In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex. The paramagnetic octahedral complex, [CoF6]3– uses outer orbital (4d) in hybridisation (sp3d2). It is thus called outer orbital or high spin or spin free complex.”