Q.Explain $[Co(NH_3)_6]^{3+}$ is an inner orbital complex whereas $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

The key idea is Valence Bond Theory (VBT): whether a complex is inner (low-spin, $d^2sp^3$ hybridisation) or outer (high-spin, $sp^3d^2$ hybridisation) depends on the metal ion's $d$-electron count and the ligand's field strength.

Reasoning:

1. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ has a $3d^6$ configuration. $NH_3$ is a strong field ligand, causing large crystal field splitting. The six $d$-electrons pair up in the three $t_{2g}$ orbitals ($t_{2g}^6 e_g^0$), leaving two $d$-orbitals empty. This allows $d^2sp^3$ hybridisation (inner orbital complex). …

The Core Idea: Valence Bond Theory and Hybridisation

Valence Bond Theory classifies complexes as inner orbital (or low-spin) and outer orbital (or high-spin) based on whether the metal uses its inner $(n-1)d$ orbitals or outer $nd$ orbitals for bonding. The deciding factor is the crystal field splitting energy ($\Delta_o$) relative to the pairing energy ($P$). When $\Delta_o > P$, electrons pair up in the inner d-orbitals, freeing an inner d-orbital for $d^2sp^3$ hybridisation — this is an inner orbital complex. When $\Delta_o < P$, electrons remain unpaired in the outer d-orbitals, forcing the use of $sp^3d^2$ hybridisation — this is an outer orbital complex.

The ligand here is ammonia ($NH_3$), a moderately strong field ligand. But the metal ion's charge and size also affect $\Delta_o$. Let's see how this plays out for Co³⁺ and Ni²⁺.

Step-by-Step Analysis

1. Determine the oxidation state and d-electron count

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

• Cobalt is in +3 oxidation state. Atomic number of Co = 27.
• Co³⁺: $[Ar] 3d^6$ (remove 4s² and one 3d electron).
• So Co³⁺ has a d⁶ configuration.

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

• Nickel is in +2 oxidation state. Atomic number of Ni = 28.
• Ni²⁺: $[Ar] 3d^8$ (remove 4s²).
• So Ni²⁺ has a d⁸ configuration.

2. Consider the ligand field and possible hybridisations

Ammonia pairs the d⁶ electrons of the highly charged Co³⁺ ion (large $\Delta_o$). For Ni²⁺ (d⁸), however, no ligand field can produce an inner orbital complex: even complete pairing of the eight 3d electrons frees only ONE 3d orbital, while $d^2sp^3$ hybridisation needs TWO.

For Co³⁺ (d⁶):

• In an octahedral field, the 3d orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy).
• With strong field NH₃, $\Delta_o$ is large. All six electrons pair up in the three $t_{2g}$ orbitals: $t_{2g}^6 e_g^0$.
• This leaves two empty 3d orbitals (the $e_g$ set). Both of these hybridise with one 4s and three 4p orbitals to form six $d^2sp^3$ hybrid orbitals.
• Since the bonding uses inner (n-1)d orbitals, it is an inner orbital complex. All electrons are paired → diamagnetic.

For Ni²⁺ (d⁸):

• In an octahedral field, d⁸ always has two unpaired electrons in the $e_g$ set regardless of field strength (because pairing would require promoting an electron to a higher energy level, which is unfavourable).
• The configuration is $t_{2g}^6 e_g^2$ — two unpaired electrons.
• All five 3d orbitals are occupied (three $t_{2g}$ full, two $e_g$ half-filled). No empty 3d orbital is available for hybridisation.
• Therefore, the metal must use its outer 4d orbitals (specifically 4d, 4s, and 4p) to form six $sp^3d^2$ hybrid orbitals.
• Since bonding uses outer (n)d orbitals, it is an outer orbital complex. Two unpaired electrons → paramagnetic. …

Method: Valence Bond Theory (VBT) Analysis of Coordination Complexes

This method uses hybridisation and magnetic behaviour to classify complexes as inner or outer orbital.

Step 1 — Determine the oxidation state and electronic configuration of the central metal ion

• For $[Co(NH_3)_6]^{3+}$

  Co atomic number = 27

  Co in +3 state: $Co^{3+}$ = $[Ar]\,3d^6$

• For $[Ni(NH_3)_6]^{2+}$

  Ni atomic number = 28

  Ni in +2 state: $Ni^{2+}$ = $[Ar]\,3d^8$

Step 2 — Identify the ligand field strength

• $NH_3$ is a strong field ligand (causes pairing of electrons in the $d$-orbitals).

Step 3 — Decide pairing and hybridisation

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

• $3d^6$ with strong field $NH_3$ → all 6 electrons pair up in the three $t_{2g}$ orbitals.
• This leaves two empty $3d$ orbitals available.
• Hybridisation: $d^2sp^3$ (inner orbital hybridisation).
• Result: Diamagnetic (no unpaired electrons).

✓ Inner orbital complex — uses $(n-1)d$ orbitals for hybridisation.

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

• $3d^8$: all five $3d$ orbitals are occupied (three doubly, two singly) — even pairing the two unpaired electrons could not empty the TWO $3d$ orbitals that $d^2sp^3$ needs.
• So hybridisation uses outer $4d$ orbitals: $sp^3d^2$ (outer orbital hybridisation). …

Here are the most common mistakes students make when explaining why $[Co(NH_3)_6]^{3+}$ is an inner orbital complex and $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex, along with how to avoid each.

Mistake 1: Confusing the oxidation state and electron count

The error: Students often forget to first determine the oxidation state of the central metal ion. They might directly use the ground state configuration of the neutral atom (Co or Ni) instead of the ion.

How to avoid:

• Always start with the oxidation state.

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

  • NH$_3$ is neutral, so the charge comes from Co.
  • Co is in +3 state.
  • Co (atomic number 27): $[Ar] 3d^7 4s^2$
  • Co$^{3+}$: remove 3 electrons → $[Ar] 3d^6$

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

  • Ni (atomic number 28): $[Ar] 3d^8 4s^2$
  • Ni$^{2+}$: remove 2 electrons → $[Ar] 3d^8$

Key takeaway: Write the d-electron count for the ion, not the atom.

Mistake 2: Forgetting that NH$_3$ is a strong field ligand

The error: Some students treat NH$_3$ as a weak field ligand (like H$_2$O) and incorrectly predict high-spin configurations.

How to avoid:

• Memorise the spectrochemical series — NH$_3$ sits on the stronger side of the series (well above H$_2$O, though en, NO$_2^-$, CN$^-$ and CO are stronger still).
• Strong field ligands cause large crystal field splitting ($\Delta_o$), favouring low-spin configurations for d$^4$ to d$^7$ ions.

For Co$^{3+}$ (d$^6$):

• Strong field → low-spin → $t_{2g}^6 e_g^0$
• All 6 electrons paired → no unpaired electrons → inner orbital complex (uses $(n-1)d$ orbitals).

For Ni$^{2+}$ (d$^8$):

• d$^8$ always has 2 unpaired electrons regardless of field strength (Hund’s rule).
• Even with NH$_3$, it remains $t_{2g}^6 e_g^2$ → outer orbital complex (uses $ns$, $np$ and $nd$ orbitals — $sp^3d^2$).

Mistake 3: Mixing up “inner” vs “outer” orbital terminology

The error: Students think “inner orbital” means the complex is low-spin, and “outer orbital” means high-spin — but that’s only true for d$^4$ to d$^7$ ions. For d$^8$, d$^9$, d$^{10}$, the spin state is fixed.

How to avoid:

• Inner orbital complex: Uses $(n-1)d$ orbitals for hybridisation (e.g., $d^2sp^3$).
• Outer orbital complex: Uses $ns$, $np$, and $nd$ orbitals (e.g., $sp^3d^2$).
• Rule of thumb:
  • If the complex has no unpaired electrons (or fewer than maximum), it’s likely inner orbital.
  • If it has the maximum possible unpaired electrons, it’s outer orbital.

For Co$^{3+}$:

• d$^6$, low-spin → 0 unpaired → $d^2sp^3$ hybridisation → inner orbital.

For Ni$^{2+}$:

• d$^8$ → 2 unpaired electrons (always) → $sp^3d^2$ hybridisation → outer orbital.

Mistake 4: Not explaining the hybridisation clearly

The error: Students state the hybridisation without showing how the orbitals are reorganised.

How to avoid:

• Show the orbital diagram step-by-step.

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

• Co$^{3+}$: $3d^6$
• Under strong field, electrons pair in $t_{2g}$ → two $3d$ orbitals become empty. …