Q.Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons?

Concept: Crystal Field Splitting – For outer orbital (high-spin) octahedral complexes, weak-field ligands (like Cl⁻, F⁻, NH₃ for Ni²⁺ here) cause minimal pairing, so electrons fill all five d-orbitals singly before pairing.

Step 1: Determine oxidation states and d-electron counts

• (i) Mn in $[MnCl_6]^{3-}$: Mn is +3 → $d^4$ (Mn: 25, Mn³⁺: 22 electrons, so 4 d-electrons).
• (ii) Fe in $[FeF_6]^{3-}$: Fe is +3 → $d^5$ (Fe: 26, Fe³⁺: 23 electrons, so 5 d-electrons).
• (iii) Co in $[CoF_6]^{3-}$: Co is +3 → $d^6$ (Co: 27, Co³⁺: 24 electrons, so 6 d-electrons).
• (iv) Ni in $[Ni(NH_3)_6]^{2+}$: Ni is +2 → $d^8$ (Ni: 28, Ni²⁺: 26 electrons, so 8 d-electrons).

Step 2: Apply weak-field (high-spin) filling for octahedral

• $d^4$: $t_{2g}^3 e_g^1$ → 4 unpaired electrons. …

1. What does "outer orbital" mean?

In octahedral complexes, ligands split the d-orbitals into two sets: the lower-energy $t_{2g}$ and the higher-energy $e_g$. The size of this splitting ($\Delta_o$) depends on the ligand.

• Strong field ligands (like $NH_3$, $CN^-$) cause a large $\Delta_o$ — electrons pair up in the $t_{2g}$ before occupying $e_g$. This gives low-spin (or inner orbital) complexes.
• Weak field ligands (like $F^-$, $Cl^-$, $H_2O$ for most 3d metals) cause a small $\Delta_o$ — electrons occupy all five d-orbitals singly first (Hund's rule). This gives high-spin (or outer orbital) complexes.

The phrase "outer orbital" specifically means the complex uses the outer $nd$ orbitals (i.e., $3d$ for first-row transition metals) and follows high-spin configuration. So we must first check the ligand: $Cl^-$ and $F^-$ are weak field; $NH_3$ is strong field.

2. Determine the oxidation state and d-electron count for each complex

We need the number of d-electrons on the central metal ion.

3. Apply the outer orbital (high-spin) configuration for each

For weak-field ligands, fill the d-orbitals according to Hund's rule: each orbital gets one electron before pairing.

• $[MnCl_6]^{3-}$ — $d^4$, high-spin:

  $t_{2g}$: ↑ ↑ ↑ (three electrons)

  $e_g$: ↑ (one electron)

  Unpaired electrons = 4

• $[FeF_6]^{3-}$ — $d^5$, high-spin:

  $t_{2g}$: ↑ ↑ ↑

  $e_g$: ↑ ↑

  Unpaired electrons = 5

• $[CoF_6]^{3-}$ — $d^6$, high-spin:

  $t_{2g}$: ↑ ↑ ↑

  $e_g$: ↑ ↑

  Then the sixth electron must pair in $t_{2g}$:

  $t_{2g}$: ↑↓ ↑ ↑

  $e_g$: ↑ ↑

  Unpaired electrons = 4 …

Method: Crystal Field Theory (CFT) — Outer Orbital (High Spin) Complexes

Step 1: Identify the oxidation state and electron configuration of the central metal ion

For each complex, find the metal’s oxidation state and its $d^n$ configuration.

• (i) $[MnCl_6]^{3-}$

  Mn (Z = 25): $[Ar]\,3d^5 4s^2$

  Oxidation state: $+3$ (since Cl⁻ is −1, total ligand charge = −6, complex charge = −3 ⇒ Mn = +3)

  $Mn^{3+}$: $3d^4$ → $d^4$ system

• (ii) $[FeF_6]^{3-}$

  Fe (Z = 26): $[Ar]\,3d^6 4s^2$

  Oxidation state: $+3$ (F⁻ is −1, total = −6, complex = −3 ⇒ Fe = +3)

  $Fe^{3+}$: $3d^5$ → $d^5$ system

• (iii) $[CoF_6]^{3-}$

  Co (Z = 27): $[Ar]\,3d^7 4s^2$

  Oxidation state: $+3$ (F⁻ is −1, total = −6, complex = −3 ⇒ Co = +3)

  $Co^{3+}$: $3d^6$ → $d^6$ system

• (iv) $[Ni(NH_3)_6]^{2+}$

  Ni (Z = 28): $[Ar]\,3d^8 4s^2$

  Oxidation state: $+2$ (NH₃ is neutral, complex charge = +2 ⇒ Ni = +2)

  $Ni^{2+}$: $3d^8$ → $d^8$ system

Step 2: Determine the ligand field strength and spin state

• Cl⁻ and F⁻ are weak field ligands → cause small Δ₀ → high spin (outer orbital) complexes.
• NH₃ is a moderate field ligand — for $Ni^{2+}$ ($d^8$), it always gives low spin (since $d^8$ has no high-spin/low-spin difference in octahedral geometry — it’s always 2 unpaired electrons). But the question specifies outer orbital — so we must check if NH₃ forces pairing. For $d^8$, pairing is complete in both cases, so it’s effectively the same.

Step 3: Fill electrons in the $t_{2g}$ and $e_g$ orbitals (high spin)

For high spin (outer orbital) octahedral complexes:

• $d^4$ (Mn³⁺): …

Common Mistakes in Crystal Field Splitting (Outer Orbital Complexes)

Mistake 1: Confusing "Outer Orbital" with "Inner Orbital"

The Error: Students often forget that outer orbital means high spin (weak field ligand, $sp^3d^2$ hybridization). They incorrectly assume all complexes are low spin.

How to Avoid:

• Outer orbital = weak field ligand = high spin = maximum unpaired electrons
• Inner orbital = strong field ligand = low spin = minimum unpaired electrons
• For $[MnCl_6]^{3-}$, $[FeF_6]^{3-}$, $[CoF_6]^{3-}$ — all have weak field ligands ($Cl^-$, $F^-$), so they are high spin

Mistake 2: Wrong Oxidation State Calculation

The Error: Students misassign oxidation states, leading to wrong $d^n$ configurations.

How to Avoid:

• For $[MnCl_6]^{3-}$: $Mn + 6(-1) = -3 \Rightarrow Mn = +3$ → $d^4$ (Mn atomic no. 25, $Mn^{3+}$ = 22 electrons = $d^4$)
• For $[FeF_6]^{3-}$: $Fe + 6(-1) = -3 \Rightarrow Fe = +3$ → $d^5$
• For $[CoF_6]^{3-}$: $Co + 6(-1) = -3 \Rightarrow Co = +3$ → $d^6$
• For $[Ni(NH_3)_6]^{2+}$: $Ni + 6(0) = +2 \Rightarrow Ni = +2$ → $d^8$

Mistake 3: Forgetting $NH_3$ is a Strong Field Ligand

The Error: Students treat $NH_3$ as weak field, making $[Ni(NH_3)_6]^{2+}$ high spin — but $NH_3$ is strong field for $3d$ metals.

How to Avoid:

• $NH_3$ is a strong field ligand → causes low spin (inner orbital) for $d^6$ and below
• For $d^8$ ($Ni^{2+}$), strong field still gives 2 unpaired electrons (no pairing possible in $t_{2g}^6 e_g^2$)

Mistake 4: Incorrect Electron Filling in High Spin vs Low Spin

The Error: Students fill $d$ orbitals without considering crystal field splitting energy ($\Delta_o$) vs pairing energy ($P$).

How to Avoid:

• High spin (weak field): Fill all $t_{2g}$ singly first, then $e_g$ singly (Hund's rule)
• Low spin (strong field): Pair in $t_{2g}$ before filling $e_g$