Q.Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

Concept: Crystal Field Splitting – the oxidation state determines the $d^n$ count, which then dictates the electronic configuration based on ligand field strength.

(i) $K_3[Co(C_2O_4)_3]$

• $K^+$ contributes $+3$, oxalate ($C_2O_4^{2-}$) is a bidentate ligand with charge $-2$ each. Let $x$ be Co oxidation state: $3 + x + 3(-2) = 0 \Rightarrow x = +3$.
• $Co^{3+}$ is $d^6$. Oxalate itself is actually a weak/intermediate-field ligand (below $H_2O$ in the series), but $Co^{3+}$'s high +3 charge raises $\Delta_o$ enough that the complex is still low-spin: $t_{2g}^6 e_g^0$.
• Coordination number = 6 (three bidentate oxalates).

(ii) cis-$[CrCl_2(en)_2]Cl$

• Outer $Cl^-$ contributes $-1$, en is neutral, $Cl$ inside is $-1$ each. Let $x$ be Cr: $x + 2(-1) + 0 = +1 \Rightarrow x = +3$.
• $Cr^{3+}$ is $d^3$. $Cl^-$ and en are weak to intermediate field; $d^3$ always gives $t_{2g}^3 e_g^0$ (half-filled $t_{2g}$).
• Coordination number = 6 (two bidentate en + two monodentate Cl).

(iii) $(NH_4)_2[CoF_4]$

• $NH_4^+$ contributes $+2$, $F^-$ is $-1$ each. Let $x$ be Co: $2 + x + 4(-1) = 0 \Rightarrow x = +2$. …

Let’s unpack each complex one by one. The three quantities we need — oxidation state, d-orbital occupation, and coordination number — are linked. Oxidation state comes from the overall charge of the complex and the known charges of ligands. The d-occupation is then just the number of d-electrons the metal has in that oxidation state (for first-row transition metals, the neutral atom’s electron configuration minus the oxidation number gives the d-count). Coordination number is simply the number of ligand donor atoms directly bonded to the metal.

(i) $K_3[Co(C_2O_4)_3]$

1. Oxidation state: The complex ion is $[Co(C_2O_4)_3]^{3-}$ because three $K^+$ ions balance it. Oxalate, $C_2O_4^{2-}$, is a bidentate ligand with a –2 charge. Three oxalates give a total ligand charge of $3 \times (-2) = -6$. Let the Co oxidation state be $x$. Then $x + (-6) = -3$, so $x = +3$. Thus Co is in the +3 oxidation state.

2. d-orbital occupation: Cobalt (atomic number 27) has the ground-state configuration $[Ar]\,3d^7 4s^2$. In the +3 state, we remove three electrons — typically the two 4s electrons and one 3d electron. So Co³⁺ has $7 - 1 = 6$ d-electrons. Hence it is a $d^6$ system.

3. Coordination number: Each oxalate ion binds through two oxygen atoms (bidentate). Three oxalates give $3 \times 2 = 6$ donor atoms. So the coordination number is 6.

(ii) cis-$[CrCl_2(en)_2]Cl$

1. Oxidation state: The complex is a salt: the cation is $[CrCl_2(en)_2]^+$ and the anion is $Cl^-$. Ethylenediamine (en) is a neutral bidentate ligand (charge 0). Each chloride ligand inside the coordination sphere has a –1 charge. Two chlorides give –2. Let Cr oxidation state be $x$. Then $x + (-2) + 0 = +1$, so $x = +3$. Chromium is in the +3 state.

2. d-orbital occupation: Chromium (atomic number 24) has $[Ar]\,3d^5 4s^1$. In the +3 state, we remove three electrons — the 4s electron and two 3d electrons. So Cr³⁺ has $5 - 2 = 3$ d-electrons. It is a $d^3$ system.

3. Coordination number: Two chlorides (monodentate) and two ethylenediamine molecules (each bidentate, donating two N atoms) give $2 + (2 \times 2) = 6$ donor atoms. Coordination number is 6.

(iii) $(NH_4)_2[CoF_4]$

1. Oxidation state: The complex ion is $[CoF_4]^{2-}$ because two $NH_4^+$ ions balance it. Fluoride is a –1 ligand. Four fluorides give –4. Let Co oxidation state be $x$. Then $x + (-4) = -2$, so $x = +2$. Cobalt is in the +2 state. …

Method: Charge Balance + Ligand Field Analysis

This method uses electroneutrality to find the metal oxidation state, then deduces d-electron count and coordination geometry from the ligand types and complex formula.

Steps for any complex

1. Assign known charges

   • Ligands: neutral ($H_2O$, $NH_3$, $en$), anionic ($Cl^-$, $C_2O_4^{2-}$, $F^-$), or cationic (rare).
   • Counter-ions: $K^+$, $(NH_4)^+$, $SO_4^{2-}$, $Cl^-$ outside brackets.

2. Solve for metal oxidation state

   Let $x$ = oxidation state of metal.

   Total charge of complex ion + counter-ions = 0.

3. Determine d-orbital occupation

   • For $3d$ metals: remove electrons equal to oxidation state from neutral atom configuration.
   • Example: $Co$ (atomic no. 27) = $[Ar]\,3d^7 4s^2$; $Co^{3+}$ = $3d^6$.

4. Find coordination number

   Count total donor atoms attached to metal (each ligand denticity matters).

(i) $K_3[Co(C_2O_4)_3]$

• Counter-ion: $3 \times K^+ = +3$
• Complex ion: $[Co(C_2O_4)_3]^{3-}$
• Ligand: oxalate ($C_2O_4^{2-}$) is bidentate, charge $-2$ each.
• Equation: $x + 3(-2) = -3 \implies x - 6 = -3 \implies x = +3$
• Oxidation state: $\mathbf{+3}$
• d-orbital occupation: $Co^{3+}$ = $3d^6$ (low-spin — driven by $Co^{3+}$'s high +3 charge, not by oxalate itself being strong-field; oxalate actually lies below $H_2O$ in the spectrochemical series)
• Coordination number: 3 bidentate ligands × 2 = $\mathbf{6}$

(ii) cis-$[CrCl_2(en)_2]Cl$

• Counter-ion: $Cl^-$ outside bracket = $-1$
• Complex ion: $[CrCl_2(en)_2]^+$
• Ligands: $Cl^-$ (monodentate, $-1$ each), $en$ (neutral bidentate)
• Equation: $x + 2(-1) + 2(0) = +1 \implies x - 2 = +1 \implies x = +3$
• Oxidation state: $\mathbf{+3}$
• d-orbital occupation: $Cr$ (24) = $[Ar]\,3d^5 4s^1$; $Cr^{3+}$ = $3d^3$
• Coordination number: 2 Cl + 2 en (each bidentate = 2×2) = $\mathbf{6}$

(iii) $(NH_4)_2[CoF_4]$

• Counter-ion: $2 \times NH_4^+ = +2$
• Complex ion: $[CoF_4]^{2-}$
• Ligand: $F^-$ (monodentate, $-1$ each)
• Equation: $x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2$ …

Here are the common mistakes students make when solving Crystal Field Splitting problems like this, along with how to avoid each.

Mistake 1: Getting the Oxidation State Wrong (Especially with Ambiguous Ligands)

The Mistake:

Students often misassign the charge of ligands like oxalate ($C_2O_4^{2-}$) or ethylenediamine (en, neutral). For example, in $K_3[Co(C_2O_4)_3]$, they might forget that oxalate is -2 or that $K$ is +1.

How to Avoid:

• Memorize common ligand charges: Oxalate ($C_2O_4^{2-}$) is -2, EDTA is -4, CN$^-$ is -1, NH$_3$ and en are 0.
• Use the "total charge = 0" rule for neutral compounds. For (i): $3(K^+) + [Co + 3(C_2O_4^{2-})] = 0 \implies 3 + x - 6 = 0 \implies x = +3$.
• For complex ions with a counter ion: In (ii) cis-$[CrCl_2(en)_2]Cl$, the outer Cl$^-$ is -1, so the complex cation $[CrCl_2(en)_2]^+$ must be +1. Then: $x + 2(-1) + 2(0) = +1 \implies x = +3$.

Key Result: Co is +3 in (i); Cr is +3 in (ii).

Mistake 2: Confusing d-orbital Occupation with Electronic Configuration

The Mistake:

Students write the full electronic configuration (e.g., $[Ar] 3d^6$) but fail to specify the d-orbital occupation in the context of the complex. The question asks for "d orbital occupation," which means how the electrons are distributed in the $t_{2g}$ and $e_g$ sets after crystal field splitting.

How to Avoid:

• Step 1: Find the oxidation state and the number of d electrons ($d^n$). For (i) Co$^{3+}$: Co atomic number = 27. Co$^{3+}$ = $27 - 3 = 24$ electrons. Configuration: $[Ar] 3d^6$. So $d^6$.
• Step 2: Determine the geometry and metal-charge effect. Oxalate actually sits on the weaker half of the spectrochemical series (below $H_2O$) — the pairing here is NOT because oxalate itself is a strong-field ligand. It is $Co^{3+}$'s high +3 oxidation state (small, highly charged ion) that raises $\Delta_o$ enough to force pairing for almost any ligand except very weak-field ones like $F^-$. So for an octahedral $d^6$ $Co^{3+}$ complex, all 6 electrons pair up in the $t_{2g}$ set.
• Step 3: Write the occupation as $t_{2g}^6 e_g^0$.

Key Result: For (i): $t_{2g}^6 e_g^0$ (low spin). For (iii) $(NH_4)_2[CoF_4]$: Co$^{2+}$ is $d^7$, F$^-$ is weak field, tetrahedral geometry. Tetrahedral splitting is small, so high spin: $e^4 t_2^3$.

Mistake 3: Forgetting to Account for Geometry (Octahedral vs. Tetrahedral)

The Mistake:

Students assume all complexes are octahedral. For example, in (iii) $(NH_4)_2[CoF_4]$, the formula $[CoF_4]^{2-}$ indicates 4 ligands, which is tetrahedral (or square planar, but for Co$^{2+}$ with weak field F$^-$, it's tetrahedral).

How to Avoid:

• Count the coordination number first. Coordination number = number of ligand donor atoms directly bonded to the metal.
• Match geometry to coordination number:
  • 4 → usually tetrahedral (for $d^7$ like Co$^{2+}$) or square planar (for $d^8$ like Ni$^{2+}$, Pt$^{2+}$).
  • 6 → octahedral.
• For tetrahedral complexes: The splitting is inverted and smaller. The $e$ set is lower energy, $t_2$ set is higher.

Key Result: (iii) is tetrahedral, so occupation is $e^4 t_2^3$ (high spin $d^7$).

Mistake 4: Misidentifying the Coordination Number from the Formula

The Mistake:

In (ii) cis-$[CrCl_2(en)_2]Cl$, students might count only the Cl atoms inside the bracket, forgetting that en (ethylenediamine) is a bidentate ligand (donates 2 atoms).

How to Avoid:

• Bidentate ligands count as 2 coordination sites. en = H$_2$N-CH$_2$-CH$_2$-NH$_2$ binds through both N atoms.
• Calculate coordination number: 2 Cl (monodentate) + 2 en (each bidentate) = $2 + 2 \times 2 = 6$.
• For (iv) $[Mn(H_2O)_6]SO_4$: The 6 water molecules are monodentate, so coordination number = 6.

Key Result: (ii) coordination number = 6; (iv) coordination number = 6.

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