Q.Specify the oxidation numbers of the metals in the following coordination entities:

Concept: Werner Coordination Theory — the oxidation number of the central metal is the charge left after removing all ligands (with their known charges) from the complex ion.

Step 1: For each complex, set the sum of (metal oxidation state) + (ligand charges) = overall charge on the complex ion.

Step 2: Use standard ligand charges: $H_2O, NH_3, en$ (ethylenediamine) are neutral; $CN^-$, $Br^-$, $Cl^-$ are $-1$ each.

Step 3: Solve for the metal’s oxidation number.

1. $[Co(H_2O)(CN)(en)_2]^{2+}$: $x + 0 + (-1) + 0 = +2 \Rightarrow x = +3$
2. $[CoBr_2(en)_2]^{+}$: $x + 2(-1) + 0 = +1 \Rightarrow x = +3$ …

Werner’s theory is the foundation here. The key insight: a coordination compound has a central metal ion with a specific oxidation state, surrounded by ligands that are neutral or carry a fixed charge. The overall charge of the complex ion is the sum of the metal’s charge and the charges of all ligands. So to find the metal’s oxidation number, you simply set up an algebraic equation.

Let’s go through each one step by step.

(i) $[Co(H_2O)(CN)(en)_2]^{2+}$

1. Identify ligand charges.

   • $H_2O$ (water) is a neutral ligand — charge = 0.
   • $CN^-$ (cyanide) carries a charge of –1.
   • $en$ (ethylenediamine, $NH_2CH_2CH_2NH_2$) is a neutral bidentate ligand — charge = 0. There are two of them.

2. Set up the equation.

   Let the oxidation number of Co be $x$. The complex ion has an overall charge of $+2$.

3. Solve.

(ii) $[CoBr_2(en)_2]^{+}$

1. Ligand charges.

   • $Br^-$ (bromide) is –1. There are two of them.
   • $en$ is neutral (0). Two of them.

2. Equation.

   Let Co oxidation number be $x$.

3. Solve.

(iii) $[PtCl_4]^{2-}$

1. Ligand charge.

   • $Cl^-$ is –1. Four chlorides.

2. Equation.

   Let Pt oxidation number be $x$.

3. Solve.

(iv) $K_3[Fe(CN)_6]$

1. Handle the counterion first.

   The compound is $K_3[Fe(CN)_6]$. Potassium ($K^+$) is always +1. Three potassium ions give a total positive charge of $+3$. The whole compound is neutral, so the complex ion $[Fe(CN)_6]^{3-}$ must have a charge of –3.

2. Ligand charge. …

Method: Charge Balance Method

This method uses the principle that the sum of all charges in a coordination entity equals the overall charge on the complex ion or compound.

Steps:

1. Identify the overall charge on the complex (given as superscript or from the compound).
2. Assign known charges to ligands (neutral ligands = 0, anionic ligands = their charge).
3. Let the oxidation number of the metal be $x$.
4. Set up equation: $x + \text{(sum of ligand charges)} = \text{overall charge}$
5. Solve for $x$.

(i) $[Co(H_2O)(CN)(en)_2]^{2+}$

• Ligands:
  • $H_2O$ = neutral → charge = 0
  • $CN^-$ = anionic → charge = $-1$
  • $en$ (ethylenediamine) = neutral → charge = 0
• Overall charge = $+2$
• Equation: $x + (0) + (-1) + 2(0) = +2$ $x - 1 = +2$ $x = +3$

Oxidation number of Co = $+3$

(ii) $[CoBr_2(en)_2]^{+}$

• Ligands:
  • $Br^-$ = anionic → charge = $-1$ each, total = $-2$
  • $en$ = neutral → charge = 0
• Overall charge = $+1$
• Equation: $x + (-2) + 2(0) = +1$ $x - 2 = +1$ $x = +3$

Oxidation number of Co = $+3$

(iii) $[PtCl_4]^{2-}$

• Ligands:
  • $Cl^-$ = anionic → charge = $-1$ each, total = $-4$
• Overall charge = $-2$
• Equation: $x + (-4) = -2$ $x = +2$

Oxidation number of Pt = $+2$

(iv) $K_3[Fe(CN)_6]$

• This is a neutral compound.
• $K^+$ = charge $+1$ each, three $K^+$ = total $+3$
• Complex ion $[Fe(CN)_6]^{3-}$ must balance the $+3$ from $K^+$
• Ligands: $CN^-$ = $-1$ each, total = $-6$ …

Here are the common mistakes students make when assigning oxidation numbers in coordination compounds, specifically for the Werner Coordination Theory context, and how to avoid each.

Mistake 1: Forgetting the Overall Charge of the Complex Ion

The Error: Students often ignore the superscript charge on the square bracket (e.g., $[Co(H_2O)(CN)(en)_2]^{2+}$) and set the sum of oxidation numbers equal to zero instead of $+2$.

How to Avoid:

• Always write the charge balance equation first. Let $x$ = oxidation number of the metal. Sum of (oxidation numbers of all ligands) + $x$ = charge on the complex ion.
• Example (i): $[Co(H_2O)(CN)(en)_2]^{2+}$ Ligands: $H_2O$ (0), $CN^-$ (-1), $en$ (0). Equation: $x + 0 + (-1) + 0 = +2 \implies x = +3$.

Mistake 2: Assigning Wrong Charges to Neutral Ligands

The Error: Treating neutral molecules like $H_2O$, $NH_3$, or $en$ (ethylenediamine) as if they carry a charge.

How to Avoid:

• Memorize the common neutral ligands: $H_2O$, $NH_3$, $CO$, $en$, $NO$ (nitrosyl, but careful — it can be neutral or charged depending on bonding).
• Rule: If the ligand is a molecule (not an ion), its oxidation number is 0 unless specified otherwise.

Mistake 3: Forgetting the Charge of Anionic Ligands

The Error: Using the wrong charge for ligands like $CN^-$, $Cl^-$, $Br^-$, or $OH^-$.

How to Avoid:

• Make a quick reference list:
  • $CN^-$: $-1$
  • $Cl^-$, $Br^-$, $I^-$: $-1$
  • $OH^-$: $-1$
  • $O^{2-}$: $-2$ (rare in simple complexes, but appears in oxo complexes)
• Example (ii): $[CoBr_2(en)_2]^{+}$ $Br^-$ = $-1$ each, $en$ = $0$. Equation: $x + 2(-1) + 0 = +1 \implies x = +3$.

Mistake 4: Ignoring the Counter-Ion in Neutral Complexes

The Error: When the complex is part of a salt (e.g., $K_3[Fe(CN)_6]$), students forget to account for the charge of the counter-ions to find the charge on the complex ion.

How to Avoid:

• Step 1: Find the charge on the complex ion using the counter-ions. $K_3[Fe(CN)_6]$: $K^+$ has charge $+1$, so $3 \times (+1) = +3$. The salt is neutral, so the complex ion must be $[Fe(CN)_6]^{3-}$.
• Step 2: Now solve for $Fe$: $x + 6(-1) = -3 \implies x = +3$.

Mistake 5: Misidentifying the Charge of Ambidentate Ligands

The Error: Ligands like $CN^-$ can bind through C or N, but the charge remains $-1$ regardless. Students sometimes assign a different charge based on bonding mode.

How to Avoid:

• The charge of a ligand is fixed by its formula, not by how it binds. $CN^-$ is always $-1$, $NO_2^-$ is always $-1$, $SCN^-$ is always $-1$.

Mistake 6: Forgetting That the Sum Must Be an Integer …