Q.Which of the following complexes show linkage isomerism?

The key idea is linkage isomerism, which occurs when an ambidentate ligand can coordinate through two different donor atoms.

Reasoning:

1. Ambidentate ligands have more than one possible donor atom. Common examples are $NO_2^-$ (N or O), $SCN^-$ (S or N), and $CN^-$ (C or N).
2. Check each complex:
   • (i) $[Co(NH_3)_5(NO_2)]^{2+}$: $NO_2^-$ is ambidentate (nitro vs nitrito) → shows linkage isomerism.
   • (ii) $[Co(H_2O)_5CO]^{3+}$: $CO$ is not ambidentate here (only C donates) → no linkage isomerism. …

Linkage isomerism is a type of structural isomerism where the same ligand can coordinate to the metal ion through two different atoms. This happens only with ambidentate ligands — ligands that have more than one potential donor site. The classic examples are:

• The nitrite ion ($NO_2^-$): can bind through nitrogen (nitro, $-NO_2$) or through oxygen (nitrito, $-ONO$).
• The thiocyanate ion ($SCN^-$): can bind through sulfur (thiocyanato, $-SCN$) or through nitrogen (isothiocyanato, $-NCS$).

So the key question for each complex is: Does it contain an ambidentate ligand? If yes, linkage isomerism is possible. If the ligand is monodentate (only one donor atom) or chelating in a fixed way, it is not.

Let’s examine each option.

1. Option (i): $[Co(NH_3)_5(NO_2)]^{2+}$

   The ligand here is $NO_2^-$. It is ambidentate — it can coordinate via N (forming the nitro isomer) or via O (forming the nitrito isomer). Both isomers are known and stable. So this complex does show linkage isomerism.

2. Option (ii): $[Co(H_2O)_5CO]^{3+}$

   The ligand here is $CO$ (carbon monoxide). CO is not ambidentate — it almost always binds through the carbon atom (carbonyl). While CO can sometimes bind through oxygen in very rare bridging modes, that is not relevant for simple mononuclear complexes like this. In standard coordination chemistry, CO is a monodentate ligand with only one effective donor atom (iii). So no linkage isomerism is possible here.

3. Option (iii): $[Cr(NH_3)_5SCN]^{2+}$

   The ligand is $SCN^-$ (thiocyanate). This is a classic ambidentate ligand — it can bind through sulfur (thiocyanato) or through nitrogen (isothiocyanato). Both bonding modes are well-documented. So this complex does show linkage isomerism.

4. Option (iv): $[Fe(en)_2Cl_2]^{+}$ …

Method: Ambidentate Ligand Identification

Concept: Linkage isomerism occurs when a ligand can coordinate to the metal through two different donor atoms. Such ligands are called ambidentate ligands.

Steps:

1. Identify the ligand in question — look for a ligand that has more than one possible donor atom.

2. Check if the ligand is ambidentate — common examples:

   • $NO_2^-$ (nitro vs nitrito): can bind through N or O
   • $SCN^-$ (thiocyanato vs isothiocyanato): can bind through S or N
   • $CN^-$ (cyano vs isocyano): can bind through C or N
   • $CO$ (carbonyl): binds only through C — not ambidentate

3. Apply to each option:

   • (i) $[Co(NH_3)_5(NO_2)]^{2+}$ — $NO_2^-$ is ambidentate → shows linkage isomerism …

Common Mistakes in Linkage Isomerism Questions

Mistake 1: Confusing Ambidentate Ligands with Any Polyatomic Ligand

The error: Students often think any ligand with multiple atoms can show linkage isomerism. They pick (ii) $[Co(H_2O)_5CO]^{3+}$ because CO has two atoms.

Why it's wrong: Linkage isomerism requires an ambidentate ligand — one that has two different donor atoms capable of bonding through either one. CO is not ambidentate: it always bonds through carbon (carbonyl). Water ($H_2O$) also bonds only through oxygen.

How to avoid: Memorise the common ambidentate ligands:

• $NO_2^-$ (nitro vs nitrito) — bonds through N or O
• $SCN^-$ (thiocyanato vs isothiocyanato) — bonds through S or N
• $CN^-$ (cyano vs isocyano) — bonds through C or N

Mistake 2: Forgetting That Both Donor Atoms Must Be Available for Bonding

The error: Students see $NO_2^-$ in (i) and $SCN^-$ in (iii) and correctly identify them as ambidentate. But they sometimes also pick (iv) $[Fe(en)_2Cl_2]^{+}$ because "en" has two N atoms.

Why it's wrong: Ethylenediamine (en) is bidentate — it uses both N atoms simultaneously to form a chelate ring. It does not have a choice between two different donor atoms; both N atoms are identical. Linkage isomerism requires the same ligand to attach through different atoms in different isomers.

How to avoid: Distinguish clearly:

• Ambidentate ligand = one ligand, two different possible donor atoms (e.g., N vs O in $NO_2^-$)
• Bidentate ligand = one ligand, two identical donor atoms that both bind at once (e.g., en, oxalate)

Mistake 3: Ignoring the Charge or Oxidation State When Checking Feasibility

The error: Students see $SCN^-$ in (iii) and immediately mark it correct without checking if the metal centre can accept bonding from both S and N.

Why it's wrong: For $SCN^-$, S-bonding is favoured by soft metals (like Pt, Pd), while N-bonding is favoured by hard metals (like Cr, Fe). In (iii), $Cr^{3+}$ is a hard acid, so SCN typically bonds through N. However, linkage isomerism is still possible — it just means one isomer is more stable. The question asks "show linkage isomerism", meaning can they exist as linkage isomers, not whether they are equally stable.

How to avoid: For exam purposes, if the ligand is ambidentate and the complex is stable, linkage isomerism is possible. Don't overthink hard/soft unless the question specifically asks about stability.

Mistake 4: Misidentifying the Ligand in (ii) …