Q.Indicate the complex ion which shows geometrical isomerism.

The key idea is that geometrical isomerism occurs in coordination compounds with a coordination number of 4 (square planar) or 6 (octahedral) when two or more different ligands are arranged in distinct spatial positions (cis/trans or fac/mer).

Reasoning:

1. Option (i): $[Cr(H_2O)_4Cl_2]^{+}$ is octahedral with two identical Cl ligands. These can be adjacent (cis) or opposite (trans), so geometrical isomers exist.
2. Option (ii): $[Pt(NH_3)_3Cl]$ is square planar with three identical NH₃ ligands — only one arrangement possible (no isomerism). …

Geometrical isomerism in coordination compounds is a form of stereoisomerism where the same set of ligands can be arranged differently in space around the central metal ion. The key condition is that the complex must have at least two different types of ligands, and the geometry must allow distinct spatial arrangements — typically seen in square planar complexes of type $[MA_2B_2]$ or $[MA_2BC]$, and octahedral complexes of type $[MA_4B_2]$ or $[MA_3B_3]$.

Let’s examine each option carefully.

1. Option (i): $[Cr(H_2O)_4Cl_2]^{+}$

   This is an octahedral complex with chromium in the +3 oxidation state (since each water is neutral and each chloride is -1, total charge = +3 - 2 = +1). The coordination sphere has four identical water ligands and two identical chloride ligands. In an octahedral geometry, two identical ligands can be placed either adjacent to each other (cis, 90° apart) or opposite each other (trans, 180° apart). These are non-superimposable mirror images? No — cis and trans are diastereomers, not enantiomers, but they are distinct geometrical isomers. So this complex does show geometrical isomerism.

2. Option (ii): $[Pt(NH_3)_3Cl]^{+}$ (the stem's printed formula omits the charge; charge balance requires it — three neutral NH$_3$ plus one Cl$^-$ on Pt$^{2+}$ gives a net +1 complex ion)

   Platinum(II) with a coordination number of 4 is essentially always square planar. For a square-planar $[MA_3B]$ complex, the three identical A ligands and the single B ligand have only one possible spatial arrangement — every position is related to every other by a simple rotation, so there is no cis/trans distinction. No geometrical isomerism.

3. Option (iii): $[Co(NH_3)_6]^{3+}$

   This is an octahedral complex with six identical ammonia ligands. All positions are equivalent — there is only one possible arrangement. No geometrical isomerism possible.

4. Option (iv): $[Co(CN)_5(NC)]^{3-}$ …

Method: Check for Coordination Number and Ligand Arrangement that Allows Cis–Trans Isomerism

Geometrical isomerism in coordination compounds occurs when ligands can occupy different spatial positions around the central metal ion. The key is to look for:

• Coordination number 4 (square planar) or 6 (octahedral)
• At least two different types of ligands (not all identical)
• Possibility of cis (adjacent) and trans (opposite) arrangements

Step-by-step analysis for each option

Option (i): $[Cr(H_2O)_4Cl_2]^{+}$

• Coordination number: 6 (octahedral)
• Ligands: 4 water molecules + 2 chloride ions → two different types
• Possible isomers:
  • Cis: both Cl⁻ adjacent
  • Trans: both Cl⁻ opposite
• ✓ Shows geometrical isomerism

Option (ii): $[Pt(NH_3)_3Cl]$

• Coordination number: 4 (square planar for Pt²⁺)
• Ligands: 3 ammonia + 1 chloride → only one chloride ligand
• No possibility of cis/trans because you need at least two identical ligands to swap positions
• ✗ No geometrical isomerism

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

• Coordination number: 6 (octahedral) …

🧠 Step 1: Understand the Core Concept First

Geometrical isomerism (cis-trans or fac-mer) occurs when:

• The coordination number is 4 (square planar or tetrahedral) or 6 (octahedral).
• There are two or more different ligands arranged around the central metal.
• The spatial arrangement of identical ligands can be different.

Key rule: If all ligands are identical, no geometrical isomerism is possible.

✗ Common Mistake #1: Forgetting that $[Pt(NH_3)_3Cl]$ is neutral and square planar

• What students do: They see 4 ligands and assume tetrahedral → no geometrical isomerism.
• Why it’s wrong: $Pt^{2+}$ is d⁸, almost always square planar, not tetrahedral.
• How to avoid: Memorise:
  • $Pt^{2+}$, $Pd^{2+}$, $Au^{3+}$ → square planar (coordination number 4).
  • Square planar complexes with formula $[MA_3B]$ do show geometrical isomerism (only one isomer exists here, but the possibility is there — actually $[Pt(NH_3)_3Cl]$ has no geometrical isomers because all three NH₃ are identical; but if formula were $[MA_2B_2]$, it would).

Correction for this exam: Option (ii) $[Pt(NH_3)_3Cl]$ has no geometrical isomers because three NH₃ are identical — only one arrangement exists.

✗ Common Mistake #2: Ignoring the charge and coordination number

• What students do: They see $[Cr(H_2O)_4Cl_2]^{+}$ and think “Cr is +3, coordination number 6” but forget to check if it’s octahedral.
• Why it’s wrong: Cr³⁺ is d³, always octahedral. With 4 H₂O and 2 Cl, it’s $[MA_4B_2]$ type.
• How to avoid:
  • Always determine oxidation state and coordination number first.
  • For octahedral $[MA_4B_2]$: cis and trans isomers exist.

✓ Option (i) shows geometrical isomerism.

✗ Common Mistake #3: Assuming $[Co(NH_3)_6]^{3+}$ can show isomerism

• What students do: They see 6 ligands and think “octahedral → maybe isomers”.
• Why it’s wrong: All six ligands are identical (NH₃). No geometrical isomerism possible.
• How to avoid:
  • Geometrical isomerism requires at least two different types of ligands.

✗ Option (iii) has no geometrical isomers.

✗ Common Mistake #4: Misidentifying linkage isomerism as geometrical isomerism

• What students do: They see $[Co(CN)_5(NC)]^{3-}$ and think “CN and NC are different → geometrical isomers possible”. …