Q.Draw structures of geometrical isomers of [Fe(NH3)2(CN)4]−
Concept understanding — Geometrical Isomerism Conditions
Geometrical Isomerism: The Intuition
Imagine you have two friends standing on opposite sides of a door. If the door is open, they can walk around and swap places easily — there's no real difference between who is on the left and who is on the right. But if the door is locked shut, they are stuck. One is permanently on the left side, the other on the right. That locked door creates two distinct arrangements: Friend A on the left, Friend B on the right versus Friend A on the right, Friend B on the left.
That locked door is the key idea behind geometrical isomerism.
In chemistry, molecules are three-dimensional. Atoms connected by a single bond can rotate freely — like an open door. But a double bond (or a ring structure) locks the atoms in place. If you have two different groups attached to each carbon of a double bond, you get two distinct spatial arrangements that cannot interconvert without breaking the bond. These are geometrical isomers (also called cis-trans or E-Z isomers).
The Precise Conditions
For a molecule to show geometrical isomerism, it must satisfy two conditions simultaneously:
Condition 1: There must be a restricted rotation around a bond — typically a carbon-carbon double bond (C=C) or a ring structure.
Condition 2: Each of the two atoms (or groups) involved in that restricted rotation must have two different substituents attached to it.
Let's unpack each.
Condition 1: Restricted Rotation
A single bond (C−C) allows free rotation — the atoms spin around the bond axis like a wheel. So no geometrical isomers exist there. A double bond (C=C) has a pi (π) bond that locks the molecule flat. Rotation would break the π bond, which requires a lot of energy (about 250–270 kJ/mol). At room temperature, this rotation simply does not happen.
Rings (like cyclopropane, cyclobutane, etc.) also restrict rotation because the ring is a closed loop — atoms cannot rotate past each other without breaking the ring.
Condition 2: Two Different Substituents on Each End
This is the "different groups" rule. Look at each carbon of the double bond (or each ring carbon involved). If both carbons have two different groups attached, geometrical isomers exist. If even one carbon has two identical groups, there is only one possible arrangement.
A common mistake: students check only one carbon. Both carbons must have two different substituents. If one carbon has two identical groups (like two hydrogens), the molecule is identical in both arrangements — no isomerism.
How to Check: A Step-by-Step Method
Take any molecule with a double bond. Follow these steps:
- Identify the double bond (or ring). Mark the two carbon atoms involved.
- List the two groups attached to the first carbon. Are they different from each other? If yes, proceed. If no → no geometrical isomerism.
- List the two groups attached to the second carbon. Are they different from each other? If yes → geometrical isomerism exists. If no → no geometrical isomerism.
If the two groups on a carbon are identical, the molecule is symmetric about that carbon. Flipping the other side gives the same molecule — no isomers.
Examples to Cement the Idea
Example 1: But-2-ene (CH3CH=CHCH3)
- Carbon 1 of the double bond: attached to CH3 and H → different ✓
- Carbon 2 of the double bond: attached to CH3 and H → different ✓
Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).
Example 2: 1,2-Dichloroethene (ClCH=CHCl)
- Carbon 1: attached to Cl and H → different ✓
- Carbon 2: attached to Cl and H → different ✓
Result: cis and trans isomers exist.
Example 3: 1,1-Dichloroethene (Cl2C=CH2)
- Carbon 1: attached to Cl and Cl → identical ✗
Result: No geometrical isomerism. Both doubly-bonded carbons carry two identical substituents (H and H on one, Cl and Cl on the other in 1,1-dichloroethene) — geometrical isomerism requires each double-bond carbon to carry two DIFFERENT groups, and neither one here does. The molecule is the same no matter how you look at it.
Example 4: Cyclopropane-1,2-dicarboxylic acid (a ring)
- The ring restricts rotation.
- Carbon 1: attached to COOH and H → different ✓
- Carbon 2: attached to COOH and H → different ✓
Result: cis and trans isomers exist (both carboxylic acid groups on same side vs opposite sides of the ring).
The Naming: cis-trans vs E-Z
For simple cases where the two identical groups are on the same side (like both methyl groups), we use cis (same side) and trans (opposite sides). But when all four substituents are different, cis-trans fails — you need the E-Z system (based on priority rules from Cahn-Ingold-Prelog). That's a separate topic, but the condition for geometrical isomerism remains the same.
Geometrical isomerism requires:
- Restricted rotation (double bond or ring)
- Two different substituents on each of the two atoms involved
If both conditions hold, the molecule exists as two distinct spatial isomers that differ in physical properties (melting point, boiling point, polarity) and often in biological activity.
The conditions required for geometrical isomerism are covered in both the NCERT/CBSE Class 11 Organic Chemistry and Class 12 Coordination Compounds chapters, and ‘conditions for geometrical isomerism’ is a commonly searched important-question topic for board exams, JEE Main and NEET. Applying these two conditions correctly to both alkenes and coordination complexes is a skill tested across multiple competitive-exam chemistry sections.
Why this formula?
Geometrical Isomerism: Why the Conditions Hold
Geometrical isomerism (also called cis-trans or E-Z isomerism) arises when atoms or groups are arranged differently in space around a rigid part of a molecule — typically a double bond or a ring. The key is that rotation is restricted, so the spatial positions become fixed and distinct.
Let’s break down why the conditions are what they are.
1. The Core Requirement: Restricted Rotation
For two molecules to be geometrical isomers, they must have the same connectivity but different spatial arrangement due to a barrier to rotation.
- Double bonds (C=C): The π-bond locks the two carbons in place — rotation requires breaking the π-bond (energy ~250 kJ/mol), so it doesn’t happen at room temperature.
- Rings (e.g., cycloalkanes): The ring structure physically prevents free rotation about C–C single bonds within the ring.
Why this matters: Without restricted rotation, the molecule would freely interconvert between arrangements — no distinct isomers exist.
2. Condition 1: Two Different Groups on Each Carbon (for C=C)
Consider a general alkene:
C=C
Each carbon must have two different substituents (not counting the other carbon of the double bond).
Why?
- If one carbon has two identical groups (e.g., both H), then swapping the groups on that carbon produces the same molecule — no isomerism.
Example:
- 1,2-dichloroethene (ClHC=CHCl): Each carbon has H and Cl (different) → geometrical isomers exist.
- 1,1-dichloroethene (Cl2C=CH2): One carbon has two Cl (identical) → no geometrical isomers.
Formal condition:
For a C=C bond, geometrical isomerism is possible iff each doubly bonded carbon bears two different substituents.
3. Condition 2: For Rings — Similar Logic
In a ring (e.g., cyclopropane, cyclohexane), the ring itself restricts rotation. Here, geometrical isomerism occurs when two substituents on different ring carbons can be on the same side (cis) or opposite sides (trans).
Why?
- The ring is a closed loop — you cannot rotate one carbon relative to another without breaking bonds.
- If the two substituents are on different carbons, their relative orientation (same side / opposite sides) is fixed.
Condition:
- The ring must have at least two substituents (could be same or different) on different carbons.
- If both substituents are on the same carbon, swapping them doesn’t change the molecule (no isomerism).
Example:
- 1,2-dimethylcyclopropane: Two methyl groups on adjacent carbons → cis and trans isomers exist.
- 1,1-dimethylcyclopropane: Both methyls on same carbon → no geometrical isomerism.
4. The E-Z Notation (Why It’s Needed)
When the four substituents on a C=C are all different, cis-trans naming fails. The Cahn-Ingold-Prelog priority rules assign E (opposite sides) or Z (same side).
Why this works:
- Priority is based on atomic number (higher = higher priority).
- Compare the two groups on each carbon — if the higher priority groups are on the same side → Z (German zusammen = together); if opposite → E (entgegen = opposite).
Key insight: The condition remains the same — each carbon must have two different groups — but the naming becomes unambiguous.
5. Summary of Conditions (Exam-Ready)
| Structure | Condition | Why? |
|---|---|---|
| C=C double bond | Each carbon must have two different substituents | Otherwise, swapping groups gives same molecule |
| Ring (e.g., cycloalkane) | At least two substituents on different carbons | Ring prevents rotation; same-side/opposite-side are distinct |
| General | Restricted rotation (double bond or ring) | Without it, free rotation makes isomers identical |
6. Common Exam Pitfall
Don’t confuse:
- Geometrical isomerism ≠ optical isomerism (chirality).
- A molecule can have geometrical isomers even if it has no chiral centre.
- For rings: cis and trans are geometrical isomers only if the substituents are on different carbons.
Final takeaway: The conditions exist because restricted rotation creates a fixed spatial arrangement, and different substituents ensure that swapping positions actually changes the molecule. Without either, you get the same compound — not an isomer.
The key idea is that geometrical isomerism in coordination compounds arises when ligands of two different types can occupy positions that are either adjacent (cis) or opposite (trans) to each other.
Reasoning:
- The complex [Fe(NH3)2(CN)4]− has a coordination number of 6, so it is octahedral.
- There are two identical ammine (NH3) ligands and four identical cyano (CN−) ligands.
- The two NH3 ligands can be placed either at 90° to each other (adjacent, cis) or at 180° to each other (opposite, trans). The four CN− ligands occupy the remaining positions.
The two geometrical isomers are:
- cis isomer: The two NH3 groups are adjacent.
- trans isomer: The two NH3 groups are opposite each other.
The geometrical isomers are the cis and trans forms of [Fe(NH3)2(CN)4]−.
[Fe(NH3)2(CN)4]− is an octahedral MA2B4 complex, so it shows two geometrical isomers: a cis form (the two NH3 groups adjacent, 90∘ apart) and a trans form (the two NH3 groups opposite, 180∘ apart).
Type of complex. Six ligands around Fe make it octahedral; with two NH3 (A) and four CN− (B) it is an MA2B4 system, which gives exactly two geometrical isomers (no optical isomerism, since both forms have a plane of symmetry).
cis isomer — the two NH3 groups occupy adjacent corners (90∘):
NH3
|
H3N --- Fe --- CN
|
CN
(The remaining two CN− lie in front of and behind the plane. The two NH3 groups share an edge of the octahedron.)
trans isomer — the two NH3 groups occupy opposite (axial) corners (180∘), with the four CN− in the equatorial plane:
NH3
|
NC --- Fe --- CN
|
NH3
(The remaining two CN− lie in front of and behind the plane; all four CN− are coplanar.)
[Fe(NH3)2(CN)4]− has two geometrical isomers: the cis isomer (the two NH3 ligands adjacent, 90∘ apart) and the trans isomer (the two NH3 ligands opposite, 180∘ apart), as drawn above.
Method: Ligand-Position Enumeration (cis/trans) for Geometrical Isomerism in Octahedral Complexes
Why this method?
Geometrical (cis-trans) isomerism in octahedral complexes arises when two identical ligands can be placed adjacent (cis) or opposite (trans) to each other. The test is positional: two arrangements are distinct isomers when no rotation of the whole complex converts one into the other — symmetry planes only matter later, for deciding optical activity.
Step-by-step solution for [Fe(NH3)2(CN)4]−
Step 1: Identify the coordination number and geometry
- Fe is the central metal.
- Ligands: 2 NH3 (neutral) and 4 CN− (anionic).
- Total coordination number = 6 → octahedral geometry.
Step 2: Determine the formula type
This is a [MA2B4] type complex (where A = NH3, B = CN−).
For such complexes, geometrical isomerism is possible.
Step 3: Draw the two possible arrangements
- Cis isomer: Both NH3 ligands are adjacent (at 90° to each other).
- Trans isomer: Both NH3 ligands are opposite (at 180° to each other).
Step 4: Check that the two arrangements are genuinely distinct
| Isomer | Structure (simplified) | Distinct? |
|---|---|---|
| Cis | NH3 groups on adjacent corners (90°) | ✓ No rotation can move them to opposite corners |
| Trans | NH3 groups on opposite corners (180°) | ✓ No rotation can move them to adjacent corners |
The two arrangements cannot be interconverted by any rotation of the octahedron, so they are distinct geometrical isomers. (Note: both forms retain a plane of symmetry, so neither is chiral — there is no optical isomerism here.)
Final Answer
Two geometrical isomers exist:
- Cis isomer — both NH3 groups on adjacent positions.
- Trans isomer — both NH3 groups on opposite positions.
Key point: No rotation converts the cis arrangement into the trans one — they are distinct, stable isomers. Both retain a plane of symmetry, so neither is optically active.
Quick Exam Tip
For [M(A)2(B)4] type complexes:
- Cis → the 2 identical A ligands at 90°.
- Trans → the 2 identical A ligands at 180°.
- Total geometrical isomers = 2 (cis and trans) — and since both forms have a plane of symmetry, there is no optical isomerism.
Here are the most common mistakes students make when tackling geometrical isomerism for coordination compounds like [Fe(NH3)2(CN)4]−, along with clear strategies to avoid them.
Mistake 1: Forgetting to Determine the Coordination Number and Geometry First
- The Mistake: Students jump straight to drawing isomers without first identifying that the complex is octahedral (coordination number 6). They might incorrectly treat it as square planar or tetrahedral.
- Why it happens: The formula [Fe(NH3)2(CN)4]− has 6 ligands, but the presence of only two types of ligands (NH3 and CN) can trick students into thinking it’s simpler.
- How to avoid: Always start by counting the total number of ligands attached to the metal. Here, 2+4=6 — six ligands mean an octahedral geometry. (Iron here is Fe3+, i.e. d5 — see Mistake 5's charge calculation.) Write this down before drawing anything.
Mistake 2: Confusing Geometrical Isomerism with Optical Isomerism
- The Mistake: Students often try to draw non-superimposable mirror images (optical isomers) for this complex, even though it is not chiral.
- Why it happens: The complex has two identical ligands (NH3) and four identical ligands (CN). In an octahedral [Ma2b4] type complex, the only possible geometrical isomers are cis and trans. There is no chiral center.
- How to avoid: Memorize the condition: For geometrical isomerism in octahedral complexes, you need at least two different types of ligands. For [Ma2b4], only cis and trans exist. Do not draw mirror images unless the complex is [M(AA)3] or [Ma2b2c2] type.
Mistake 3: Drawing Only One Isomer (Usually the Trans)
- The Mistake: Students draw only the trans isomer (with NH3 groups opposite each other) and forget the cis isomer (with NH3 groups adjacent).
- Why it happens: The trans isomer looks more symmetric and is often the first one that comes to mind.
- How to avoid: Systematically place the two identical ligands (NH3) in all possible distinct positions. In an octahedron:
- Trans: NH3 at positions 1 and 6 (opposite ends of an axis).
- Cis: NH3 at positions 1 and 2 (adjacent corners).
- Any other arrangement is equivalent to one of these two by rotation.
Mistake 4: Drawing Incorrect Bond Angles or Ligand Positions
- The Mistake: Students draw the cis isomer with the two NH3 groups at a 90° angle but then place the CN groups incorrectly, making the structure look like a square planar complex or with distorted octahedral geometry.
- Why it happens: Lack of practice in drawing 3D octahedral structures on paper.
- How to avoid: Use the standard wedge-dash notation:
- Draw a square plane (four CN groups in the cis isomer; two CN and two NH3 in the trans isomer).
- Place the remaining two ligands above and below the plane (axial positions).
- For the cis isomer: the two NH3 groups are on adjacent corners of the square plane.
- For the trans isomer: the two NH3 groups are at 180° — e.g. both on the axial positions. (An axial + equatorial pair is only 90° apart — that is cis, not trans.)
Mistake 5: Forgetting the Charge and Oxidation State
- The Mistake: Students draw the complex without considering the charge, leading to incorrect ligand placement or ignoring that CN− is a strong field ligand.
- Why it happens: The focus is only on the shape.
- How to avoid: Always note the overall charge (−1). Here, Fe is in the +3 oxidation state (since NH3 is neutral and CN− is −1 each: x+0+4(−1)=−1⇒x−4=−1⇒x=+3). This affects the electronic configuration and stability of isomers, but for drawing geometrical isomers, it ensures you use the correct number of each ligand.
Correct Structures (for reference)
Trans isomer:
- Two NH3 groups are opposite each other (e.g., axial positions).
- Four CN groups occupy the equatorial square plane.
Cis isomer:
- Two NH3 groups are adjacent (e.g., both in the equatorial plane, next to each other).
- Four CN groups occupy the remaining four positions (two in the plane, two axial).
Key takeaway: For [Fe(NH3)2(CN)4]−, there are exactly two geometrical isomers: cis and trans. No optical isomers exist. Always verify the coordination number and ligand arrangement before drawing.
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.Which of the following complexes do not exhibit geometrical isomerism? I. K[Cr(H2O)2(C2O4)2] II. [Co(en)3]Cl3 III. [Co(NH3)5(NO2)](NO3)2 The correct answer is (A) I, III only (B) II, III only (C) I, II only (D) I, II, III
›Reveal solutionSolution
Geometrical isomerism requires at least two different possible spatial arrangements of ligands. M(AA)2B2 (complex I) has cis/trans forms; M(AA)3 (II) and MA5B (III) each have only one possible arrangement, so they do not show geometrical isomerism.
Concept and Intuition
Whether an octahedral complex shows geometrical (cis-trans) isomerism depends on its ligand-substitution pattern, not just on having a mix of ligands:
- M(AA)2B2 (two bidentate symmetric chelates + two monodentate ligands): the two B ligands can be adjacent (cis) or opposite (trans) — genuine geometrical isomerism exists (and the cis form is additionally chiral).
- M(AA)3 (three identical bidentate chelates, e.g. [Co(en)3]3+): by symmetry there is only one way to arrange three identical chelate rings around the octahedron — no cis/trans distinction is possible. Only optical isomerism (mirror-image Δ and Λ forms) exists.
- MA5B (five identical monodentate ligands + one different one): since all six octahedral positions are equivalent by symmetry when five ligands are identical, placing the lone B ligand at "any" position gives the same single structure — no geometrical isomerism is possible.
Step-by-Step Solution
- I: K[Cr(H2O)2(C2O4)2] = [Cr(C2O4)2(H2O)2]−, an M(AA)2B2 complex → cis and trans forms exist → does show geometrical isomerism.
- II: [Co(en)3]Cl3 = M(AA)3 → only one spatial arrangement (differs only by optical handedness) → does not show geometrical isomerism.
- III: [Co(NH3)5(NO2)](NO3)2 = MA5B → only one possible position for the lone NO2− ligand → does not show geometrical isomerism.
- Complexes NOT showing geometrical isomerism: II and III.
Common Mistakes
- Assuming any complex with more than one type of ligand automatically shows geometrical isomerism — the pattern (MA5B vs MA4B2 vs MA3B3 vs M(AA)3, etc.) determines this, not just ligand diversity.
- Confusing optical isomerism (which M(AA)3 does show) with geometrical isomerism (which it does not).
✓Final answerThe correct option is (B) — II, III only.
ANSWER: B
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.Which of the following exhibit cis-trans isomerism? (I) 2-Methylpent-2-ene (II) Styrene (III) 2-Chlorobut-2-ene (IV) 1-Phenylprop-1-ene The correct answer is (A) I & II only (B) III & IV only (C) I & III only (D) II & IV only
›Reveal solutionSolution
This tests the basic criterion for geometrical (cis–trans) isomerism: each alkene carbon must bear two different substituents. Only 2-chlorobut-2-ene and 1-phenylprop-1-ene qualify.
Concept and Intuition
Geometric (cis-trans) isomerism about a C=C double bond arises only when restricted rotation is combined with each sp2 carbon of the double bond carrying two non-identical groups. If either carbon has two identical substituents, the molecule and its "other geometry" are actually the same compound — no isomerism.
Step-by-Step Solution
- 2-Methylpent-2-ene: CH3−C(CH3)=CH−CH2−CH3. The C2 (left alkene carbon) bears two methyl groups — identical substituents — so no cis-trans isomerism, regardless of what's on C3.
- Styrene: C6H5−CH=CH2. The terminal alkene carbon (=CH2) carries two hydrogens — identical — so no cis-trans isomerism.
- 2-Chlorobut-2-ene: CH3−CCl=CH−CH3. C2 carries CH3 and Cl (different); C3 carries H and CH3 (different). Both alkene carbons have two different groups → geometric isomerism exists (cis and trans forms).
- 1-Phenylprop-1-ene: C6H5−CH=CH−CH3. C1 carries phenyl and H (different); C2 carries H and CH3 (different). Both carbons qualify → geometric isomerism exists.
- So only (III) and (IV) show cis-trans isomerism — option (B).
Common Mistakes
- Overlooking that a trisubstituted alkene carbon with two identical alkyl groups (as in 2-methylpent-2-ene) blocks isomerism even though the other carbon has different groups.
- Forgetting that a terminal vinylic =CH2 group (as in styrene) always kills cis-trans isomerism at that end.
✓Final answerThe correct option is (B) — III & IV only.
ANSWER: B
- AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQQ.Which of the following exhibit only geometrical isomerism? (A) Diaquadioxalatochromate (III) ion (B) Dichloridobis(ethane-1, 2-diamine)platinum (IV) ion (C) Triamminetrinitrito – N cobalt (III) (D) Tris(ethane-1, 2-diamine)cobalt (III) ion
›Reveal solutionSolution
Comparing the isomerism possibilities of each octahedral coordination-compound type (M(AA)2B2, MA3B3, M(AA)3) shows that only the MA3B3-type complex is restricted to geometrical (fac/mer) isomerism alone. The answer is (C).
Concept and Intuition
For octahedral complexes, the type and combination of ligands determines what kinds of isomerism are possible. A complex of type M(AA)2B2 (two bidentate chelating ligands plus two monodentate ligands) can exist as cis and trans geometrical isomers; critically, the cis isomer of this type lacks any plane of symmetry and is chiral (shows optical isomerism), while the trans isomer is not chiral. A complex of type M(AA)3 (three identical bidentate chelating ligands) has only one possible geometric arrangement (no cis/trans distinction exists), so it shows only optical isomerism, always as a pair of non-superimposable mirror-image (Δ/Λ) forms. A complex of type MA3B3 can arrange its ligands as facial (fac, three of one type on one triangular face) or meridional (mer, three of one type in a plane through the metal) — both of these arrangements possess a mirror plane of symmetry and are therefore achiral, so this type shows geometrical isomerism only, with no optical activity.
Step-by-Step Solution
- (A) Diaquadioxalatochromate(III) ion, [Cr(C2O4)2(H2O)2]−: type M(AA)2B2. Its cis isomer is chiral (optically active) and its trans isomer is not — so this complex shows both geometrical and optical isomerism, not geometrical isomerism alone.
- (B) Dichloridobis(ethane-1,2-diamine)platinum(IV) ion, [Pt(en)2Cl2]2+: also type M(AA)2B2, with the same situation as (A) — cis form chiral, trans form not, so both geometrical and optical isomerism are present.
- (C) Triamminetrinitrito–N cobalt(III), [Co(NH3)3(NO2)3]: type MA3B3. This shows fac and mer geometrical isomers. Both the fac isomer (C3v symmetry, has mirror planes) and the mer isomer (Cs symmetry, has a mirror plane) are achiral. So this complex shows only geometrical isomerism, with no optical isomerism at all — this is the answer.
- (D) Tris(ethane-1,2-diamine)cobalt(III) ion, [Co(en)3]3+: type M(AA)3. With three identical bidentate ligands arranged around an octahedron, there is only a single possible geometric arrangement (no cis/trans distinction), but the whole complex lacks a plane of symmetry, so it exists as non-superimposable Δ and Λ enantiomers — this shows only optical isomerism, not geometrical isomerism.
Common Mistakes
- Assuming any chelate complex with two different ligand types automatically shows only geometrical isomerism, without checking whether the cis form is chiral.
- Forgetting that M(AA)3 complexes have no cis/trans forms at all, so they cannot show geometrical isomerism, only optical.
- Not recognising that both fac and mer isomers of an MA3B3 complex are achiral, which is what makes option (C) the one showing geometrical isomerism exclusively.
✓Final answerThe correct option is (C) — Triamminetrinitrito–N cobalt (III).
ANSWER: C
- AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQQ.Which of the following complexes exhibit geometrical isomerism? (only) I) [Co(en)(NH3)2Cl2]Cl II) [Co(NH3)4Cl2]Cl III) [Co(en)3]Cl3 IV) [Co(en)2Cl2]Br (A) I, II & III only (B) II, III & IV only (C) I, II & IV only (D) II & III only
›Reveal solutionSolution
Tests which octahedral complex types show geometrical (cis-trans) isomerism; the answer is (C) I, II & IV only since the tris-chelate [Co(en)3]3+ has no cis/trans forms.
Concept and Intuition
Geometrical isomerism (cis-trans) in octahedral complexes arises when two or more identical ligands (or ligating groups) can occupy either adjacent (cis) or opposite (trans) positions. Complexes of type MA4B2, MA3B3, and mixed-ligand types with two identical monodentate ligands like M(AA)B2C2 or M(AA)2B2 show this. However, a complex where all three ligand positions are filled by the same symmetric bidentate chelate, i.e. M(AA)3, has only ONE possible geometric arrangement (the chelate rings are geometrically forced into one shape) — such complexes show only optical isomerism (as non-superimposable mirror images), never geometrical isomerism.
Step-by-Step Solution
- I) [Co(en)(NH3)2Cl2]+: ligand set is one en (bidentate, counts as occupying 2 cis sites) + 2 NH3 + 2 Cl. The two Cl's can be cis or trans to each other → geometrical isomerism exists.
- II) [Co(NH3)4Cl2]+: type MA4B2 — the two Cl ligands can be cis (adjacent) or trans (opposite) → geometrical isomerism exists.
- III) [Co(en)3]3+: type M(AA)3, three identical symmetric bidentate ligands — geometrically only one arrangement is possible (octahedral tris-chelate); only optical (Δ/Λ) isomers exist, no cis/trans forms → NO geometrical isomerism.
- IV) [Co(en)2Cl2]+: type M(AA)2B2 — the two Cl's can be cis or trans → geometrical isomerism exists (and the cis form is additionally chiral).
- So complexes I, II, and IV show geometrical isomerism; III does not.
Common Mistakes
- Assuming any complex with a bidentate ligand automatically shows geometrical isomerism — it depends on whether a genuine cis/trans choice exists, which fails for the symmetric M(AA)3 case.
✓Final answerThe correct option is (C) — I, II & IV only.
ANSWER: C
- AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQQ.The total number of possible four membered ring cis and trans isomers for the molecular formula C4H6Cl2 is ________ (A) 3 (B) 4 (C) 5 (D) 2
›Reveal solutionSolution
Dichlorocyclobutane (C4H6Cl2) has three substitution patterns (1,1-, 1,2-, 1,3-); only the 1,2- and 1,3- patterns show cis/trans isomerism, each contributing a cis and a trans form, giving 4 cis/trans isomers in total.
Concept and Intuition
Cyclobutane (C4H8) with two hydrogens replaced by chlorine gives dichlorocyclobutane, C4H6Cl2. Because the ring holds the carbon skeleton rigid (no free rotation around the ring bonds the way there is in an open chain), placing two substituents on ring carbons in different relative positions can create genuine, non-interconvertible geometric (cis/trans) isomers — exactly analogous to cis/trans isomerism in cyclic compounds generally. The key is that cis/trans isomerism requires two different substituents on each of two ring carbons that are directly compared (i.e., no ring carbon carrying two identical Cl's), so a substitution pattern with both Cl atoms on the same carbon cannot show cis/trans isomerism at all.
Step-by-Step Solution
- Enumerate the possible relative positions of two Cl atoms on a four-membered ring: 1,1- (geminal, same carbon), 1,2- (adjacent carbons), and 1,3- (opposite/across the ring).
- 1,1-dichlorocyclobutane: both Cl's are on one carbon; that carbon has no distinguishable "up/down" substituent pair to compare across the ring, so no cis/trans isomerism is possible here — it is a single compound.
- 1,2-dichlorocyclobutane: the two Cl's are on adjacent ring carbons, each of which also bears an H; the two Cl's can be on the same face of the ring (cis) or on opposite faces (trans) — 2 distinct isomers.
- 1,3-dichlorocyclobutane: the two Cl's are on carbons across the ring from each other, again each carbon bearing an H; same-face (cis) and opposite-face (trans) arrangements are both possible and are non-superimposable — 2 distinct isomers.
- Total genuine cis/trans (geometrical) isomers = 2 (from 1,2-) + 2 (from 1,3-) = 4. (The 1,1- isomer is a real compound too, but it doesn't participate in cis/trans isomerism, so it isn't counted among the "cis and trans isomers.")
Common Mistakes
- Including the 1,1-dichloro compound in the cis/trans count (it cannot show cis/trans isomerism at all, since both substituents are on one carbon).
- Forgetting the 1,3- (across-the-ring) substitution pattern also shows cis/trans isomerism, not just the 1,2- pattern.
✓Final answerThe correct option is (B) — 4.
ANSWER: B
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