Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molec

NCERT Exemplar · Q22

Q. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature?

(i) 2-Bromobutane

(ii) 1-Bromobutane

(iii) 2-Bromopropane

(iv) 2-Bromopropan-2-ol

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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:

C=C

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 ( $\pi$ ) bond that locks the molecule flat. Rotation would break the $\pi$ 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.

Watch out

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.

Tip

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 ( $CH_3CH=CHCH_3$ )

Carbon 1 of the double bond: attached to $CH_3$ and $H$ → different ✓
Carbon 2 of the double bond: attached to $CH_3$ and $H$ → different ✓

Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).

cis-But-2-ene, with both methyl groups on the same side of the C=C double bond

trans-But-2-ene, with the two methyl groups on opposite sides of the C=C double bond

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 ( $Cl_2C=CH_2$ )

Carbon 1: attached to $Cl$ and $Cl$ → identical ✗ …

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:

\text{C} = \text{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 ( $\text{ClHC=CHCl}$ ): Each carbon has H and Cl (different) → geometrical isomers exist.
1,1-dichloroethene ( $\text{Cl}_2\text{C=CH}_2$ ): 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). …

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A molecule is chiral if it has a carbon atom bonded to four different groups (a chiral centre). Among the given options, only 2-Bromobutane has such a carbon, making it the chiral molecule.

Why chirality matters — and how to spot it

Chirality is a property of molecular handedness: a chiral molecule and its mirror image cannot be superimposed, like your left and right hands. For most organic molecules at the JEE/NEET level, chirality arises from a stereogenic centre — typically a carbon atom with four different substituents. If any two groups on that carbon are identical, the molecule is achiral (it has a plane of symmetry).

So the task is simple: check each molecule for a carbon with four distinct attachments.

Step-by-step analysis

1. 2-Bromobutane

Structure: $\mathrm{CH_3-CHBr-CH_2-CH_3}$

Number the carbons:

C1: $\mathrm{CH_3-}$ (three H's, one C — not a chiral centre)
C2: $\mathrm{-CHBr-}$ $- CHBr -$ — this carbon is bonded to:
- a hydrogen (H)
- a bromine (Br)
- a methyl group ( $\mathrm{CH_3-}$ )
- an ethyl group ( $\mathrm{-CH_2CH_3}$ )

All four groups are different. Therefore, C2 is a chiral centre. The molecule exists as a pair of enantiomers.

Tip

A quick check: if the carbon is attached to four different atoms or groups (count the atoms directly attached, then look at the next sphere if needed), it's chiral. Here, H, Br, CH₃, and CH₂CH₃ are all distinct.

2. 1-Bromobutane

Structure: $\mathrm{CH_2Br-CH_2-CH_2-CH_3}$

C1: $\mathrm{-CH_2Br}$ — two hydrogens, one bromine, one carbon. Two H's are identical → not chiral.
C2, C3, C4: each has at least two identical substituents (e.g., two H's on a $\mathrm{CH_2}$ group). No chiral centre.

The molecule is achiral. …

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This is a classic trap in stereochemistry. Let's break down the common mistakes students make when tackling this exact problem, and how to avoid each.

✗ Mistake 1: Confusing "chiral" with "having a chiral centre"

Many students think: If a molecule has a chiral carbon, it must be chiral.

But that's not always true — a molecule can have chiral centres and still be achiral if it has a plane of symmetry (meso compound).

✓ How to avoid:

Always check for internal symmetry (plane of symmetry) before concluding chirality.
A chiral centre is a necessary but not sufficient condition for chirality.

✗ Mistake 2: Forgetting to check for symmetry in the whole molecule

Students often look only at one carbon and ignore the rest of the molecule.

For example, in 2-Bromopropan-2-ol (iv), the central carbon has four different groups? Let's check:

Carbon: attached to –Br, –OH, –CH₃, and –CH₃ → Two identical methyl groups → not a chiral centre.

✓ How to avoid:

Draw the full structure.
Check every substituent on the carbon in question — if any two are identical, it's not a chiral centre.

✗ Mistake 3: Assuming all halogenated alkanes are chiral

Just because a molecule has a bromine atom doesn't make it chiral.

Example: 1-Bromobutane (ii) has Br at the end — the carbon with Br is attached to two H atoms → achiral.

✓ How to avoid:

A carbon must have four different groups to be a chiral centre.
Count groups carefully: –H counts as a group!

✗ Mistake 4: Misidentifying the chiral centre in 2-Bromobutane

2-Bromobutane (i) has the structure:

CH₃–CHBr–CH₂–CH₃

The carbon with Br is attached to:

–H
–Br
–CH₃
–CH₂CH₃

All four are different → chiral centre exists.

No plane of symmetry → molecule is chiral.

✓ How to avoid:

Write the full condensed formula.
List the four groups explicitly.
Check for symmetry in the whole molecule.

✗ Mistake 5: Overlooking that 2-Bromopropane is symmetric

2-Bromopropane (iii):

CH₃–CHBr–CH₃

The central carbon is attached to:

–H
–Br
–CH₃
–CH₃

Two identical methyl groups → not a chiral centre. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. Draw the structure of geometrical isomers of [Co(NH3)4Cl2].$
›Reveal solutionSolution
[Co(NH3)4Cl2]+ is an octahedral complex of the type [MA4B2], which shows cis-trans geometrical isomerism depending on the relative positions of the two identical Cl ligands.

The complex [Co(NH3)4Cl2]+ has an octahedral geometry with 4 NH3 and 2 Cl- ligands around the central Co(III) ion. For an [MA4B2] type octahedral complex, two arrangements of the two B (Cl) ligands are possible:

cis-isomer: the two Cl- ligands occupy adjacent positions on the octahedron, with a Cl-Co-Cl bond angle of 90 degrees. (Structure: picture an octahedron with NH3 on four positions and the two Cl ligands on two adjacent corners.) …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which complexes do not show geometrical isomerism?$
(a) Square planar complexes
(b) Tetrahedral complexes
(c) Octahedral complexes
(d) All of the above
›Reveal solutionSolution
Geometrical (cis/trans, fac/mer) isomerism requires ligand positions that are not all equivalent/adjacent; a tetrahedral geometry has no such distinction, so it alone among these never shows geometrical isomerism.

(a) Square planar complexes (e.g. $[Pt(NH_3)_2Cl_2]$ , type $MA_2B_2$ ) do show cis–trans geometrical isomerism, since two positions can be adjacent (cis, $90^\circ$ ) or opposite (trans, $180^\circ$ ).

(c) Octahedral complexes (types $MA_4B_2$ , $MA_3B_3$ , etc.) do show both cis–trans and facial–meridional (fac/mer) geometrical isomerism, since some positions are adjacent and some are directly opposite. …
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CBSE 2025Set ANNUAL1 mark
$[Fe(NH_3)_2(CN)_4]^-$
›Reveal solutionSolution

This octahedral $MA_2B_4$ complex can arrange its two identical $NH_3$ ligands either adjacent to each other (cis) or directly opposite each other (trans), giving two geometrical isomers.

Identifying the isomerism

$[Fe(NH_3)_2(CN)_4]^-$ is an octahedral complex of the general type $[MA_2B_4]$ , where $A = NH_3$ (2 ligands) and $B = CN^-$ (4 ligands). This type of complex shows cis–trans (geometrical) isomerism depending on the relative positions of the two $A$ ( $NH_3$ ) ligands.

cis-isomer: Picture an octahedron with the six positions labelled $+x,-x,+y,-y,+z,-z$ . In the cis isomer, the two $NH_3$ ligands occupy adjacent positions, i.e. at $90^\circ$ to each other (e.g. one $NH_3$ at $+z$ and the other at $+x$ ), with the four $CN^-$ ligands occupying the remaining four positions ( $-z, -x, +y, -y$ ).

trans-isomer: In the trans isomer, the two $NH_3$ ligands occupy diametrically opposite positions, i.e. at $180^\circ$ to each other (e.g. one at $+z$ and the other at $-z$ ), with all four $CN^-$ ligands occupying the equatorial plane ( $+x, -x, +y, -y$ ).

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which kinds of isomerism are exhibited by octahedral Co(NH3)4Br2Cl ?$
(a) Geometrical and ionization
(b) Geometrical and Optical
(c) Optical and ionization
(d) Geometrical only
›Reveal solutionSolution

Co(NH3)4Br2Cl is written as [Co(NH3)4Br2]Cl; being an octahedral MA4B2-type complex it shows cis-trans (geometrical) isomerism, and because Cl and Br can swap places inside/outside the coordination sphere it also shows ionization isomerism.

Formula analysis: cobalt is in the +3 oxidation state; 4 NH3 (neutral) and 2 Br- occupy the coordination sphere (charge = 3 - 2 = +1), balanced by one Cl- as the counter ion outside the sphere: [Co(NH3)4Br2]+ Cl-.

Geometrical isomerism: this is an octahedral complex of type MA4B2 (4 identical NH3 and 2 identical Br in the sphere). The two Br ligands can be mutually cis (adjacent, 90 degrees apart) or trans (opposite, 180 degrees apart), giving cis- and trans-tetraamminedibromidocobalt(III) chloride. (Note: MA4B2 does not show optical isomerism, because both cis and trans forms possess a plane of symmetry.)

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Assertion [A] : Complexes of MX6 and MX5L type [X and L are unidentate] do not show geometrical isomerism. Reason [R] : Geometrical isomerism is not shown by the complexes of coordination number 6.$
(a) Both [A] and [R] are true and [R] is the correct explanation of [A].
(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].
(c) [A] is true, but [R] is false.
(d) [A] is false, but [R] is true.
›Reveal solutionSolution

$MX_6$ and $MX_5L$ complexes genuinely show no geometrical isomerism, but that is NOT because coordination number 6 in general excludes geometrical isomerism — many other CN-6 complexes (e.g. $MX_4L_2$ , $MX_3L_3$ ) do show it.

[A] For an octahedral complex $MX_6$ (all six ligands identical) there is only one possible spatial arrangement, and for $MX_5L$ (five identical + one different) the single different ligand can occupy any of the six equivalent octahedral positions — again only one distinct structure results. So neither shows geometrical isomerism. [A] is TRUE.

…

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CBSE 2024Set ANNUAL1 markMCQ
$[Co(NH_3)_4Cl_2]^+$
(a) linkage isomerism
(b) geometrical isomerism
(c) coordination isomerism
(d) ionization isomerism
›Reveal solutionSolution
$[Co(NH_3)_4Cl_2]^+$ has two possible spatial arrangements of the two $Cl^-$ ligands on the octahedron (cis and trans) — geometrical isomerism — giving violet (cis) and green (trans) forms.

The complex $[Co(NH_3)_4Cl_2]^+$ is octahedral with formula type $[MA_4B_2]$ (4 $NH_3$ + 2 $Cl$ around $Co^{3+}$ ). Two distinct, non-interconvertible spatial arrangements are possible:

cis- $[Co(NH_3)_4Cl_2]^+$ : the two $Cl$ ligands occupy adjacent (90°) positions — this isomer is violet.

trans- $[Co(NH_3)_4Cl_2]^+$ : the two $Cl$ ligands occupy opposite (180°) positions — this isomer is green.

Both isomers have the same molecular formula, the same donor atoms, and the same metal oxidation state ( $+3$ ) — only the spatial arrangement of ligands differs. Because the ligand geometry around the metal differs, the crystal-field splitting and hence the d–d transition energies (and so the colour absorbed/observed) differ between the two forms. This is the defining signature of geometrical (cis–trans) isomerism, not a difference in connectivity or ionisation.

…
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CBSE 2023Set ANNUAL1 mark
$[Co(NO_2)_3(NH_3)_3]$
›Reveal solutionSolution

$[Co(NO_2)_3(NH_3)_3]$ is an octahedral complex of the type $MA_3B_3$ , which exhibits two geometrical isomers — facial (fac) and meridional (mer) — depending on how the two sets of three identical ligands are arranged relative to each other.

In an octahedral complex $MA_3B_3$ (here $M=Co^{3+}$ , $A = NO_2^-$ , $B=NH_3$ ), simple cis–trans naming does not apply since there are three ligands of each type; instead the isomers are called facial (fac) and meridional (mer):

fac- $[Co(NO_2)_3(NH_3)_3]$ : Picture an octahedron with vertices labelled 1–6 (1,2,3 forming the top triangular face; 4,5,6 the bottom face). All three $NO_2^-$ ligands occupy one triangular face (positions 1, 2, 3 — mutually cis, each at $90^\circ$ to the other two), while all three $NH_3$ ligands occupy the opposite triangular face (positions 4, 5, 6). The three like ligands thus form a triangular "face" of the octahedron on each side.

…

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CBSE 2020Set 56/2/11 mark
$[Co(NH_3)_5NO_2]Cl_2$
›Reveal solutionSolution
The complex $[Co(NH_3)_5NO_2]Cl_2$ exhibits linkage isomerism because the $NO_2^-$ ligand can coordinate through either the nitrogen atom ( $-NO_2$ , nitro) or the oxygen atom ( $-ONO$ , nitrito), giving two distinct isomers.

Why This Question Tests a Key Concept

This problem isn't just about memorising a name — it's about recognising that a ligand can bind in more than one way. The $NO_2^-$ ion is an ambidentate ligand: it has two different donor atoms (N and O) that can form a coordinate bond with the central metal ion. That single fact is the entire foundation of linkage isomerism.

Important

Linkage isomerism arises only when an ambidentate ligand coordinates through different atoms. The complex must have the same molecular formula but differ in which atom of the ligand is bonded to the metal.

Step-by-Step Reasoning

Identify the coordination sphere and counter ions

The formula is $[Co(NH_3)_5NO_2]Cl_2$ . The square brackets enclose the coordination sphere: $Co^{3+}$ (cobalt in +3 oxidation state) surrounded by five $NH_3$ ligands and one $NO_2^-$ ligand. The two $Cl^-$ ions are outside the brackets — they are counter ions, not directly bonded to cobalt.

Recognise the ambidentate nature of $NO_2^-$

The nitrite ion can bind through:

Nitrogen atom: forming a nitro complex, $[Co(NH_3)_5(NO_2)]^{2+}$

Oxygen atom: forming a nitrito complex, $[Co(NH_3)_5(ONO)]^{2+}$

Both have the same overall formula $[Co(NH_3)_5NO_2]Cl_2$ , but the connectivity differs.

Check for other isomerism types

Geometrical isomerism requires different spatial arrangements of ligands (e.g., cis/trans in square planar or octahedral complexes). Here, all five $NH_3$ are identical, and the sixth position is occupied by $NO_2^-$ — there is no possibility of different geometric arrangements.

Optical isomerism requires chirality (non-superimposable mirror images). This complex has no chiral centre or plane of asymmetry. …
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CBSE 2018Set ANNUAL1 mark
$Q. Explain with an example the ionisation isomerism in complex compounds.$
›Reveal solutionSolution
Ionisation isomers have identical formulae but produce different ions in solution by interchanging a ligand inside the coordination sphere with the counter-ion; e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br.

Ionisation isomerism occurs when the counter-ion (the ion outside the coordination sphere) can itself act as a ligand and thus exchange places with a ligand inside the coordination sphere. The two isomers have the same overall formula but ionise to give different ions in solution.

Example:

[Co(NH3)5Br]SO4 -> [Co(NH3)5Br]2+ + SO4^2- (gives sulphate ion; gives white ppt with BaCl2). …
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CBSE 2018Set 56/11 mark
$[Cu(NH_3)_4][PtCl_4]$
›Reveal solutionSolution

Interchanging ligands between the two complex ions gives the coordination isomer $[Pt(NH_3)_4][CuCl_4]$ .

Concept. Coordination isomerism (a CBSE Class-12 coordination-compounds topic) occurs in salts where both the cation and the anion are complex ions; the ligands can be distributed differently between the two metal centres.

…

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CBSE 2017Set ANNUAL1 markMCQ
$Q. Which complex exhibit geometrical isomerism?$
(a) [MnBr4]2+
(b) [Pt(NH3)3Cl]+
(c) [PtCl2(P(C2H5)3)2]
(d) [Fe(H2O)5NO]2+
›Reveal solutionSolution
Geometrical (cis-trans) isomerism requires at least two different pairs of ligands arranged around a square planar or octahedral centre in more than one distinguishable way; only the MA2B2 square planar complex among the options qualifies.

Checking each option:

[MnBr4]²⁻: tetrahedral geometry (Mn²⁺, d⁵, weak field with 4 identical Br⁻ ligands). Tetrahedral complexes of the type MA4 do not show geometrical isomerism, since all four positions are equivalent.

[Pt(NH3)3Cl]⁺: square planar, type MA3B (3 identical NH3 + 1 Cl). With only one different ligand, there is only one possible spatial arrangement — no cis-trans isomerism possible. …
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Q.Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature?

Concept understanding — Geometrical Isomerism Conditions

Geometrical Isomerism: The Intuition

The Precise Conditions

Condition 1: Restricted Rotation

Condition 2: Two Different Substituents on Each End

How to Check: A Step-by-Step Method

Examples to Cement the Idea

Example 1: But-2-ene ( $CH_3CH=CHCH_3$ )

Example 2: 1,2-Dichloroethene ( $ClCH=CHCl$ )

Example 3: 1,1-Dichloroethene ( $Cl_2C=CH_2$ )

Why this formula?

Geometrical Isomerism: Why the Conditions Hold

1. The Core Requirement: Restricted Rotation

2. Condition 1: Two Different Groups on Each Carbon (for C=C)

Why?

3. Condition 2: For Rings — Similar Logic

Why?

4. The E-Z Notation (Why It’s Needed)

Why chirality matters — and how to spot it

Step-by-step analysis

1. 2-Bromobutane

2. 1-Bromobutane

Method: Chirality Detection via Asymmetric Carbon (Stereocenter) Analysis

Concept First — Why This Works

Steps

Applying to the Options

(i) 2-Bromobutane

(ii) 1-Bromobutane

✗ Mistake 1: Confusing "chiral" with "having a chiral centre"

✓ How to avoid:

✗ Mistake 2: Forgetting to check for symmetry in the whole molecule

✓ How to avoid:

✗ Mistake 3: Assuming all halogenated alkanes are chiral

✓ How to avoid:

✗ Mistake 4: Misidentifying the chiral centre in 2-Bromobutane

✓ How to avoid:

✗ Mistake 5: Overlooking that 2-Bromopropane is symmetric

Why This Question Tests a Key Concept

Step-by-Step Reasoning

Q.Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature?

Why chirality matters — and how to spot it

Step-by-step analysis

1. 2-Bromobutane

2. 1-Bromobutane

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