Q.Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic.
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 the magnetic property of a complex depends on the number of unpaired electrons, which is determined by the hybridisation and geometry of the central metal ion under Valence Bond Theory.
Step 1: Electronic configuration of Ni²⁺
Ni (Z = 28): [Ar]3d84s2
Ni²⁺: [Ar]3d8 — three 3d orbitals are doubly occupied and two are singly occupied (two unpaired electrons).
Step 2: [Ni(CN)4]2− — square planar, diamagnetic
CN⁻ is a strong field ligand. It forces pairing of the two unpaired 3d electrons, leaving one 3d orbital empty. The hybridisation is dsp2 (one 3d, one 4s, two 4p orbitals). All eight electrons are paired → no unpaired electrons → diamagnetic.
Step 3: [NiCl4]2− — tetrahedral, paramagnetic
Cl⁻ is a weak field ligand. It does not cause pairing. The hybridisation is sp3 (one 4s, three 4p orbitals), using the original 3d⁸ configuration with two unpaired electrons. Hence, the complex is paramagnetic.
[Ni(CN)4]2− is diamagnetic (all electrons paired, dsp2 hybridisation) while [NiCl4]2− is paramagnetic (two unpaired electrons, sp3 hybridisation).
On valence bond theory: Ni2+ is 3d8 — three electron pairs plus two unpaired electrons. In [Ni(CN)4]2−, the strong field CN− ligand forces the two unpaired 3d electrons to pair up, emptying one 3d orbital — that orbital joins one 4s and two 4p orbitals in dsp2 hybridisation, giving a square planar geometry with no unpaired electrons (diamagnetic). In [NiCl4]2−, the weak field Cl− causes no pairing, so no 3d orbital is freed and bonding uses sp3 hybridisation — a tetrahedral geometry that retains two unpaired electrons (paramagnetic).
Why This Happens: The Concept
Valence Bond Theory (VBT) explains bonding in terms of hybridization of atomic orbitals. But to understand magnetism and geometry, we need to see how the ligand field affects the d-orbital energies.
For a Ni2+ ion, the electronic configuration is [Ar]3d8. In a free ion, all five d-orbitals are degenerate (same energy). When ligands approach, they split these orbitals into different energy levels depending on the geometry.
The nature of the ligand (strong field vs weak field) decides whether electrons pair up or remain unpaired. This pairing directly determines:
- The hybridization scheme (and thus geometry)
- The magnetic property (diamagnetic = all paired, paramagnetic = unpaired electrons)
Step-by-Step Analysis
1. Identify the central metal ion and its d-electron count
Nickel in both complexes is in the +2 oxidation state.
Ni atomic number = 28.
Ni2+: loses two 4s electrons → configuration: 3d8.
So we have 8 electrons in the 3d orbitals.
2. Consider the ligand strength
- CN− is a strong field ligand (high up in the spectrochemical series). It causes a large crystal field splitting (Δ).
- Cl− is a weak field ligand (low in the spectrochemical series). It causes a small crystal field splitting (Δ).
This difference is the entire reason for the different outcomes.
3. Case 1: [Ni(CN)4]2− — Strong field, square planar
Because CN− is a strong field ligand, the splitting between the d-orbitals is large. In a square planar geometry, the d-orbital splitting pattern (from highest to lowest energy) is approximately:
dx2−y2≫dxy>dz2>dxz=dyz
The energy gap is so large that it is energetically favourable for electrons to pair up in the lower orbitals rather than occupy the high-energy dx2−y2 orbital.
So the 8 d-electrons fill as:
- dxz,dyz: 2 electrons each (paired)
- dz2: 2 electrons (paired)
- dxy: 2 electrons (paired)
- dx2−y2: empty
This leaves zero unpaired electrons — the complex is diamagnetic.
Now, for bonding: the empty dx2−y2 orbital, along with one 4s and two 4p orbitals, undergoes dsp2 hybridization (one d, one s, two p). This gives a square planar geometry.
4. Case 2: [NiCl4]2− — Weak field, tetrahedral
Cl− is a weak field ligand. The splitting Δ is small. In a tetrahedral geometry, the d-orbital splitting is inverted compared to octahedral:
- Lower energy set: e (dx2−y2,dz2)
- Higher energy set: t2 (dxy,dyz,dzx)
The splitting Δt is much smaller than in octahedral complexes (roughly 94 of Δo). So the energy cost of pairing electrons is greater than the energy gained by occupying the lower e set.
Thus, the 8 d-electrons fill from the bottom up:
- e set (lower): 4 electrons — both orbitals doubly occupied
- t2 set (higher): 4 electrons — the first three occupy the three orbitals singly (Hund's rule), and the fourth pairs up in one of them
A common mistake is to think that d8 in a weak field always gives two unpaired electrons — but this is only true for tetrahedral geometry. In an octahedral weak field, d8 would have two unpaired electrons in the eg set, but the geometry would be different.
This gives two unpaired electrons — the complex is paramagnetic.
For bonding: since the d-orbitals are all occupied (or partially occupied), the metal uses sp3 hybridization (one s, three p orbitals) — no d-orbital is empty for dsp2. This yields a tetrahedral geometry.
Summary Table
| Property | [Ni(CN)4]2− | [NiCl4]2− |
|---|---|---|
| Ligand type | Strong field (CN−) | Weak field (Cl−) |
| Geometry | Square planar | Tetrahedral |
| Hybridization | dsp2 | sp3 |
| Unpaired electrons | 0 | 2 |
| Magnetic nature | Diamagnetic | Paramagnetic |
A quick way to remember: Strong field + d8 → square planar + diamagnetic. Weak field + d8 → tetrahedral + paramagnetic. The ligand decides the pairing, and the pairing decides the geometry.
[Ni(CN)4]2− is diamagnetic (no unpaired electrons) due to strong field CN− causing pairing in a square planar dsp2 geometry, while [NiCl4]2− is paramagnetic (two unpaired electrons) due to weak field Cl− leaving electrons unpaired in a tetrahedral sp3 geometry.
Method: Valence Bond Theory (VBT) Analysis of Geometry and Magnetism
Concept: Valence bond theory links ligand field strength → electron pairing → hybridisation → geometry and magnetic behaviour.
Step 1: Determine the oxidation state and electron configuration of Ni
-
For [Ni(CN)4]2−:
Let Ni oxidation state be x.
x+4(−1)=−2⇒x=+2
Ni in +2 state: Atomic number 28, configuration [Ar]3d84s2
Remove 2 electrons → 3d8
-
For [NiCl4]2−:
Same calculation → Ni is also in +2 state → 3d8
Step 2: Identify ligand field strength and decide hybridization
-
CN⁻ is a strong field ligand → causes pairing of electrons
- In 3d8, one 3d orbital is emptied by pairing
- Hybridization: dsp2 (one d, one s, two p orbitals)
- Geometry: Square planar
-
Cl⁻ is a weak field ligand → no pairing
- All 3d orbitals remain singly occupied as far as possible
- Hybridization: sp3 (one s, three p orbitals)
- Geometry: Tetrahedral
Step 3: Count unpaired electrons and determine magnetism
| Complex | Hybridization | Geometry | Unpaired electrons | Magnetic nature |
|---|---|---|---|---|
| [Ni(CN)4]2− | dsp2 | Square planar | 0 | Diamagnetic |
| [NiCl4]2− | sp3 | Tetrahedral | 2 | Paramagnetic |
Step 4: Final explanation
- In [Ni(CN)4]2−, strong CN⁻ forces pairing of 3d electrons → all electrons paired → diamagnetic (repelled by magnetic field)
- In [NiCl4]2−, weak Cl⁻ cannot cause pairing → two unpaired electrons remain → paramagnetic (attracted by magnetic field)
Key takeaway: The same metal ion (Ni2+) with 3d8 configuration gives different geometries and magnetic properties depending on ligand strength — a direct consequence of valence bond theory.
Here are the common mistakes students make on this question, along with how to avoid each one.
Mistake 1: Forgetting to check the oxidation state of the central metal ion first.
- The Mistake: Students jump straight to the geometry or the ligand without first calculating the oxidation state of Nickel (Ni). This leads to the wrong d-electron count (dn configuration).
- Why it’s wrong: The number of d-electrons determines how the electrons will fill the orbitals, which directly controls the magnetic property (diamagnetic vs. paramagnetic).
- How to Avoid:
- Always start with the charge balance.
- For [Ni(CN)4]2−:
- Let the oxidation state of Ni be x.
- Charge of CN⁻ is −1 each. Total ligand charge = 4×(−1)=−4.
- Overall complex charge = −2.
- Equation: x+(−4)=−2⟹x=+2.
- Result: Ni is in the +2 oxidation state.
- For [NiCl4]2−:
- Charge of Cl⁻ is −1 each. Total ligand charge = 4×(−1)=−4.
- Overall complex charge = −2.
- Equation: x+(−4)=−2⟹x=+2.
- Result: Ni is also in the +2 oxidation state.
- Key fact: Ni in the ground state is [Ar]3d84s2. In the +2 state, it loses the two 4s electrons, giving a 3d8 configuration.
Mistake 2: Assuming the same geometry leads to the same magnetic property.
- The Mistake: Students think that because both complexes have the same metal ion (Ni²⁺) and the same coordination number (4), they will have the same magnetic behavior.
- Why it’s wrong: The geometry (square planar vs. tetrahedral) and the strength of the ligand (strong field CN⁻ vs. weak field Cl⁻) completely change how the d-orbitals split and how electrons fill them.
- How to Avoid:
- Remember the rule: Strong field ligands (like CN⁻, CO, NH₃) cause pairing of electrons. Weak field ligands (like Cl⁻, Br⁻, H₂O) cause no pairing (Hund's rule is followed).
- Visualize the splitting:
- Tetrahedral (Td): Splitting is small. Electrons fill all orbitals singly first (Hund's rule).
- Square planar (D4h): Splitting is large. Electrons pair up in the lower energy orbitals.
Mistake 3: Drawing the wrong orbital filling diagram for square planar geometry.
- The Mistake: Students use the tetrahedral or octahedral splitting diagram for the square planar complex.
- Why it’s wrong: Square planar geometry has a very specific d-orbital splitting pattern. The energy order is: dx2−y2≫dxy>dz2>dxz=dyz.
- How to Avoid:
- Draw the correct diagram:
- For [Ni(CN)4]2− (Square planar, strong field):
- The dx2−y2 orbital is very high in energy (empty).
- The dxy orbital is next highest.
- The dz2, dxz, and dyz are lower.
- Filling for d8: All 8 electrons pair up in the four lower orbitals (dxy, dz2, dxz, dyz). The dx2−y2 is empty.
- Result: No unpaired electrons → Diamagnetic.
- For [Ni(CN)4]2− (Square planar, strong field):
- Draw the correct diagram:
Mistake 4: Drawing the wrong orbital filling diagram for tetrahedral geometry.
- The Mistake: Students forget that in tetrahedral geometry, the dxy, dxz, and dyz orbitals are higher in energy than the dx2−y2 and dz2 orbitals.
- Why it’s wrong: The splitting is inverted compared to octahedral.
- How to Avoid:
- Draw the correct diagram:
- For [NiCl4]2− (Tetrahedral, weak field):
- The dxy, dxz, dyz (the t2 set) are higher.
- The dx2−y2, dz2 (the e set) are lower.
- Filling for d8: First, fill the lower e set with 4 electrons (paired). Then, place the remaining 4 electrons in the higher t2 set. According to Hund's rule, they will occupy all three orbitals singly before pairing.
- Result: Two unpaired electrons (in the t2 set) → Paramagnetic.
- For [NiCl4]2− (Tetrahedral, weak field):
- Draw the correct diagram:
Mistake 5: Confusing the terms "diamagnetic" and "paramagnetic".
- The Mistake: Students write the correct electron configuration but then state the wrong magnetic property.
- Why it’s wrong: It's a direct loss of marks for a simple definition.
- How to Avoid:
- Memorize:
- Diamagnetic: All electrons are paired. The substance is repelled by a magnetic field.
- Paramagnetic: Has one or more unpaired electrons. The substance is attracted by a magnetic field.
- Check your final diagram: Count the unpaired electrons. If the count is zero, it's diamagnetic. If non-zero, it's paramagnetic.
- Memorize:
Summary Table to Avoid Mistakes
| Feature | [Ni(CN)4]2− | [NiCl4]2− |
|---|---|---|
| Oxidation State of Ni | +2 | +2 |
| d-electron count | d8 | d8 |
| Ligand | CN⁻ (Strong field) | Cl⁻ (Weak field) |
| Geometry | Square planar | Tetrahedral |
| Orbital Splitting | Large (Δ is large) | Small (Δ is small) |
| Electron Filling | Pairing occurs | Hund's rule (no pairing) |
| Unpaired Electrons | 0 | 2 |
| Magnetic Property | Diamagnetic | Paramagnetic |
- CBSE 2026Set ANNUAL1 markQ.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.)
- trans-isomer: the two Cl- ligands occupy opposite positions on the octahedron, directly across from each other, with a Cl-Co-Cl bond angle of 180 degrees. (Structure: the two Cl ligands sit at the top and bottom apex positions, with all four NH3 ligands in the square equatorial plane.)
These two forms are geometrical (cis-trans) isomers - same connectivity/formula, but different spatial arrangement of ligands, and are not interconvertible without breaking bonds.
✓Final answerTwo geometrical isomers exist: cis-[Co(NH3)4Cl2]+ (Cl ligands at 90 degrees to each other) and trans-[Co(NH3)4Cl2]+ (Cl ligands at 180 degrees, opposite each other).
- CBSE 2026Set ANNUAL1 markMCQQ.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(NH3)2Cl2], type MA2B2) do show cis–trans geometrical isomerism, since two positions can be adjacent (cis, 90∘) or opposite (trans, 180∘).
-
(c) Octahedral complexes (types MA4B2, MA3B3, etc.) do show both cis–trans and facial–meridional (fac/mer) geometrical isomerism, since some positions are adjacent and some are directly opposite.
-
(b) Tetrahedral complexes: all four coordination positions are geometrically equivalent and mutually adjacent — there is no pair of ligand sites at 180∘ to each other. Any two ligand arrangements that look different on paper are actually superimposable by rotation, so a genuine cis/trans (or fac/mer) distinction cannot exist. (Tetrahedral complexes with four different unidentate ligands can show optical isomerism instead, which is a separate phenomenon from geometrical isomerism.)
-
(d) 'All of the above' is therefore false, since square planar and octahedral complexes do show geometrical isomerism.
✓Final answer(b) Tetrahedral complexes
-
- CBSE 2025Set ANNUAL1 markQ.Draw structures of geometrical isomers of [Fe(NH3)2(CN)4]−.
›Reveal solutionSolution
This octahedral MA2B4 complex can arrange its two identical NH3 ligands either adjacent to each other (cis) or directly opposite each other (trans), giving two geometrical isomers.
Identifying the isomerism
[Fe(NH3)2(CN)4]− is an octahedral complex of the general type [MA2B4], where A=NH3 (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 (NH3) ligands.
cis-isomer: Picture an octahedron with the six positions labelled +x,−x,+y,−y,+z,−z. In the cis isomer, the two NH3 ligands occupy adjacent positions, i.e. at 90∘ to each other (e.g. one NH3 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 NH3 ligands occupy diametrically opposite positions, i.e. at 180∘ 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).
(In a 2-D sketch, both isomers are drawn as an octahedron outline with Fe at the centre: the cis form shows the two NH3 labels on two adjacent vertices, and the trans form shows them on two vertices directly across from each other, with CN labelling the remaining four vertices in both cases.)
✓Final answerTwo geometrical isomers exist: cis-[Fe(NH3)2(CN)4]− (the two NH3 ligands mutually adjacent, 90∘ apart) and trans-[Fe(NH3)2(CN)4]− (the two NH3 ligands mutually opposite, 180∘ apart), with CN− occupying the remaining coordination positions in each case.
- CBSE 2024Set ANNUAL1 markMCQQ.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.)
Ionization isomerism: this arises when a ligand and the counter ion can exchange places, giving isomers that ionise to give different ions in solution. Here, [Co(NH3)4Br2]Cl (which gives Cl- in solution, tests positive with AgNO3) is an ionization isomer of [Co(NH3)4BrCl]Br (which gives Br- in solution, tests positive with AgNO3 differently) - both have the same overall composition Co(NH3)4Br2Cl but different ions in solution.
Since both types of isomerism apply here (and optical does not, since cis-MA4B2 has a mirror plane), the correct choice is geometrical and ionization.
✓Final answer(a) Geometrical and ionization.
- CBSE 2024Set ANNUAL1 markMCQQ.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
MX6 and MX5L 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. MX4L2, MX3L3) do show it.
[A] For an octahedral complex MX6 (all six ligands identical) there is only one possible spatial arrangement, and for MX5L (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.
[R] However, it is false to generalise that coordination number 6 as such never shows geometrical isomerism. Octahedral complexes of type MX4L2 (cis/trans isomers), MX3L3 (facial/meridional isomers), and MX2L2L2 (various geometrical arrangements) are classic, well-known examples of geometrical isomerism at coordination number 6. So [R] as a blanket statement is FALSE.
✓Final answer(c) [A] is true, but [R] is false.
- CBSE 2024Set ANNUAL1 markMCQQ.The existence of two different coloured complexes with composition of [Co(NH3)4Cl2]+ is due to(a) linkage isomerism(b) geometrical isomerism(c) coordination isomerism(d) ionization isomerism
›Reveal solutionSolution
[Co(NH3)4Cl2]+ 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(NH3)4Cl2]+ is octahedral with formula type [MA4B2] (4 NH3 + 2 Cl around Co3+). Two distinct, non-interconvertible spatial arrangements are possible:
- cis-[Co(NH3)4Cl2]+: the two Cl ligands occupy adjacent (90°) positions — this isomer is violet.
- trans-[Co(NH3)4Cl2]+: 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.
Why the other options are wrong: (a) Linkage isomerism needs an ambidentate ligand (e.g. −NO2/−ONO) — none is present here. (c) Coordination isomerism requires both a complex cation and complex anion exchanging ligands — this is a single cationic complex with a simple counter-ion. (d) Ionisation isomerism arises when a ligand and the counter-ion exchange places (giving different ions in solution) — here what's inside vs. outside the coordination sphere is fixed; only the spatial layout of the two Cl ligands changes.
✓Final answer(b) Geometrical isomerism (cis-violet vs trans-green forms)
- CBSE 2023Set ANNUAL1 markQ.Draw the geometrical isomers of [Co(NO2)3(NH3)3]. (½+½=1)
›Reveal solutionSolution
[Co(NO2)3(NH3)3] is an octahedral complex of the type MA3B3, 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 MA3B3 (here M=Co3+, A=NO2−, B=NH3), 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(NO2)3(NH3)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 NO2− ligands occupy one triangular face (positions 1, 2, 3 — mutually cis, each at 90∘ to the other two), while all three NH3 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.
mer-[Co(NO2)3(NH3)3]: The three NO2− ligands instead occupy three positions that lie in one plane passing through the metal centre — e.g. positions 1 (top), 2 (equatorial), and 4 (bottom), so that two of the three NO2− groups are mutually trans (180∘ apart) while the third is cis to both. The three NH3 ligands occupy the remaining three positions, also lying along that same meridian plane on the other side.
(Since a text/JSON answer cannot render a diagram directly: in fac-, looking down any one of the C3 axes of the octahedron shows all three NO2− on the near face; in mer-, one NO2−–Co–NO2− angle is 180∘ and the third NO2− sits at 90∘ to both.)
✓Final answerTwo geometrical isomers exist: fac-isomer (three NO2− on one triangular face, three NH3 on the opposite face) and mer-isomer (two NO2− groups mutually trans, the third cis to both, with NH3 similarly arranged on the meridian).
- CBSE 2020Set 56/2/11 markQ.What type of isomerism is shown by the complex [Co(NH3)5NO2]Cl2?
›Reveal solutionSolution
The complex [Co(NH3)5NO2]Cl2 exhibits linkage isomerism because the NO2− ligand can coordinate through either the nitrogen atom (−NO2, 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 NO2− 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.
ImportantLinkage 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
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Identify the coordination sphere and counter ions
The formula is [Co(NH3)5NO2]Cl2. The square brackets enclose the coordination sphere: Co3+ (cobalt in +3 oxidation state) surrounded by five NH3 ligands and one NO2− ligand. The two Cl− ions are outside the brackets — they are counter ions, not directly bonded to cobalt.
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Recognise the ambidentate nature of NO2−
The nitrite ion can bind through:
- Nitrogen atom: forming a nitro complex, [Co(NH3)5(NO2)]2+
- Oxygen atom: forming a nitrito complex, [Co(NH3)5(ONO)]2+
Both have the same overall formula [Co(NH3)5NO2]Cl2, but the connectivity differs.
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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 NH3 are identical, and the sixth position is occupied by NO2− — 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.
- Ionisation isomerism would involve exchange of ligands with counter ions (e.g., [Co(NH3)5Cl]NO2), but here the Cl− ions are clearly outside the coordination sphere and the NO2− is inside — no such exchange is possible without changing the formula.
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Confirm linkage isomerism
Since the only variable is which atom of NO2− bonds to cobalt, and both possibilities yield distinct compounds with different properties (e.g., colour, reactivity), this is a textbook case of linkage isomerism.
TipA quick way to spot linkage isomerism: look for a ligand with two different donor atoms (like NO2−, SCN−, CN−, or CO). If the ligand is inside the coordination sphere and the rest of the complex is symmetric, linkage isomerism is almost certainly the answer.
Watch outA common mistake is to confuse linkage isomerism with ionisation isomerism. Remember: linkage isomerism changes which atom of the same ligand is bonded; ionisation isomerism swaps a ligand with a counter ion. Here, Cl− is outside the bracket and NO2− is inside — no swap occurs.
✓Final answerThe complex [Co(NH3)5NO2]Cl2 shows linkage isomerism due to the ambidentate NO2− ligand.
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- CBSE 2018Set ANNUAL1 markQ.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).
- [Co(NH3)5SO4]Br -> [Co(NH3)5SO4]+ + Br- (gives bromide ion; gives pale-yellow ppt with AgNO3).
Because the free ions in solution differ, the two are distinct compounds.
✓Final answerIonisation isomers: same formula, different ions in solution; e.g. [Co(NH3)5Br]SO4 (free SO4^2-) vs [Co(NH3)5SO4]Br (free Br-).
- CBSE 2018Set 56/11 markQ.Write the coordination isomer of [Cu(NH3)4][PtCl4].
›Reveal solutionSolution
Interchanging ligands between the two complex ions gives the coordination isomer [Pt(NH3)4][CuCl4].
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.
Why. In [Cu(NH3)4][PtCl4], the NH3 ligands are on Cu and the Cl− ligands are on Pt. Exchanging the ligand sets produces a genuinely different compound with the same overall formula.
Steps. Move all four NH3 to platinum and all four Cl− to copper.
✓Final answer[Pt(NH3)4][CuCl4]
- CBSE 2017Set ANNUAL1 markMCQQ.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:
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[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.
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[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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[PtCl2(P(C2H5)3)2]: square planar, type MA2B2 (2 Cl + 2 PEt3). The two identical Cl ligands can be adjacent (cis) or opposite (trans) to each other, giving two distinct geometrical isomers — this does show geometrical isomerism.
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[Fe(H2O)5NO]²⁺: octahedral, type MA5B (5 identical H2O + 1 NO). With only one different ligand, there is only one possible position for it — no geometrical isomerism.
✓Final answer(c) [PtCl2(P(C2H5)3)2]
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