Isomerism In Co-ordination Compounds

ChemistryCoordination CompoundsFor NEET aspirants

Isomerism in coordination compounds means that two or more complexes have the same formula but a different arrangement of atoms. It comes in two families: structural isomerism (ionisation, solvate, linkage, coordination), where different atoms are bonded to the metal, and stereoisomerism (geometrical and optical), where the same bonds point in different directions. This page draws every case in 3D and gives computer-checked counts for isomerism in coordination compounds of all common formula types. It is a favourite of JEE Main, JEE Advanced and NEET.

On this page1Two families2Structural isomers3cis/trans and fac/mer4Counting isomers5Optical isomers6Isomer table7Solved examples

Key Formulas - Quick Reference

Structural isomers: different bonds; stereoisomers: same bonds, different arrangement in space
Ionisation: $[M A_{5} X] Y$ and $[M A_{5} Y] X$ give different ions in water (test with $AgN O_{3}$ or $BaC l_{2}$ )
★ Must learn Linkage isomerism needs an ambidentate ligand: $N O_{2}^{-}$ , $SC N^{-}$ , $C N^{-}$ , $S_{2} O_{3}^{2 -}$
Coordination isomerism needs both the cation and the anion to be complex ions
★ Must learn Tetrahedral: no geometrical isomers; square planar: no optical isomers
★ Must learn Square planar: $M A_{2} B_{2}$ = 2, $M A_{2} BC$ = 2, $MABCD$ = 3, $M (AB)_{2}$ = 2
★ Must learn Octahedral: $M A_{4} B_{2}$ = 2 (cis, trans), $M A_{3} B_{3}$ = 2 (fac, mer), $M A_{2} B_{2} C_{2}$ = 5 geometrical (6 in all), $MABCDEF$ = 15 (30 in all)
Chelates: $M (AA)_{3}$ = 1 geometrical, 2 optical; $M (AA)_{2} B_{2}$ = 2 geometrical, 3 in all (cis is chiral)
★ Must learn Optically active only if the complex has no mirror plane and no centre of symmetry

1. What Is Isomerism?

Isomers are compounds with the same composition (same molecular formula) but different structures, and therefore different physical and chemical properties such as colour, solubility, conductivity and reactivity. Coordination compounds show isomerism much more often than simple salts because a metal holds several ligands in a fixed three-dimensional pattern.

Structural isomers have the same formula but different bonds: a different ion, water molecule, donor atom or ligand set is attached to the metal.
Stereoisomers (space isomers) have exactly the same bonds; only the positions of the ligands around the metal differ.

Figure 1: The two families of isomerism in complexes. Ask what changed between the isomers: the bonds (structural) or only their directions (stereo).

2. Structural Isomerism

In structural isomers something that is bonded to the metal in one isomer is not bonded (or is bonded differently) in the other. Figure 2 shows the four main types side by side with the part that moves coloured.

Figure 2: Structural isomers differ in what is bonded to the metal. The coloured part is what changes places, and each pair can be told apart by a simple test.

2.1 Ionisation Isomerism

Ionisation isomers arise when an ion inside the coordination sphere swaps places with the counter ion outside it. They have the same formula but give different ions in solution, because only the ion outside the bracket is released.

$[Co (N H_{3})_{5} Br] S O_{4}$ (violet) and $[Co (N H_{3})_{5} (S O_{4})] Br$ (red): the first gives $S O_{4}^{2 -}$ ions, the second gives $B r^{-}$ ions
$[Co (N H_{3})_{4} ClBr] Cl$ and $[Co (N H_{3})_{4} C l_{2}] Br$
$[Pt (N H_{3})_{3} Cl] N O_{2}$ and $[Pt (N H_{3})_{3} (N O_{2})] Cl$
$[PtB r_{2} (N H_{3})_{4}] C l_{2}$ and $[PtC l_{2} (N H_{3})_{4}] B r_{2}$

They are identified by precipitation tests. With barium chloride only the first cobalt isomer gives a white precipitate, and with silver nitrate only the second gives a pale yellow one:

[Co (N H_{3})_{5} Br] S O_{4} + BaC l_{2} ⟶ [Co (N H_{3})_{5} Br] C l_{2} + BaS O_{4} ↓

[Co (N H_{3})_{5} (S O_{4})] Br + AgN O_{3} ⟶ [Co (N H_{3})_{5} (S O_{4})] N O_{3} + AgBr ↓

2.2 Solvate (Hydrate) Isomerism

Solvate isomers differ in whether solvent molecules are bonded to the metal as ligands or sit in the crystal as free solvent. When the solvent is water this is called hydrate isomerism. The classic case is $CrC l_{3} \cdot 6 H_{2} O$ , which exists as three isomers:

Isomer	Colour	Ionisable $C l^{-}$ (mol AgCl per mol)	Ions in water	Free $H_{2} O$ lost with conc. $H_{2} S O_{4}$
$[Cr (H_{2} O)_{6}] C l_{3}$	violet	3	4	0
$[CrCl (H_{2} O)_{5}] C l_{2} \cdot H_{2} O$	light (grey) green	2	3	1
$[CrC l_{2} (H_{2} O)_{4}] Cl \cdot 2 H_{2} O$	dark green	1	2	2

A second example is $[Co (N H_{3})_{5} (H_{2} O)] C l_{3}$ and $[CoCl (N H_{3})_{5}] C l_{2} \cdot H_{2} O$ . Hydrate isomerism is really ionisation isomerism in which one of the swapping groups is a water molecule.

2.3 Linkage Isomerism

Linkage isomers contain an ambidentate ligand that is bonded through a different donor atom in each isomer (see Introduction and Nomenclature). The ligand has two donor atoms but uses only one at a time.

$[Co (N H_{3})_{5} (N O_{2})] C l_{2}$ (yellow, bonded through N, nitrito-N) and $[Co (N H_{3})_{5} (ONO)] C l_{2}$ (red, bonded through O, nitrito-O). The red nitrito-O isomer slowly changes into the more stable yellow nitrito-N isomer.
$[Cr (H_{2} O)_{5} (CN)]^{2 +}$ and $[Cr (H_{2} O)_{5} (NC)]^{2 +}$ : cyanide bonded through C or N
$[Mn (CO)_{5} (SCN)]$ and $[Mn (CO)_{5} (NCS)]$ : thiocyanate bonded through S or N

Linkage isomers have the same ions in solution, so precipitation tests cannot tell them apart. They are distinguished by colour and by infrared (IR) spectroscopy, because an N-O stretch in M- $N O_{2}$ absorbs at a different frequency from the O-N=O group in M-ONO. Ligands such as $N H_{3}$ , $H_{2} O$ and $C l^{-}$ have only one kind of donor atom and can never give linkage isomers.

2.4 Coordination Isomerism

Coordination isomers are possible only when both the cation and the anion are complex ions. The isomers differ in how the ligands are shared between the two metal centres.

$[Co (N H_{3})_{6}] [Cr (CN)_{6}]$ and $[Cr (N H_{3})_{6}] [Co (CN)_{6}]$
$[Cr (N H_{3})_{6}] [Co (C_{2} O_{4})_{3}]$ and $[Co (N H_{3})_{6}] [Cr (C_{2} O_{4})_{3}]$
$[Cu (N H_{3})_{4}] [PtC l_{4}]$ and $[Pt (N H_{3})_{4}] [CuC l_{4}]$
$[Pt (N H_{3})_{4}] [PtC l_{4}]$ (Magnus's green salt) and $[Pt (N H_{3})_{3} Cl] [Pt (N H_{3}) C l_{3}]$ : here both metals are Pt(II) and the ligands are simply redistributed

Exam Trick

Ask "what moved?" An ion moved across the bracket: ionisation (test with $AgN O_{3}$ or $BaC l_{2}$ ). A water molecule moved: hydrate (count the AgCl, or the water lost to conc. $H_{2} S O_{4}$ ). The donor atom changed: linkage (IR or colour). Ligands moved between two complex ions: coordination. Nothing moved, only the directions changed: stereoisomerism.

2.5 Polymerisation Isomerism

Compounds with the same empirical formula but molecular formulas that are whole-number multiples of it are called polymerisation isomers. For $Pt (N H_{3})_{2} C l_{2}$ :

Compound	Multiple of $Pt (N H_{3})_{2} C l_{2}$	Molar mass (g/mol)
$[Pt (N H_{3})_{2} C l_{2}]$	1	300
$[Pt (N H_{3})_{4}] [PtC l_{4}]$	2	600
$[Pt (N H_{3})_{3} Cl]_{2} [PtC l_{4}]$	3	900
$[Pt (N H_{3})_{4}] [Pt (N H_{3}) C l_{3}]_{2}$	3	900

Strictly these are not true isomers, because their molecular masses differ. The term is kept because the compounds have the same percentage composition.

2.6 Coordination Position Isomerism

In a bridged polynuclear complex, ligands can be distributed differently between the two metal atoms that share the bridges. Both isomers below contain two Co(III) atoms joined by two $O H^{-}$ bridges, 8 $N H_{3}$ and 2 $C l^{-}$ :

$[(N H_{3})_{4} Co (OH)_{2} Co (N H_{3})_{2} C l_{2}] S O_{4}$ : both Cl on the same cobalt
$[Cl (N H_{3})_{3} Co (OH)_{2} Co (N H_{3})_{3} Cl] S O_{4}$ : one Cl on each cobalt

Charge check: $2 (+ 3) + 2 (- 1) + 2 (- 1) = + 2$ , balanced by one $S O_{4}^{2 -}$ .

Figure 3: Coordination position isomers have the same bridged skeleton; only the positions of the non-bridging ligands on the two metal atoms change.

Quick Recall: tap to check

[Co (N H_{3})_{5} Br] S O_{4}

and

[Co (N H_{3})_{5} (S O_{4})] Br

are which type of isomers?

Ionisation isomers:

BaC l_{2}

precipitates sulphate only from the first,

AgN O_{3}

precipitates bromide only from the second.

What must a ligand be for linkage isomerism?

Ambidentate, such as

N O_{2}^{-}

SC N^{-}

When is coordination isomerism possible?

Only when both the cation and the anion are complex ions.

Key idea

Structural isomers differ in what is bonded to the metal, so a simple test (a precipitate, conductivity or IR) always tells them apart.

3. Geometrical Isomerism

Geometrical isomers have the same ligands bonded to the metal, but the ligands occupy different relative positions. When two identical ligands are next to each other (at 90°) the isomer is cis; when they are opposite (at 180°) it is trans. For this reason geometrical isomerism is also called cis-trans isomerism. It is common in complexes of coordination number 4 (square planar) and 6 (octahedral).

3.1 Square Planar Complexes (CN = 4)

In a square every corner has two cis neighbours and one trans partner, so geometrical isomers appear as soon as there are two kinds of ligand arranged as $M A_{2} B_{2}$ :

$M A_{2} B_{2}$ , e.g. $[Pt (N H_{3})_{2} C l_{2}]$ : 2 isomers. The cis isomer is cisplatin, an anticancer drug; the trans isomer is inactive.
$M A_{2} BC$ , e.g. $[Pt (N H_{3})_{2} Cl (N O_{2})]$ or $[PtCl (N H_{3}) (py)_{2}]$ : 2 isomers (the two A ligands cis or trans).
$MABCD$ , e.g. $[Pt (N H_{3}) (py) (Cl) (Br)]$ : 3 isomers, one for each ligand that can be trans to A.
$M (AB)_{2}$ with an unsymmetrical chelate, e.g. bis(glycinato)platinum(II), $[Pt (gly)_{2}]$ : 2 isomers (the two N donors cis or trans).
$M A_{4}$ , $M A_{3} B$ and $M (AA) B_{2}$ : no geometrical isomers.

Figure 4: In a square every position has one trans partner, so cis/trans isomers exist; in a tetrahedron no position is trans to another, so they cannot.

Tetrahedral complexes never show geometrical isomerism. In a tetrahedron all four positions are at 109.5° to each other; no position is "opposite" another. Any two ligands are always neighbours, so every arrangement of $M A_{2} B_{2}$ is the same.

3.2 Octahedral Complexes: cis and trans

An octahedron has six positions: each position has four cis neighbours and one trans partner. The simplest case is $M A_{4} B_{2}$ . In $[Co (N H_{3})_{4} C l_{2}]^{+}$ the two chloride ions can be cis or trans, and the two isomers even have different colours:

Figure 5: Geometrical isomers of

[Co (N H_{3})_{4} C l_{2}]^{+}

(type

M A_{4} B_{2}

): same ligands, different positions, so different colours and properties.

The same happens with chelating ligands. In $M (AA)_{2} B_{2}$ , such as $[CoC l_{2} (en)_{2}]^{+}$ , the two Cl atoms can be cis or trans (Figure 8). Other examples of cis-trans isomerism are $M A_{4} BC$ ( $[CoCl (N H_{3})_{4} (N O_{2})]^{+}$ ) and $M (AA)_{2} BC$ ( $[CoCl (N H_{3}) (en)_{2}]^{2 +}$ ).

3.3 Octahedral Complexes: fac and mer

For $M A_{3} B_{3}$ , such as $[Co (N H_{3})_{3} (N O_{2})_{3}]$ or $[CoC l_{3} (N H_{3})_{3}]$ , "cis" and "trans" are not enough to describe the three B ligands. Two isomers exist:

fac (facial): the three B ligands occupy the corners of one triangular face of the octahedron, all at 90° to each other.
mer (meridional): the three B ligands lie on a meridian, a plane through the metal, like a letter T; two of them are trans (180°).

Figure 6: The two isomers of

[Co (N H_{3})_{3} (N O_{2})_{3}]

(type

M A_{3} B_{3}

). The shaded triangle is the face (fac) or the meridian plane through Co (mer).

fac (facial)Three identical ligands on one triangular face; every angle between them is 90°.

mer (meridional)Three identical ligands on a meridian through the metal; two are trans (180°).

3.4 Counting Geometrical Isomers

Drawing every arrangement is slow and error-prone. A faster method is to number the six corners, remember the three trans pairs (1-6, 2-4, 3-5), and sort the isomers by which identical ligands are trans to each other:

Figure 7: Count octahedral isomers by listing which identical ligands are trans. For

M a_{2} b_{2} c_{2}

this gives 5 geometrical isomers; the all-cis one is chiral, so 6 in all.

Exam Trick

Tetrahedral = no cis/trans; square planar = no d/l. Before counting anything, check the shape. For octahedral $M a_{2} b_{2} c_{2}$ remember "5 and 6": 5 geometrical isomers and 6 stereoisomers, because only the all-cis form has a mirror image partner.

Key idea

Geometrical isomers need a trans position: square planar and octahedral complexes have them, tetrahedral complexes never do.

4. Optical Isomerism

4.1 Chirality and Enantiomers

A complex that cannot be superimposed on its mirror image is chiral. The complex and its mirror image form a pair of optical isomers, also called enantiomers. Key points:

Enantiomers have the same geometrical arrangement of ligands; they differ only as a left hand differs from a right hand.
They rotate the plane of plane-polarised light by equal amounts in opposite directions. The one that rotates it to the right is dextrorotatory (d or +) and the other is laevorotatory (l or -).
Their other physical and chemical properties are the same, except in reactions with other chiral substances.
A 1:1 mixture of the two, called a racemic mixture, shows no rotation.
To test for chirality, draw the mirror image and try to superimpose it on the original by rotating it. If it never fits, the complex is optically active.

A quick rule: a complex that has a mirror plane (or a centre of symmetry) always coincides with its mirror image, so it is optically inactive. Look for such a plane first.

4.2 Tetrahedral and Square Planar Complexes

Tetrahedral $Mabcd$ , with four different ligands, is chiral, just like a carbon atom with four different groups. Such complexes are usually too labile to separate, but an unsymmetrical chelate makes a stable example: bis(benzoylacetonato)beryllium(II), a tetrahedral $M (AB)_{2}$ complex, has been separated into d and l forms.
Square planar complexes are optically inactive. The plane of the molecule is itself a mirror plane, so the mirror image is identical (unless a ligand is chiral in its own right).

4.3 Octahedral Complexes with Chelating Ligands

Optical isomerism is most common in octahedral complexes of bidentate ligands. The chelate rings twist around the metal like the blades of a propeller, and a left-handed propeller is the mirror image of a right-handed one.

$M (AA)_{2} B_{2}$ : $[CoC l_{2} (en)_{2}]^{+}$ . The trans isomer has a mirror plane (the plane of the four N atoms) and is optically inactive. The cis isomer has no mirror plane and exists as a d and l pair. So this ion has 3 stereoisomers in all:

Figure 8:

[Co (en)_{2} C l_{2}]^{+}

has 3 stereoisomers: the cis form is chiral (d and l), the trans form has a mirror plane and is optically inactive.

$M (AA)_{3}$ : $[Co (en)_{3}]^{3 +}$ , $[Cr (ox)_{3}]^{3 -}$ . Three identical chelates can be placed in only one geometrical way, so there are no geometrical isomers. But the arrangement is always chiral, giving a d and l pair (2 stereoisomers). There is no optically inactive "meso" form.

Figure 9:

[Co (en)_{3}]^{3 +}

(type

M (AA)_{3}

) is a three-bladed propeller: its mirror image (

Λ

) is a propeller of the opposite twist to

Δ

, so it is optically active.

$M (AA)_{2} BC$ , such as $[CoCl (N H_{3}) (en)_{2}]^{2 +}$ , behaves like $M (AA)_{2} B_{2}$ : trans is inactive and cis is a d/l pair. $M (AA) B_{2} C_{2}$ , such as $[CoC l_{2} (en) (N H_{3})_{2}]^{+}$ , has 3 geometrical isomers, and only the all-cis one is chiral (4 stereoisomers). With an unsymmetrical chelate, $M (AB)_{3}$ such as $[Cr (gly)_{3}]$ has fac and mer isomers, and both are chiral, so there are 4 stereoisomers.

4.4 Octahedral Complexes without Chelates

$M a_{2} b_{2} c_{2}$ , e.g. $[Pt (py)_{2} (N H_{3})_{2} C l_{2}]^{2 +}$ : of its 5 geometrical isomers, only the one with all three pairs cis is chiral, giving 6 stereoisomers.
$Mabcdef$ , e.g. $[Pt (py) (N H_{3}) (N O_{2}) ClBrI]$ : all 15 geometrical isomers are chiral, giving 15 pairs of enantiomers (30 stereoisomers).
$M A_{4} B_{2}$ and $M A_{3} B_{3}$ (cis, trans, fac, mer) always contain a mirror plane, so they are optically inactive.

Exam Trick

Trans kills chirality; chelates create it. A trans pair of identical ligands almost always gives a mirror plane, so trans isomers are inactive. Three identical chelates ( $M (AA)_{3}$ ) are always chiral. Two chelates are chiral only in the cis arrangement.

JEE Advanced

d/l versus $Δ$ / $Λ$ . The labels d and l describe an experiment (the direction of rotation of polarised light) and cannot be read off a drawing. The shape itself is described by $Δ$ (right-handed propeller) and $Λ$ (left-handed propeller). The link between the two is found only by measurement. Enantiomers of complexes are separated by forming salts with a chiral counter ion such as d-tartrate. In 1914 Werner separated the carbon-free complex "hexol", $[Co {(OH)_{2} Co (N H_{3})_{4}}_{3}]^{6 +}$ , into optical isomers, proving that optical activity does not need carbon.

Figure 10: Shape first, then trans pairs, then mirror planes. This order gives the isomer count of any complex in the table below.

Quick Recall: tap to check

Is trans-

[Co (en)_{2} C l_{2}]^{+}

optically active?

No. The plane of Co and the four N atoms is a mirror plane.

How many stereoisomers does

[Cr (ox)_{3}]^{3 -}

have?

Two: the

Δ

and

Λ

enantiomers.

Why can a tetrahedral

M A_{2} B_{2}

complex not show cis-trans isomerism?

Every corner of a tetrahedron is next to every other corner, so no two positions are trans.

Key idea

Chirality needs the absence of a mirror plane:

M (AA)_{3}

is always chiral, cis-

M (AA)_{2} B_{2}

is chiral, trans isomers are not.

5. Isomer Count Table

All counts below were checked by generating every arrangement on the octahedron, square or tetrahedron and removing duplicates. A and B are unidentate ligands; AA is a symmetrical chelate such as en or ox; AB is an unsymmetrical chelate such as gly.

Octahedral type	Example	Geometrical isomers	Total stereoisomers
$M A_{5} B$	$[CoCl (N H_{3})_{5}]^{2 +}$	1	1
$M A_{4} B_{2}$	$[CoC l_{2} (N H_{3})_{4}]^{+}$	2 (cis, trans)	2
$M A_{3} B_{3}$	$[Co (N H_{3})_{3} (N O_{2})_{3}]$	2 (fac, mer)	2
$M A_{4} BC$	$[CoCl (N H_{3})_{4} (N O_{2})]^{+}$	2	2
$M A_{3} B_{2} C$	general type	3	3
$M A_{3} BCD$	general type	4	5
$M A_{2} B_{2} C_{2}$	$[Pt (py)_{2} (N H_{3})_{2} C l_{2}]^{2 +}$	5	6
$M A_{2} B_{2} CD$	general type	6	8
$MABCDEF$	$[Pt (py) (N H_{3}) (N O_{2}) ClBrI]$	15	30
$M (AA)_{3}$	$[Co (en)_{3}]^{3 +}$ , $[Cr (ox)_{3}]^{3 -}$	1	2
$M (AA)_{2} B_{2}$	$[CoC l_{2} (en)_{2}]^{+}$	2	3
$M (AA)_{2} BC$	$[CoCl (N H_{3}) (en)_{2}]^{2 +}$	2	3
$M (AA) B_{2} C_{2}$	$[CoC l_{2} (en) (N H_{3})_{2}]^{+}$	3	4
$M (AB)_{3}$	$[Cr (gly)_{3}]$	2 (fac, mer)	4

For coordination number 4 the shape decides everything:

CN 4 type	Square planar: isomers	Tetrahedral: isomers
$M A_{2} B_{2}$	2 (cis, trans)	1
$M A_{2} BC$	2	1
$MABCD$	3	2 (d and l)
$M (AB)_{2}$	2 (cis, trans)	2 (d and l)

Figure 11: The whole topic on one page. Structural isomers change the bonds; stereoisomers keep the bonds and change only their directions.

6. Solved Examples

Solved Example 1

Name the type of isomerism shown by each pair: (a)

[Cr (N H_{3})_{6}] [Cr (CN)_{6}]

and

[Cr (N H_{3})_{4} (CN)_{2}] [Cr (N H_{3})_{2} (CN)_{4}]

; (b)

[Cr (H_{2} O)_{6}] C l_{3}

and

[CrCl (H_{2} O)_{5}] C l_{2} \cdot H_{2} O

; (c)

[Pt (N H_{3})_{4} B r_{2}] C l_{2}

and

[Pt (N H_{3})_{4} C l_{2}] B r_{2}

; (d)

[Co (N H_{3})_{5} (N O_{2})] C l_{2}

and

[Co (N H_{3})_{5} (ONO)] C l_{2}

Solution:

(a) Both ions are complexes and the ligands are redistributed between them: coordination isomerism.
(b) One water molecule and one chloride swap between the sphere and the crystal: hydrate (solvate) isomerism.
(c) $B r^{-}$ and $C l^{-}$ swap between inside and outside the bracket: ionisation isomerism.
(d) Nitrite binds through N in one and through O in the other: linkage isomerism.

Solved Example 2

How would you distinguish

[Co (N H_{3})_{5} Br] S O_{4}

from

[Co (N H_{3})_{5} (S O_{4})] Br

in the laboratory?

Solution:

They are ionisation isomers, so they release different ions in water.

Add $BaC l_{2}$ solution: $[Co (N H_{3})_{5} Br] S O_{4}$ gives a white precipitate of $BaS O_{4}$ ; the other gives none, because its sulphate is bonded to Co.
Add $AgN O_{3}$ solution: $[Co (N H_{3})_{5} (S O_{4})] Br$ gives a pale yellow precipitate of AgBr; the other gives none, because its bromide is bonded to Co.

Solved Example 3

Complexes A, B and C are three chromium(III) complexes with the empirical formula

H_{12} O_{6} C l_{3} Cr

. All contain water and chloride as ligands. Complex A does not lose weight with concentrated

H_{2} S O_{4}

, while B and C lose 6.75% and 13.5% of their weight. Identify them. (Molar mass of

CrC l_{3} \cdot 6 H_{2} O

= 266.5 g/mol.)

Solution:

Concentrated $H_{2} S O_{4}$ removes only water of crystallisation, not coordinated water.

One $H_{2} O$ : $\frac{18}{266.5} \times 100 = 6.75%$ ; two $H_{2} O$ : $\frac{36}{266.5} \times 100 = 13.5%$
A has no free water: $[Cr (H_{2} O)_{6}] C l_{3}$ (violet)
B has one free water: $[CrCl (H_{2} O)_{5}] C l_{2} \cdot H_{2} O$ (light green)
C has two free water molecules: $[CrC l_{2} (H_{2} O)_{4}] Cl \cdot 2 H_{2} O$ (dark green)

These three are hydrate isomers. With $AgN O_{3}$ they give 3, 2 and 1 mol of AgCl per mole.

Solved Example 4

How many geometrical isomers does the square planar complex

[Pt (N H_{3}) (py) (Cl) (Br)]

have, and is any of them optically active?

Solution:

This is an $MABCD$ square planar complex. Fix $N H_{3}$ at one corner; the corner trans to it can hold py, Cl or Br, and each choice fixes the other two positions (Figure 4):

$N H_{3}$ trans to py
$N H_{3}$ trans to Cl
$N H_{3}$ trans to Br

Answer: 3 geometrical isomers, none optically active, because every square planar complex has the molecular plane as a mirror plane.

Solved Example 5

How many geometrical isomers are possible for (a)

[Cr (C_{2} O_{4})_{3}]^{3 -}

and (b)

[CoC l_{3} (N H_{3})_{3}]

? Which of them are optically active?

Solution:

(a) $M (AA)_{3}$ : three identical chelates can be placed in only one way, so there are no geometrical isomers. The arrangement is chiral, so it exists as a d and l pair (optical isomerism).
(b) $M A_{3} B_{3}$ : 2 geometrical isomers, fac and mer. Both contain a mirror plane, so neither is optically active.

Solved Example 6

A complex of coordination number six has the composition

M a_{2} b_{2} c_{2}

(a, b, c are unidentate). The total number of its geometrical isomers is:
(A) 3
(B) 5
(C) 7
(D) 9

Solution:

Answer: (B). Sort the arrangements by which identical pairs are trans (Figure 7):

all three pairs trans: 1
only a-a trans: 1
only b-b trans: 1
only c-c trans: 1
no pair trans (all cis): 1

That gives 5 geometrical isomers. The all-cis isomer has no mirror plane, so it has a non-superimposable mirror image: the total number of stereoisomers is 6.

Solved Example 7

Which of the following is optically active?
(A) trans-

[CoC l_{2} (en)_{2}]^{+}

(B) cis-

[CoC l_{2} (en)_{2}]^{+}

(C)

[Pt (N H_{3})_{2} C l_{2}]

(D) cis-

[CoC l_{2} (N H_{3})_{4}]^{+}

Solution:

Answer: (B). The cis bis-chelate has no mirror plane (Figure 8). (A) has a mirror plane through the four N atoms; (C) is square planar; (D) has a mirror plane containing both Cl atoms and the metal.

Solved Example 8

Draw (describe) all the isomers of

[CoC l_{2} (en) (N H_{3})_{2}]^{+}

and state which are optically active.

Solution:

This is $M (AA) B_{2} C_{2}$ . Place en on two cis corners. Of the four remaining corners, only one pair is mutually trans. So there are three cases:

the two Cl trans (the two $N H_{3}$ cis): has a mirror plane, inactive
the two $N H_{3}$ trans (the two Cl cis): has a mirror plane, inactive
all cis (both Cl cis and both $N H_{3}$ cis): no mirror plane, chiral, so a d and l pair

Answer: 3 geometrical isomers and 4 stereoisomers in total.

Solved Example 9

Which kind of isomerism is shown by the octahedral compound

Co (N H_{3})_{4} B r_{2} Cl

?
(A) geometrical and ionisation
(B) geometrical and optical
(C) optical and ionisation
(D) geometrical only

Solution:

Answer: (A). The compound can be written in two ways:

$[Co (N H_{3})_{4} B r_{2}] Cl$ and $[Co (N H_{3})_{4} BrCl] Br$ , which are ionisation isomers
each of these is an octahedral $M A_{4} B_{2}$ or $M A_{4} BC$ ion with cis and trans forms

Both types have a mirror plane in every form, so there is no optical isomerism. In all there are 4 isomers.

Solved Example 10

The complex salt

[Co (en)_{2} (SCN) (N O_{2})] Br

can exist in many isomeric forms. How many isomers of all types are possible?

Solution:

Build the count step by step.

The cation $[Co (en)_{2} XY]^{+}$ is $M (AA)_{2} BC$ : trans, d-cis and l-cis, so 3 stereoisomers.
Linkage: $SC N^{-}$ can bind through S or N, and $N O_{2}^{-}$ through N or O: $2 \times 2 = 4$ combinations. So the given composition has $3 \times 4 = 12$ isomers.
Ionisation: $B r^{-}$ can go inside and push out $SC N^{-}$ or $N O_{2}^{-}$ . $[Co (en)_{2} (N O_{2}) Br] SCN$ has $3 \times 2 = 6$ forms (only nitrite is left inside to show linkage isomerism), and $[Co (en)_{2} (SCN) Br] N O_{2}$ also has 6.
Total: $12 + 6 + 6 = 24$ .

Answer: 24 isomers.

Practice Questions: Source Exercise 2

How many stereoisomers are possible for the complex $[Co (en)_{2} C l_{2}]^{+}$ ?Answer: 3: one trans (inactive) and a d/l pair of the cis form.
What type of isomerism is shown by $[Cr (en)_{3}]^{3 +}$ ?Answer: Optical isomerism (d and l); it has no geometrical isomers.
What type of isomerism is shown by $[Pt (N H_{3})_{4}] [PtC l_{4}]$ ?Answer: Coordination isomerism (with $[Pt (N H_{3})_{3} Cl] [Pt (N H_{3}) C l_{3}]$ ) and polymerisation isomerism (with $[Pt (N H_{3})_{2} C l_{2}]$ ).

Practice Questions: MCQ and Quick Checks

$[Co (N H_{3})_{5} (N O_{2})] C l_{2}$ and $[Co (N H_{3})_{5} (ONO)] C l_{2}$ are related as (A) geometrical isomers (B) linkage isomers (C) coordination isomers (D) ionisation isomersAnswer: (B)
The isomerism shown by $[Cr (C_{2} O_{4})_{3}]^{3 -}$ is (A) linkage (B) optical (C) geometrical (D) ionisationAnswer: (B)
Cis-trans isomerism is found in square planar complexes of formula (A) $M A_{4}$ (B) $M A_{3} B$ (C) $M A_{2} B_{2}$ (D) $MA B_{3}$ Answer: (C)
Which of these can show geometrical isomerism? (A) $M A_{3} B$ square planar (B) $M A_{2} B_{2}$ tetrahedral (C) $MABCD$ square planar (D) $MABCD$ tetrahedralAnswer: (C)
fac-mer isomerism is associated with (A) $M (A A^{'})_{2}$ (B) $M (AA)_{3}$ (C) $MABCD$ (D) $M A_{3} X_{3}$ Answer: (D)
Optical isomerism is NOT shown by (A) $[Cr (ox)_{3}]^{3 -}$ (B) cis- $[Co (en)_{2} C l_{2}]^{+}$ (C) trans- $[Co (en)_{2} C l_{2}]^{+}$ (D) $[Cr (en)_{3}]^{3 +}$ Answer: (C)
Which ligand cannot give linkage isomers? (A) $N O_{2}^{-}$ (B) $SC N^{-}$ (C) $C N^{-}$ (D) $N H_{3}$ Answer: (D): $N H_{3}$ has only one donor atom.
Which isomerism is shown by $[Cr (N H_{3})_{2} (OH)_{2} C l_{2}]^{-}$ ? (A) ionisation and optical (B) hydrate and coordination (C) coordination and geometrical (D) geometrical and opticalAnswer: (D): it is $M a_{2} b_{2} c_{2}$ , with 5 geometrical isomers, one of them chiral.
The two complexes $[PtC l_{2} (PE t_{3})]_{2}$ (chloride bridged, with the two terminal Cl on the same side or on opposite sides) show (A) optical (B) coordination (C) geometrical (D) bridged isomerismAnswer: (C)
Which pairs are correctly matched? 1. $[Co (N H_{3})_{5} (N O_{2})] C l_{2}$ / $[Co (N H_{3})_{5} (ONO)] C l_{2}$ : linkage; 2. $[Cu (N H_{3})_{4}] [PtC l_{4}]$ / $[Pt (N H_{3})_{4}] [CuC l_{4}]$ : coordination; 3. $[Pt (N H_{3})_{4} C l_{2}] B r_{2}$ / $[Pt (N H_{3})_{4} B r_{2}] C l_{2}$ : ionisationAnswer: All three (1, 2 and 3).
True or false: (i) octahedral complexes show geometrical isomerism; (ii) tetrahedral complexes show geometrical isomerism; (iii) square planar complexes show geometrical isomerism; (iv) complexes of CN 2 do not show geometrical isomerism.Answer: T, F, T, T
True or false: $[Cr (H_{2} O)_{6}] C l_{3}$ and $[CrC l_{2} (H_{2} O)_{4}] Cl \cdot 2 H_{2} O$ are ionisation isomers.Answer: False: they are hydrate (solvate) isomers.
Fill in the blank: the trans form of an $[M (AA)_{2} a_{2}]^{n +}$ complex does not show ____ isomerism.Answer: optical

Practice Questions: Counting and Drawing

How many geometrical isomers are possible for: (a) $[PtCl (py) (N H_{3}) Br]$ (square planar); (b) $[CoC l_{3} (H_{2} O)_{3}]$ ; (c) $[CoC l_{2} (en)_{2}]^{+}$ ; (d) $[CrC l_{3} (en) (N H_{3})]$ ; (e) $[Pt (py) (N H_{3}) (N O_{2}) ClBrI]$ ?Answer: (a) 3; (b) 2 (fac, mer); (c) 2 (cis, trans); (d) 2; (e) 15
Which structures can show geometrical isomerism: linear, square planar, tetrahedral, octahedral?Answer: Square planar and octahedral only.
How many isomers does $[CoCl (en)_{2} (N H_{3})]^{2 +}$ have, and which are optically active?Answer: 3: trans (inactive) and a d/l pair of the cis form.
Which isomers of $[PtC l_{2} (en)_{2}]^{2 +}$ and $[CrC l_{2} (en) (N H_{3})_{2}]^{+}$ are optically active?Answer: Only cis- $[PtC l_{2} (en)_{2}]^{2 +}$ ; only the all-cis form of $[CrC l_{2} (en) (N H_{3})_{2}]^{+}$ .
Describe the structures of cis- $[CrC l_{2} (CN)_{4}]^{3 -}$ , mer- $[CoC l_{3} (N H_{3})_{3}]$ and fac- $[Co (N O_{2})_{3} (H_{2} O)_{3}]$ .Answer: Octahedral in each case: the two Cl at 90°; the three Cl on a meridian (two of them trans); the three $N O_{2}$ on one face (all at 90°).
Give one example each of ionisation, linkage and coordination isomerism.Answer: $[PtC l_{2} (N H_{3})_{4}] B r_{2}$ / $[PtB r_{2} (N H_{3})_{4}] C l_{2}$ ; $[Co (N H_{3})_{5} (N O_{2})] C l_{2}$ / $[Co (N H_{3})_{5} (ONO)] C l_{2}$ ; $[Co (H_{2} O)_{6}] [Cr (CN)_{6}]$ / $[Cr (H_{2} O)_{6}] [Co (CN)_{6}]$ .
How many geometrical isomers does $[Co (N H_{3})_{4} C l_{2}]^{+}$ have? Name them and give their colours.Answer: 2: cis (violet) and trans (green).
How many stereoisomers are possible for tris(glycinato)chromium(III), $[Cr (gly)_{3}]$ ?Answer: 4: fac and mer, each as a d/l pair.
Does bis(glycinato)platinum(II), $[Pt (gly)_{2}]$ , show geometrical or optical isomerism?Answer: Geometrical only (cis and trans); being square planar, it is optically inactive.
Why is bis(benzoylacetonato)beryllium(II) optically active although it has no carbon stereocentre?Answer: It is a tetrahedral $M (AB)_{2}$ complex with no mirror plane, so the complex as a whole is chiral.

Common Mistakes to Avoid

Watch out

Looking for cis and trans isomers in a tetrahedral complex. All four positions are adjacent, so tetrahedral complexes have no geometrical isomers.
Calling a square planar complex optically active. Its molecular plane is a mirror plane.
Marking trans- $[CoC l_{2} (en)_{2}]^{+}$ as chiral. Only the cis form is chiral.
Drawing a "meso" form of $[Co (en)_{3}]^{3 +}$ or $[Cr (ox)_{3}]^{3 -}$ . $M (AA)_{3}$ has only a d and an l form, and no geometrical isomers.
Counting a pair of enantiomers as two geometrical isomers. $M a_{2} b_{2} c_{2}$ has 5 geometrical isomers but 6 stereoisomers; read the question carefully.
Mixing up ionisation and hydrate isomers. If water moves in or out of the sphere, it is hydrate (solvate) isomerism.
Expecting linkage isomers with $N H_{3}$ , $H_{2} O$ or $C l^{-}$ . Only ambidentate ligands ( $N O_{2}^{-}$ , $SC N^{-}$ , $C N^{-}$ ) give them.
Proposing coordination isomers when only one ion is complex. Both the cation and the anion must be complex ions.
Deciding from a drawing which enantiomer is d and which is l. The sign of rotation is found only by experiment.

Frequently Asked Questions

What are the types of isomerism in coordination compounds?

There are two families. Structural isomerism, where different atoms are bonded to the metal, includes ionisation, solvate (hydrate), linkage, coordination, polymerisation and coordination position isomerism. Stereoisomerism, where the same bonds point in different directions, includes geometrical (cis-trans, fac-mer) and optical isomerism.

Why do tetrahedral complexes not show geometrical isomerism?

In a tetrahedron every ligand position is at 109.5 degrees to each of the other three, so no two positions are opposite each other. Any two ligands are always neighbours, and every arrangement of an MA2B2 complex can be rotated into every other. So cis and trans forms cannot exist.

Why are square planar complexes optically inactive?

A square planar complex lies flat, so the plane containing the metal and the four donor atoms is a mirror plane. Reflecting the complex in a mirror gives exactly the same complex, which can be superimposed on the original. It therefore cannot exist as a pair of enantiomers.

What is the difference between fac and mer isomers?

Both are isomers of octahedral MA3B3 complexes such as triamminetrinitrito-N-cobalt(III). In the fac (facial) isomer the three identical ligands sit on one triangular face, all at 90 degrees. In the mer (meridional) isomer they lie in a plane through the metal, so two of them are trans at 180 degrees.

How can ionisation isomers be distinguished?

Ionisation isomers release different ions in water, so simple precipitation tests work. For the two pentaamminecobalt(III) salts, barium chloride gives white barium sulphate only with the one that has sulphate outside the bracket, and silver nitrate gives pale yellow silver bromide only with the one that has bromide outside.

Which complexes show optical isomerism?

A complex shows optical isomerism when it has no mirror plane and no centre of symmetry. Common examples are tetrahedral Mabcd, octahedral M(AA)3 such as tris(ethane-1,2-diamine)cobalt(III), the cis forms of M(AA)2B2 and M(AA)2BC, and the all-cis form of Ma2b2c2. Square planar complexes are never chiral.

How are isomer counting questions asked in JEE Main and JEE Advanced?

JEE Main usually asks for the number of geometrical isomers of a given complex or which complex is optically active. JEE Advanced often asks for the total number of isomers, combining stereoisomers with linkage and ionisation isomers, or gives the answer as an integer, so every enantiomer pair must be counted.

Which isomerism questions are common in NEET?

NEET mostly asks you to identify the type of isomerism between two given formulas, to pick the complex that shows optical or geometrical isomerism, and to recall facts such as cisplatin being the cis isomer and tetrahedral complexes having no geometrical isomers. Ionisation, linkage and hydrate pairs appear most often.

Previous year questions on Isomerism In Co-ordination Compounds

17 questions from past papers, each with a step-by-step solution.

Show all 17 questions

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