Coordination Compounds: Class 12 Chemistry · Chapter 5

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

100 min
46 practice questions
36 MCQs

What you will learn in Coordination Compounds

Use Werner's primary and secondary valences to predict how many ions a coordination compound gives in water
Identify the coordination entity, central atom, ligands and their denticity, coordination number, coordination sphere, polyhedron and oxidation state of a complex
Write formulas and IUPAC names of mononuclear coordination compounds
Recognise geometrical, optical, linkage, coordination, ionisation and solvate isomers and count the stereoisomers of simple complexes
Explain the shapes and magnetic behaviour of complexes with valence bond theory and calculate spin-only magnetic moments
Use crystal field theory to split the d orbitals, choose between high-spin and low-spin arrangements and explain the colours of complexes
Describe the synergic bonding that holds carbon monoxide to a metal in metal carbonyls
Give examples of coordination compounds at work in analysis, metallurgy, living things, industry and medicine

Original companion text aligned to NCERT Chemistry Class 12, Chapter 5, pp. 118–140. Not affiliated with or endorsed by NCERT.

Silver chloride is one of the least soluble salts you will meet, yet a curdy white precipitate of it vanishes when you shake it with aqueous ammonia. Gold shrugs off almost every acid, yet mining companies dissolve specks of it out of crushed rock with a dilute cyanide solution and air. In both cases a metal ion has gathered a definite set of molecules or ions around itself and become a new, soluble species.

Such species are the subject of this unit. A coordination compound (उपसहसंयोजन यौगिक) contains a metal atom or ion joined to a fixed number of neighbours that each donate a pair of electrons. You will see how chemists first made sense of them and how to name them. You will find out why some come in several forms with one formula, and why many are brightly coloured or magnetic. Finally you will meet them at work, from the green of leaves to anticancer drugs.

5.1 Werner's Theory of Coordination Compounds

By the late 1800s chemists had made dozens of crystalline solids by mixing cobalt(III) chloride with ammonia. They wrote them with a dot, as $C o C l_{3} \cdot 6 N H_{3}$ , $C o C l_{3} \cdot 5 N H_{3}$ and $C o C l_{3} \cdot 4 N H_{3}$ , which hid a puzzle. Both $C o C l_{3}$ and $N H_{3}$ are stable compounds with every valence already used, so what holds them together, and does the chlorine sit in the same place in each solid?

A simple test gives the answer. Silver nitrate added to a cold solution instantly removes free chloride ions as white $A g C l$ , but it cannot pull off a chlorine atom that is bonded directly to the metal. Counting the moles of $A g C l$ per mole of compound therefore counts the free chloride. Measuring the electrical conductivity of the solution gives a second, independent count: the total number of ions each formula unit releases.

2/3 $C o C l_{3} \cdot 5 N H_{3}$ : one chlorine takes the sixth place, leaving two free chloride ions and 2 mol $A g C l$ .

Fig. 5.1Six groups always surround the cobalt; only the chloride left outside the sphere reaches the silver ions.NCERT Table 5.1 · p. 119

A cobalt ion sits inside a large dashed circle labelled coordination sphere, with six bonded positions around it. On the right, free chloride ions float outside, an arrow labelled add silver nitrate points down, and the moles of silver chloride formed are given. In step 1 all six positions hold ammonia, three chloride ions are free and 3 mol of silver chloride forms; the formula reads Co(NH3)6 in brackets with Cl3 outside. In step 2 one chlorine has moved inside, two chloride ions remain free and 2 mol of silver chloride forms. In step 3 two chlorines are inside, one chloride ion is free and only 1 mol of silver chloride forms.

The pattern is striking. Whenever an ammonia molecule is missing, a chlorine atom moves in to take its place next to the cobalt, and that chlorine stops behaving as a free ion. The total of ammonia molecules and bonded chlorines is always six. Conductivity agrees: the first compound gives four ions per formula unit, the second three and the third two.

Alfred Werner explained these results in 1893. He proposed that a metal ion has two kinds of combining power. The primary valence is what we now call the oxidation state; it is satisfied by negative ions and is ionisable. The secondary valence is the number of groups attached directly to the metal, today called the coordination number. In modern terms Werner's ideas read as follows.

Every metal ion in such a compound has two separate combining powers, a primary one and a secondary one.
The primary valence is met by anions only, and those not needed elsewhere float free when the solid dissolves.
The secondary valence counts the groups that stay bonded in solution: molecules such as $N H_{3}$ , anions such as $C l^{-}$ , or both. Each metal ion has its own characteristic number, six for cobalt(III).
Secondary bonds point in fixed directions, so a complex has a definite shape, and each coordination number favours particular shapes.

Today the bracket in a formula such as $[C o (N H_{3})_{6}] C l_{3}$ marks off everything held by secondary valences: the coordination entity. The ions left outside are counter ions, and the shape traced by the bonded atoms is the coordination polyhedron. Counting isomers led Werner to conclude that six-coordinate entities such as $[C o (N H_{3})_{6}]^{3 +}$ are octahedral. Four-coordinate entities turned out to be of two kinds: tetrahedral, like $[Z n (N H_{3})_{4}]^{2 +}$ , and square planar, like $[P t (N H_{3})_{4}]^{2 +}$ .

Example 5.1NCERT Example 5.1 · p. 120

Secondary valence from a silver nitrate test

Each compound below was dissolved in water and treated with excess silver nitrate. Assume every neutral molecule in the formula is bonded to the metal. Find the secondary valence and write Werner's formula.

Table 5.1

Compound	mol AgCl per mol
violet $C r C l_{3} \cdot 6 H_{2} O$	3
$C r C l_{3} \cdot 5 N H_{3}$	2
$C u C l_{2} \cdot 4 N H_{3}$	2
$I r C l_{3} \cdot 3 N H_{3}$	0
$P t C l_{4} \cdot 4 N H_{3}$	2

Try it yourself first. Then reveal the solution one step at a time.

Compare

Double salt or complex?

Double salt

Exists as one compound only in the crystal.
Falls apart completely into its simple ions in water.
Ammonium alum, $N H_{4} A l (S O_{4})_{2} \cdot 12 H_{2} O$ , gives the normal tests for $N H_{4}^{+}$ , $A l^{3 +}$ and $S O_{4}^{2 -}$ .

Complex

The coordination entity survives in solution.
Only the counter ions are released.
$K_{3} [F e (C N)_{6}]$ gives $K^{+}$ and $[F e (C N)_{6}]^{3 -}$ ; thiocyanate shows no blood-red colour, so no free $F e^{3 +}$ is present.

Quick check

1 of 4

One mole of

[R h C l_{2} (N H_{3})_{4}] C l

is treated with excess silver nitrate. How many moles of

A g C l

form?

1230

Show answer

Answer (A) 1

Only the chloride outside the bracket is free, so 1 mol

A g C l

forms. Counting all three chlorines (answer 3) ignores the two bonded to rhodium.

5.2 Definitions of Some Important Terms Pertaining to Coordination Compounds

A handful of words lets us describe any complex precisely. We will lean on a few laboratory examples. One is the deep-blue $[C u (N H_{3})_{4}]^{2 +}$ ion that forms when ammonia is added to copper sulphate solution. Others are the $[A g (N H_{3})_{2}]^{+}$ ion in Tollens' reagent and $K_{3} [C o (N O_{2})_{6}]$ , the yellow precipitate whose appearance reveals potassium ions.

(a) Coordination entity and (b) central atom

Definition

Coordination entity

The metal at the heart of a complex together with every group attached directly to it, written inside square brackets. The whole unit may carry a positive charge, a negative charge or none.

Example $[A g (N H_{3})_{2}]^{+}$ (cation), $[Z n (O H)_{4}]^{2 -}$ (anion), $[F e (C O)_{5}]$ (neutral).

The metal at the centre is the central atom or ion. It holds its neighbours in a definite arrangement and accepts an electron pair from each, so it behaves as a Lewis acid. In $[A g (N H_{3})_{2}]^{+}$ the central ion is $A g^{+}$ ; in $[Z n (O H)_{4}]^{2 -}$ it is $Z n^{2 +}$ ; in $[F e (C O)_{5}]$ it is an iron atom with oxidation state zero.

(c) Ligands

Definition

Ligand

An ion or molecule bonded to the central atom through one or more donor atoms, each of which supplies a lone pair. Ligands act as Lewis bases.

Example $C l^{-}$ , $O H^{-}$ , $H_{2} O$ , $N H_{3}$ , $C O$ , ethane-1,2-diamine (en).

Size does not decide how a ligand binds; the number of donor atoms it uses does, and this number is its denticity. Ammonia uses one nitrogen, so it is unidentate. Ethane-1,2-diamine, $H_{2} N C H_{2} C H_{2} N H_{2}$ (abbreviated en), grips with both nitrogen atoms, and the glycinate ion, $H_{2} N C H_{2} C O O^{-}$ , with one nitrogen and one oxygen, so both are didentate. A polydentate ligand uses more: the ethylenediaminetetraacetate ion, $E D T A^{4 -}$ , holds a metal through two nitrogen and four oxygen atoms, so it is hexadentate (Fig. 5.18). Even a protein can act as a ligand, gripping a metal ion through several side-chain atoms at once.

2/3en binds through both nitrogens and closes a five-membered chelate ring.

Fig. 5.2Donor atoms decide how a ligand binds: one atom, two atoms closing a ring, or a choice of two different atoms.NCERT p. 121

A metal atom M sits at the left. In step 1 an ammonia molecule, a nitrogen with three hydrogens, donates its lone pair to M along a single arrow, labelled unidentate, one donor atom. In step 2 an ethane-1,2-diamine molecule binds through both nitrogen atoms; the metal, the two nitrogens and the two carbons enclose a shaded five-membered ring, labelled didentate. In step 3 the single metal is replaced by two small panels: in the top one a nitrite ion bonds through its nitrogen (nitrito-N), and in the bottom one it bonds through an oxygen (nitrito-O).

Definition

Chelate

The ring closed when one ligand grips the same metal ion with two or more of its donor atoms; such a ligand is a chelating ligand, from the Greek word for a crab's claw.

Example $[N i (e n)_{3}]^{2 +}$ has three five-membered chelate rings.

Chelate complexes are usually much more stable than similar complexes made from unidentate ligands with the same donor atoms; this is the chelate effect. Breaking one metal–nitrogen bond in $[N i (e n)_{3}]^{2 +}$ does not release the ligand, because its other end still holds on, so the free end is likely to reattach.

Some ligands offer two different donor atoms and can use either one. These are ambidentate ligands. Nitrite, $N O_{2}^{-}$ , can bond through nitrogen or through oxygen, and thiocyanate, $S C N^{-}$ , through sulphur or nitrogen. Each choice gives a different compound, as Section 5.4.3 shows.

(d) Coordination number and (e) coordination sphere

Definition

Coordination number (CN)

How many donor atoms are joined directly to the central atom by $σ$ bonds. Any $π$ bonds between metal and ligand do not add to the count.

ExampleCN = 2 in $[A g (N H_{3})_{2}]^{+}$ , 4 in $[C u (N H_{3})_{4}]^{2 +}$ , 6 in $[C r (C_{2} O_{4})_{3}]^{3 -}$ .

Count donor atoms, not ligands. Three oxalate ions in $[C r (C_{2} O_{4})_{3}]^{3 -}$ supply six oxygen donors, so the coordination number is 6, and one $E D T A^{4 -}$ alone gives a coordination number of 6 in $[C a (E D T A)]^{2 -}$ .

Everything inside the square brackets, the central atom with its ligands, forms the coordination sphere. Dissolve $K_{3} [C o (N O_{2})_{6}]$ and the bracketed unit $[C o (N O_{2})_{6}]^{3 -}$ stays intact, while its three $K^{+}$ counter ions drift off on their own.

(f) Coordination polyhedron

Join the donor atoms around the metal with imaginary lines and you get the coordination polyhedron. The commonest are octahedral (CN 6), tetrahedral and square planar (both CN 4); trigonal bipyramidal and square pyramidal shapes occur for CN 5. Fig. 5.3 builds each one in turn.

4/4Six ligands along $\pm x$ , $\pm y$ and $\pm z$ , the most common shape of all.

Fig. 5.3Each coordination number has one or two typical shapes.NCERT Fig. 5.1 · p. 122

A metal atom M with ligands L drawn in perspective, with dashed lines outlining the polyhedron. Step 1 shows four ligands at the corners of a tetrahedron, coordination number 4, angles 109.5 degrees, example tetraamminezinc(II). Step 2 shows four ligands at the corners of a square in one plane, coordination number 4, angles 90 degrees, example tetrachloridoplatinate(II). Step 3 shows a trigonal bipyramid, three ligands round the middle and one above and below, coordination number 5, example pentacarbonyliron(0). Step 4 shows an octahedron, four ligands in a square plus one above and one below, coordination number 6, example hexaaquachromium(III).

(g) Oxidation number of the central atom and (h) homoleptic and heteroleptic complexes

Hand every shared electron pair back to the ligand that donated it: the charge the metal is left holding is its oxidation number. In practice you find it from charge balance: oxidation number of the metal + total charge of the ligands = charge on the entity. It is written as a Roman numeral in brackets after the metal's name: cobalt in $[C o C l_{4}]^{2 -}$ is cobalt(II), since $x + 4 (- 1) = - 2$ .

A complex with only one kind of ligand, such as $[C u (N H_{3})_{4}]^{2 +}$ or $[F e (C O)_{5}]$ , is homoleptic. One with two or more kinds, such as $[C r C l (N H_{3})_{5}]^{2 +}$ , is heteroleptic.

Example

Oxidation number, coordination number and ligand types

For each entity find the oxidation number of the metal and its coordination number, and say whether it is homoleptic or heteroleptic: (a) $K_{3} [C o (N O_{2})_{6}]$ (b) $[F e (C O)_{5}]$ (c) $[C r C l (e n)_{2} (H_{2} O)]^{2 +}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

What is the coordination number of chromium in

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

3496

Show answer

Answer (D) 6

Each oxalate is didentate, so three of them supply 6 donor atoms. Answering 3 counts ligands instead of donor atoms.

5.3 Nomenclature of Coordination Compounds

Two different compounds can share a formula, and one compound can be drawn in several ways, so chemists need names and formulas that leave no room for doubt. IUPAC, the international body that standardises chemical names, sets the rules. Learn them as two short checklists, one for writing formulas and one for saying names.

5.3.1 Formulas of Mononuclear Coordination Entities

Most complexes in this unit are mononuclear: one metal atom sits at the centre. Their formulas follow a fixed recipe.

Open a square bracket and write the symbol of the central atom.
Add the ligands in alphabetical order of their formulas or abbreviations, whatever their charges: Cl before $N H_{3}$ , en before $N O_{2}$ . An abbreviation such as en or py is filed under its first letter.
Put each polyatomic ligand and each abbreviation in round brackets, and leave no spaces anywhere inside the square brackets.
Close the square bracket. A charged entity written on its own carries its charge as a right superscript, number first: $[C u (N H_{3})_{4}]^{2 +}$ .
For a salt, write the cation before the anion and take enough of each to balance the charges.

5.3.2 Naming of Mononuclear Coordination Compounds

A name is assembled like a train: the ligand names in front, the metal behind them and its oxidation state at the end. Chemists call this additive nomenclature. The steps:

Give each ligand its ligand name. Anions end in -ido, -ato or -ito (chlorido, cyanido, oxalato, nitrito); neutral molecules keep their own names, except aqua ( $H_{2} O$ ), ammine ( $N H_{3}$ ), carbonyl ( $C O$ ) and nitrosyl ( $N O$ ).
Arrange the ligand names alphabetically. Multiplying prefixes do not count: tetraammine is filed under a.
Show how many of each ligand with di, tri, tetra …; switch to bis, tris, tetrakis … with the ligand name in brackets when the name already contains a number or is long, as in bis(ethane-1,2-diamine).
Write the metal straight after the ligands. In an anionic entity its name ends in -ate (cobaltate, nickelate, zincate) and some metals use their Latin roots (see the table); in a cation or a neutral molecule it keeps its ordinary name.
End with the oxidation state of the metal in Roman numerals in brackets, written (0) for zero.
In a salt, say the cation before the anion, whichever of them is the complex. A neutral entity has no counter ion and is named on its own.

Table 5.2Names of common ligands and their donor atoms (the formula shows each charge)

Ligand	Name	Donor
$H_{2} O$	aqua	O
$N H_{3}$	ammine	N
$C O$	carbonyl	C
$N O$	nitrosyl	N
$C_{5} H_{5} N$ (py)	pyridine	N
en	ethane-1,2-diamine	2 N
$C l^{-}$ , $B r^{-}$ , $I^{-}$	chlorido, bromido, iodido	halogen
$C N^{-}$	cyanido	C
$O H^{-}$	hydroxido	O
$N O_{2}^{-}$	nitrito-N or nitrito-O	N or O
$S C N^{-}$	thiocyanato-S or thiocyanato-N	S or N
$S_{2} O_{3}^{2 -}$	thiosulphato	S
$C_{2} O_{4}^{2 -}$ (ox)	oxalato	2 O
$C O_{3}^{2 -}$	carbonato	O
$E D T A^{4 -}$	ethylenediaminetetraacetato	2 N, 4 O

Table 5.3Metal names inside anionic entities

Metal	Cation or neutral	Anion
Fe	iron	ferrate
Cu	copper	cuprate
Ag	silver	argentate
Au	gold	aurate
Pb	lead	plumbate
Sn	tin	stannate
Co	cobalt	cobaltate
Ni	nickel	nickelate
Pt	platinum	platinate
Zn	zinc	zincate

Fig. 5.4 applies the rules to sodium cobaltinitrite, $N a_{3} [C o (N O_{2})_{6}]$ , the reagent that precipitates potassium ions. The same metal can even appear twice under different names. Magnus's green salt, $[P t (N H_{3})_{4}] [P t C l_{4}]$ , is tetraammineplatinum(II) tetrachloridoplatinate(II): platinum keeps its plain name in the cation and becomes platinate in the anion.

4/4 $x + 6 (- 1) = - 3$ gives cobalt(III): sodium hexanitrito-N-cobaltate(III).

Fig. 5.4Every part of the formula maps onto a part of the name.NCERT p. 124

The formula Na3 Co NO2 six appears at the top in a shaded frame. Below it, a left column lists the pieces of the formula: Na3, (NO2)6, Co in an anion, and the charge sum x plus 6 times minus 1 equals minus 3. Arrows lead to the matching name pieces on the right. Step 1 shows sodium, with the note that the cation is named first. Step 2 adds hexanitrito-N for six nitrogen-bonded nitrite ions. Step 3 adds cobaltate, because the metal is in an anion. Step 4 adds (III) from the charge balance, and the full name sodium hexanitrito-N-cobaltate(III) appears at the bottom.

Example 5.2NCERT Example 5.2 · p. 124

From names to formulas

Write formulas for (a) tetraamminecopper(II) sulphate (b) potassium dicyanidoaurate(I) (c) pentaamminechloridochromium(III) chloride (d) triammineaquadichloridocobalt(III) chloride (e) pentacarbonyliron(0).

Try it yourself first. Then reveal the solution one step at a time.

Example 5.3NCERT Example 5.3 · p. 124

From formulas to names

Name (a) $K_{2} [H g I_{4}]$ (b) $[C r C l_{2} (H_{2} O)_{4}] C l$ (c) $N a_{3} [A g (S_{2} O_{3})_{2}]$ (d) $[C o (N H_{3})_{6}] [F e (C N)_{6}]$ (e) $[C r (H_{2} O)_{5} (N C S)]^{2 +}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which is the correct name of

K_{4} [F e (C N)_{6}]

potassium hexacyanidoiron(II)potassium hexacyanidoferrate(II)potassium hexacyanidoferrate(III)tetrapotassium hexacyanoiron

Show answer

Answer (B) potassium hexacyanidoferrate(II)

The entity is an anion, so iron becomes ferrate;

x + 6 (- 1) = - 4

gives (II). Plain 'iron' is used only in cations and neutral entities.

Practice · 2 questions

Practice: formulas and names: questions with answers

1
Short answerEasyNCERT Intext Question 5.1
Give the formula of each compound.
1. hexaaquairon(II) sulphate
  Show solution
  Fe(II) with six neutral water ligands gives a 2+ entity, balanced by one sulphate: $[F e (H_{2} O)_{6}] S O_{4}$ .
2. sodium hexafluoridoaluminate(III)
  Show solution
  $3 + 6 (- 1) = - 3$ , so three $N a^{+}$ : $N a_{3} [A l F_{6}]$ , the mineral cryolite used in the extraction of aluminium.
3. bis(ethane-1,2-diamine)oxalatocobalt(III) chloride
  Show solution
  Oxalate is −2 and en is neutral, so the entity is $3 - 2 = + 1$ , balanced by one chloride. $C_{2} O_{4}$ (C) is listed before en (e): $[C o (C_{2} O_{4}) (e n)_{2}] C l$ .
4. potassium amminetrichloridoplatinate(II)
  Show solution
  $2 + 3 (- 1) + 0 = - 1$ , so one $K^{+}$ : $K [P t C l_{3} (N H_{3})]$ .
5. hexaamminenickel(II) hexacyanidocobaltate(III)
  Show solution
  The cation is $[N i (N H_{3})_{6}]^{2 +}$ and the anion $[C o (C N)_{6}]^{3 -}$ . Three cations (6+) balance two anions (6−), so the formula is $[N i (N H_{3})_{6}]_{3}$ $[C o (C N)_{6}]_{2}$ , written without a space.
6. copper(II) hexacyanidoferrate(II)
  Show solution
  The anion $[F e (C N)_{6}]^{4 -}$ needs two $C u^{2 +}$ : $C u_{2} [F e (C N)_{6}]$ , the brown precipitate used to test for copper ions.
2
Short answerEasyNCERT Intext Question 5.2
Name each compound by the IUPAC rules.
1. $[A g (N H_{3})_{2}] C l$
  Show solution
  Diamminesilver(I) chloride; the same cation is the active part of Tollens' reagent.
2. $[C o B r (N H_{3})_{5}] S O_{4}$
  Show solution
  $x - 1 = + 2$ gives Co(III): pentaamminebromidocobalt(III) sulphate.
3. $L i [A l H_{4}]$
  Show solution
  Hydride, $H^{-}$ , is named hydrido. $x + 4 (- 1) = - 1$ gives Al(III), and aluminium in an anion becomes aluminate: lithium tetrahydridoaluminate(III), the reducing agent known as lithium aluminium hydride.
4. $[C u (e n)_{2}] S O_{4}$
  Show solution
  en is neutral, so $x = + 2$ : bis(ethane-1,2-diamine)copper(II) sulphate.
5. $K_{2} [P t C l_{6}]$
  Show solution
  $x - 6 = - 2$ gives Pt(IV): potassium hexachloridoplatinate(IV).
6. $[P t (N H_{3})_{3} (p y)] C l_{2}$
  Show solution
  Ammine before pyridine: triammine(pyridine)platinum(II) chloride.

5.4 Isomerism in Coordination Compounds

Two compounds that share a formula yet differ in how their atoms are put together are isomers (समावयवी). The rearrangement shows up as a different colour, solubility or reactivity, and sometimes as a completely different effect on living cells. Complexes show isomerism in two broad ways.

Stereoisomerism: the same atoms are bonded to the same partners, but they point in different directions in space.
- geometrical isomerism
- optical isomerism
Structural isomerism: the bonds themselves differ, because a ligand uses another donor atom or groups swap places.
- linkage isomerism
- coordination isomerism
- ionisation isomerism
- solvate isomerism

5.4.1 Geometric Isomerism

Two compounds share the formula $[P t C l_{2} (N H_{3})_{2}]$ . One, cisplatin, is among the most widely used anticancer drugs; the other has no useful effect on tumours. The only difference is where the ligands sit around the square planar platinum. In the cis isomer the two chloride ions are neighbours, 90° apart; in the trans isomer they are opposite each other, 180° apart. This kind of isomerism needs a heteroleptic complex and is common for coordination numbers 4 and 6.

For a square planar complex $[M A_{2} B_{2}]$ , where A and B are unidentate, there are exactly two arrangements. With four different ligands, $[M A B C D]$ , count by asking which ligand sits opposite A: it can be B, C or D, so there are three geometrical isomers.

2/3Swap one A with one B and the A ligands end up opposite each other, 180° apart.

Fig. 5.5Cis and trans differ in the angle between like ligands; a tetrahedron has no such choice.NCERT Fig. 5.2 · p. 125

Step 1 shows a top view of a square planar complex: a metal Pt at the centre with two ligands A on adjacent corners (right and top) and two ligands B on the other two corners; the text says cis isomer, A–M–A angle 90 degrees, the drug cisplatin. In step 2 the top A and the left B swap places so the two A ligands lie opposite each other; the text says trans isomer, angle 180 degrees. Step 3 replaces the square with a tetrahedron around a zinc atom carrying two A and two B ligands; the text says every angle is 109.5 degrees, so there is only one form.

Tetrahedral complexes never show geometrical isomerism, because every corner of a tetrahedron is adjacent to all the others. Octahedral complexes do. In $[M A_{4} B_{2}]$ , such as $[C r C l_{2} (N H_{3})_{4}]^{+}$ , the two B ligands can be cis or trans. A didentate ligand AA must span two cis positions, so $[M (A A)_{2} B_{2}]$ , such as $[R h C l_{2} (e n)_{2}]^{+}$ , also has cis and trans forms. With three of each ligand, $[M A_{3} B_{3}]$ , there are again two forms. In the facial (fac) isomer the three B ligands occupy one triangular face of the octahedron. In the meridional (mer) isomer they lie in a plane that passes through the metal.

3/4 $[M A_{3} B_{3}]$ with the three B ligands on one face.

Fig. 5.6The same six positions give cis and trans forms of

[M A_{4} B_{2}]

, and fac and mer forms of

[M A_{3} B_{3}]

.NCERT Fig. 5.3, 5.5 · p. 126

An octahedral complex drawn in perspective: a metal M with six ligand balls on dashed octahedron edges. Balls marked A are blue and balls marked B are orange, so letters as well as colours tell them apart. Step 1, cis: two B ligands sit on neighbouring corners, 90 degrees apart, example tetraamminedichloridochromium(III). Step 2, trans: the two B ligands are at the top and bottom, 180 degrees apart. Step 3, fac: three B ligands occupy the three corners of one face. Step 4, mer: three B ligands lie in one plane through the metal, two of them opposite each other, example triamminetrichloridochromium(III).

Example 5.4NCERT Example 5.4 · p. 126

Testing a claim about tetrahedral isomers

A student reports that a four-coordinate zinc complex $[Z n C l_{2} (N H_{3})_{2}]$ exists as two geometrical isomers. Use the geometry of a tetrahedron to check the claim, and suggest what the result would mean if two isomers really were found for a four-coordinate complex of this type.

Try it yourself first. Then reveal the solution one step at a time.

Example 5.5NCERT Example 5.5 · p. 127

Geometrical isomers of an octahedral MA4B2 complex

Draw the geometrical isomers of $[C r C l_{2} (H_{2} O)_{4}]^{+}$ and say which one has no overall dipole moment.

Try it yourself first. Then reveal the solution one step at a time.

5.4.2 Optical Isomerism

Your left and right hands are mirror images, yet no amount of turning makes one fit exactly over the other. An object like this is chiral. A chiral complex exists as two enantiomers, mirror images that cannot be superimposed. Enantiomers have identical melting points, solubilities and colours, but they rotate the plane of plane-polarised light by equal angles in opposite directions, which a polarimeter measures. The form that rotates the plane to the right is dextro (d), the other laevo (l). A 50 : 50 mixture of the two, a racemic mixture, shows no rotation at all.

Didentate ligands are the usual source of chirality in six-coordinate complexes. Every $[M (A A)_{3}]$ complex, such as $[N i (e n)_{3}]^{2 +}$ or $[C o (C_{2} O_{4})_{3}]^{3 -}$ , is chiral: looked at down its threefold axis, its three rings form a propeller that twists either one way or the other. With two chelate rings, $[M (A A)_{2} B_{2}]$ , only the cis isomer is chiral. The trans isomer has a mirror plane through the metal and both B ligands, so its mirror image is the same molecule.

2/3The mirror image twists anticlockwise (Λ).

Fig. 5.7A tris-chelate complex and its mirror image twist in opposite senses, so they are enantiomers.NCERT Fig. 5.6 · p. 126

The complex is viewed down its threefold axis: six donor atoms form a hexagon around the metal M, solid circles for the three nearer atoms and dashed circles for the three farther ones. Three thick orange bands, one per didentate ligand, each join a near atom to a far atom like the blades of a propeller, and a curved arrow shows the twist. Step 1 shows only the left-hand form, labelled delta, twisting clockwise. Step 2 adds a dashed mirror line and the mirror-image form, labelled lambda, twisting anticlockwise. Step 3 slides the mirror image on top of the original: the bands cross instead of matching, and the text says the blades never line up.

InteractiveIsomer explorer

An interactive octahedron drawn in perspective with a metal at the centre and six ligand balls marked with letters A, B and C, or a curved band for a didentate ligand AA. A menu chooses a formula type such as MA4B2, MA3B3, MA2B2C2, M(AA)3, M(AA)2B2 or M(AA)B2C2. The explorer works out every distinct stereoisomer by trying all 24 rotations of the octahedron, lists them as buttons with names such as cis, trans, fac, mer or chiral pair, and shows a summary of how many geometrical isomers and how many stereoisomers there are. Arrow buttons and dragging turn the model, and a Mirror button swaps the selected isomer for its mirror image, with a status line saying whether the mirror image is the same molecule or a different enantiomer.

Fig. 5.8Pick a formula type to list every stereoisomer, then turn the octahedron to test it against its mirror image.

Example 5.6NCERT Example 5.6 · p. 127

Which entity is chiral?

Which of these is optically active: (a) cis- $[C o B r C l (e n)_{2}]^{+}$ or (b) trans- $[C o B r C l (e n)_{2}]^{+}$ ?

Try it yourself first. Then reveal the solution one step at a time.

5.4.3 Linkage Isomerism

An ambidentate ligand can attach through either of its two donor atoms, and each choice gives a different compound. Thiocyanate bonded through sulphur gives $[C o (N H_{3})_{5} (S C N)]^{2 +}$ , while bonded through nitrogen it gives $[C o (N H_{3})_{5} (N C S)]^{2 +}$ . Nitrite behaves the same way, and its pentaamminecobalt(III) complexes were the first linkage isomers to be recognised, in the 1890s. On standing, the O-bonded (nitrito-O) form slowly rearranges into the more stable N-bonded (nitrito-N) one.

5.4.4 Coordination Isomerism

When both the cation and the anion of a salt are complexes, the ligands can be shared out between the two metals in different ways. In $[C u (N H_{3})_{4}] [P t C l_{4}]$ the ammonia molecules are bonded to copper and the chloride ions to platinum; in its coordination isomer $[P t (N H_{3})_{4}] [C u C l_{4}]$ the partners are exchanged. The two salts have the same overall formula but different colours and chemistry.

5.4.5 Ionisation Isomerism

Sometimes the ion outside the bracket could perfectly well bind to the metal too. Swap it with one of the bonded ligands and you get a second compound with exactly the same formula. $[P t C l_{2} (N H_{3})_{4}] B r_{2}$ and $[P t B r_{2} (N H_{3})_{4}] C l_{2}$ are such ionisation isomers. They release different ions in water, so silver nitrate gives cream $A g B r$ with the first and white $A g C l$ with the second.

5.4.6 Solvate Isomerism

Solvate isomers contain the same solvent molecules placed differently: a solvent molecule that is a ligand in one isomer sits in the crystal lattice, not touching the metal, in the other. When the solvent is water this is called hydrate isomerism. Chromium(III) chloride hexahydrate, $C r C l_{3} \cdot 6 H_{2} O$ , is known in three forms: violet $[C r (H_{2} O)_{6}] C l_{3}$ , pale green $[C r C l (H_{2} O)_{5}] C l_{2} \cdot H_{2} O$ and dark green $[C r C l_{2} (H_{2} O)_{4}] C l \cdot 2 H_{2} O$ . Silver nitrate precipitates 3, 2 and 1 mol of $A g C l$ from them, and only the lattice water is easily driven off by gentle drying.

Table 5.4The four kinds of structural isomerism

Kind	Difference	Test
Linkage	donor atom of an ambidentate ligand	infrared spectrum; colour
Coordination	which metal holds which ligands	colour; ions formed
Ionisation	a ligand swaps with the counter ion	silver nitrate gives AgBr or AgCl
Solvate	water inside or outside the sphere	moles of AgCl; mass lost on drying

Practice · 2 questions

Practice: isomerism: questions with answers

1
Short answerMediumNCERT Intext Question 5.3
Name the kinds of isomerism each complex can show, and describe the isomers.
1. $N a [C o (C_{2} O_{4})_{2} (N H_{3})_{2}]$
  Show solution
  Geometrical: the two $N H_{3}$ ligands can be cis or trans. The cis isomer has no mirror plane, so it also exists as an optical pair. Three stereoisomers in all: trans, cis-d and cis-l.
2. $[F e (C_{2} O_{4})_{3}]^{3 -}$
  Show solution
  Optical only: like every $[M (A A)_{3}]$ complex it is a propeller that twists either way, giving a pair of enantiomers. There are no geometrical isomers.
3. $[C o (N H_{3})_{5} (S C N)] C l_{2}$
  Show solution
  Linkage: thiocyanate bonded through S or through N. Ionisation: a chloride can swap with the thiocyanate, giving $[C o C l (N H_{3})_{5}]^{2 +}$ with chloride and thiocyanate as the counter ions. Combining these gives further isomers.
4. square planar $[P d (N H_{3})_{2} (S C N)_{2}]$
  Show solution
  Geometrical: cis and trans. Linkage: each thiocyanate can be S- or N-bonded, so each geometrical form has linkage isomers too.
Short answerMediumNCERT Intext Question 5.4
How would you show by simple tests that $[C r (N H_{3})_{5} (N O_{3})] S O_{4}$ and $[C r (N H_{3})_{5} (S O_{4})] N O_{3}$ are ionisation isomers?
Show solution
In water the first gives $[C r (N H_{3})_{5} (N O_{3})]^{2 +}$ and free $S O_{4}^{2 -}$ ; the second gives $[C r (N H_{3})_{5} (S O_{4})]^{+}$ and free $N O_{3}^{-}$ .
Barium chloride gives a white precipitate of $B a S O_{4}$ with the first solution only. The brown-ring test detects free nitrate in the second only. Their molar conductivities also differ, because one is a 2 : 2 electrolyte and the other a 1 : 1 electrolyte. Same formula, different free ions: ionisation isomers.

5.5 Bonding in Coordination Compounds

Werner's picture tells us which groups are attached to the metal, but not why. A good bonding theory should also explain the shapes of complexes and why many are drawn into a magnetic field. It should say why so many are coloured, and why the transition metals in particular form so many of them. Chemists answer these questions with a ladder of models. This unit climbs the first two rungs, valence bond theory (VBT) and crystal field theory (CFT). Two more advanced models, ligand field theory and molecular orbital theory, come later in your studies.

5.5.1 Valence Bond Theory

Valence bond theory pictures each ligand offering a lone pair, so the metal must provide one empty orbital aimed at each ligand. It makes them by mixing a set of its own empty orbitals into the same number of identical hybrid orbitals that point to the corners of a polyhedron. The set is built either from inner $(n - 1) d$ orbitals with $n s$ and $n p$ , or from $n s$ and $n p$ with outer $n d$ orbitals. Each lone pair then fills one hybrid, forming a coordinate $σ$ bond, and the kind of hybridisation fixes the shape.

Table 5.5Hybrid orbital sets and the shapes they give (CN = coordination number)

Hybrid set	Shape (CN)	Example
$s p^{3}$	tetrahedral (4)	$[Z n (N H_{3})_{4}]^{2 +}$
$d s p^{2}$	square planar (4)	$[N i (C N)_{4}]^{2 -}$
$s p^{3} d$	trigonal bipyramidal (5)	$[C u C l_{5}]^{3 -}$
$s p^{3} d^{2}$	octahedral, outer orbital (6)	$[F e (H_{2} O)_{6}]^{2 +}$
$d^{2} s p^{3}$	octahedral, inner orbital (6)	$[F e (C N)_{6}]^{4 -}$

Iron(II) shows how the same ion can choose between the two octahedral sets (Fig. 5.9). A free $F e^{2 +}$ ion has six 3d electrons, four of them unpaired. Cyanide ions push the 3d electrons together into three orbitals, which frees two 3d orbitals. These join 4s and the three 4p orbitals in $d^{2} s p^{3}$ hybridisation, and $[F e (C N)_{6}]^{4 -}$ ends up with no unpaired electrons: it is diamagnetic. Because it uses the inner 3d orbitals it is called an inner orbital, low spin or spin-paired complex.

Water molecules do not force the electrons to pair. The ion then reaches outwards to two 4d orbitals, giving $s p^{3} d^{2}$ hybrids, and $[F e (H_{2} O)_{6}]^{2 +}$ keeps all four unpaired electrons. It is an outer orbital, high spin or spin-free complex, and it is paramagnetic.

3/3No pairing; two 4d orbitals are used instead: $s p^{3} d^{2}$ , outer orbital, four unpaired.

Fig. 5.9Strong-field cyanide pairs the 3d electrons and gives an inner orbital complex; weak-field water does not.NCERT p. 129

Rows of orbital boxes labelled 3d, 4s, 4p and 4d, with electrons drawn as up and down arrows; electron pairs given by ligands are drawn in shaded boxes. Step 1 shows the free iron(II) ion: one pair and four single electrons in the five 3d boxes, 4 unpaired, everything else empty. Step 2 adds the cyanide complex: the six 3d electrons are paired in three boxes, and six shaded ligand pairs fill the last two 3d boxes, 4s and the three 4p boxes, bracketed as d2sp3, 0 unpaired. Step 3 adds the water complex: the 3d electrons stay as in the free ion, and six ligand pairs fill 4s, the three 4p boxes and two 4d boxes, bracketed as sp3d2, 4 unpaired.

Four-coordinate complexes follow the same logic (Fig. 5.10). Nickel(II) is $3 d^{8}$ with two unpaired electrons. With chloride ions it uses 4s and the three 4p orbitals; the $s p^{3}$ hybrids point to the corners of a tetrahedron, and $[N i C l_{4}]^{2 -}$ stays paramagnetic with two unpaired electrons. Cyanide pairs the 3d electrons, emptying one 3d orbital, so $d s p^{2}$ hybrids form; they lie in one plane, and $[N i (C N)_{4}]^{2 -}$ is square planar and diamagnetic. In $[N i (C O)_{4}]$ nickel has oxidation state zero ( $3 d^{8} 4 s^{2}$ ). Carbon monoxide drives the two 4s electrons into 3d, giving $3 d^{10}$ ; the empty 4s and 4p orbitals form $s p^{3}$ hybrids, so the molecule is tetrahedral and diamagnetic.

3/3Pairing empties one 3d orbital: $d s p^{2}$ , square planar $[N i (C N)_{4}]^{2 -}$ , diamagnetic.

Fig. 5.10The same nickel(II) ion forms a paramagnetic tetrahedral chloride complex and a diamagnetic square planar cyanide complex.NCERT p. 130

Boxes for 3d, 4s and 4p with electron arrows; shaded boxes hold electron pairs from ligands. Step 1 shows the free nickel(II) ion with three pairs and two single electrons in 3d, 2 unpaired. Step 2 adds the tetrahedral chloride complex: 3d unchanged, four ligand pairs in 4s and the three 4p boxes, bracketed sp3, 2 unpaired. Step 3 adds the square planar cyanide complex: the eight 3d electrons are paired in four boxes, and four ligand pairs fill the last 3d box, 4s and two 4p boxes, bracketed dsp2, 0 unpaired, with one 4p box left empty.

5.5.2 Magnetic Properties of Coordination Compounds

A paramagnetic sample is pulled into a magnetic field, a diamagnetic one is pushed out weakly. Weighing a sample with and without a strong magnet near it (a magnetic balance) gives its magnetic moment, and the moment reveals the number of unpaired electrons $n$ . To a good approximation for first-row complexes, only the electron spins contribute:

spin-only magnetic moment

μ = \sqrt{n (n + 2)} B M

(5.1)

where

$μ$: magnetic moment in Bohr magnetons (BM)
$n$: number of unpaired electrons

Table 5.6Spin-only magnetic moments

Unpaired electrons, n	μ (BM)
1	1.73
2	2.83
3	3.87
4	4.90
5	5.92

Now compare the pairs in the table below. The same ion keeps more unpaired electrons with water than with cyanide, so some ligands force d electrons to pair and others do not. Counting orbitals shows why this matters for valence bond theory. $d^{2} s p^{3}$ hybrids need two empty 3d orbitals. An ion with three or fewer d electrons, such as $C r^{3 +}$ , has them already, so $[C r (N H_{3})_{6}]^{3 +}$ keeps all three of its unpaired electrons. An ion with four, five or six d electrons can empty two 3d orbitals only by pairing some of them.

Table 5.7Same ion, different ligands: outer orbital (water) and inner orbital (cyanide) complexes

Ion	With water	With cyanide
$C r^{2 +}$ ( $d^{4}$ )	$[C r (H_{2} O)_{6}]^{2 +}$ : 4 unpaired	$[C r (C N)_{6}]^{4 -}$ : 2 unpaired
$M n^{2 +}$ ( $d^{5}$ )	$[M n (H_{2} O)_{6}]^{2 +}$ : 5 unpaired	$[M n (C N)_{6}]^{4 -}$ : 1 unpaired
$F e^{2 +}$ ( $d^{6}$ )	$[F e (H_{2} O)_{6}]^{2 +}$ : 4 unpaired	$[F e (C N)_{6}]^{4 -}$ : 0 unpaired

Valence bond theory describes the cyanide complexes as inner orbital ( $d^{2} s p^{3}$ ) and the aqua complexes as outer orbital ( $s p^{3} d^{2}$ ), but it has to be told which ligands cause pairing. A $d^{8}$ ion such as $N i^{2 +}$ can never empty two 3d orbitals, so its octahedral complexes, like $[N i (N H_{3})_{6}]^{2 +}$ , are always outer orbital with two unpaired electrons.

Example 5.7NCERT Example 5.7 · p. 131

Geometry from magnetism

Magnetic measurements show three unpaired electrons in the four-coordinate ion $[C o C l_{4}]^{2 -}$ . Is it tetrahedral or square planar? Find its spin-only moment.

Try it yourself first. Then reveal the solution one step at a time.

5.5.3 Limitations of Valence Bond Theory

Valence bond theory explains the shapes and the magnetism of many complexes, but it has clear weak points:

It says nothing about why complexes are coloured.
It cannot tell weak-field ligands from strong-field ones in advance; it has to be told which ligands cause pairing.
It cannot predict whether a four-coordinate complex will be tetrahedral or square planar.
It only counts unpaired electrons and cannot account for the exact size of measured magnetic moments.
It gives no numbers for how stable a complex is, either thermodynamically or kinetically.
It rests on assumptions, such as which orbitals hybridise, that are chosen to fit known facts rather than predicted.

Quick check

1 of 3

[N i (C N)_{4}]^{2 -}

is diamagnetic. Which hybridisation and shape does valence bond theory give it?

s p^{3}

, tetrahedral

d s p^{2}

, square planar

s p^{3} d^{2}

, octahedral

d^{2} s p^{3}

, octahedral

Show answer

Answer (B) $d s p^{2}$ , square planar

Pairing the eight 3d electrons empties one 3d orbital, which mixes with 4s and two 4p orbitals:

d s p^{2}

, square planar. A tetrahedral

s p^{3}

complex would keep two unpaired electrons.

5.5.4 Crystal Field Theory

Valence bond theory had to be told which ligands pair electrons. Crystal field theory predicts it, by asking how the energies of the five d orbitals change when ligands crowd round the metal. Picture each d orbital as a cloud of negative charge with its own shape and direction. In an isolated ion the five clouds have equal energy: they are degenerate. Now bring up the ligands, modelled as nothing more than negative point charges (anions) or dipoles with their negative ends inwards ( $H_{2} O$ , $N H_{3}$ ). Repulsion raises every orbital, but not equally: a cloud aimed straight at a ligand is pushed up more than one that slips between ligands. Charge smeared evenly over a sphere would raise all five together. The model ignores covalent bonding entirely, a weakness we return to in Section 5.5.6.

1/2 $d_{x^{2} - y^{2}}$ points straight at the ligands. $d_{z^{2}}$ does the same along $z$ .

Fig. 5.11In an octahedral field, orbitals aimed at the ligands are raised; those aimed between them are lowered.NCERT p. 131

A metal nucleus sits where dashed x and y axes cross, with a ligand L on each axis at the same distance. Step 1 shows the d x squared minus y squared orbital: four blue lobes lie along the axes and point straight at the four ligands; the text says strong repulsion, energy rises, and that this orbital belongs to the e g set. Step 2 rotates the lobes by 45 degrees and colours them green to show the d x y orbital, whose lobes point between the ligands; the text says weaker repulsion, energy falls, and that it belongs to the t 2 g set.

(a) Splitting in an octahedral field

Place six ligands along $\pm x$ , $\pm y$ and $\pm z$ . The $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ orbitals point at them, feel the strongest repulsion and rise; they form the $e_{g}$ set. The $d_{x y}$ , $d_{x z}$ and $d_{y z}$ orbitals point between the axes and fall; they form the $t_{2 g}$ set. This is crystal field splitting (क्रिस्टल क्षेत्र विपाटन), and the energy gap is written $Δ_{o}$ (o for octahedral). The average energy does not change, so the two $e_{g}$ orbitals rise by as much in total as the three $t_{2 g}$ orbitals fall:

\begin{aligned} E (e_{g}) & = + \frac{3}{5} Δ_{o} \\ E (t_{2 g}) & = - \frac{2}{5} Δ_{o} \end{aligned}

(5.2)

where

$Δ_{o}$: octahedral crystal field splitting energy
$E$: energy relative to the spherical-field average

3/4Six ligands: $e_{g}$ up by $\frac{3}{5} Δ_{o}$ , $t_{2 g}$ down by $\frac{2}{5} Δ_{o}$ .

Fig. 5.12The same five orbitals split one way around an octahedron and the opposite way, by less, around a tetrahedron.NCERT Fig. 5.8, 5.9 · p. 132

An energy axis points up on the left. Five short bars stand for the five d orbitals: three blue (dxy, dxz, dyz) and two orange (dx2−y2, dz2). Step 1: all five sit at the same low level, labelled free ion. Step 2: all five rise together to a dashed line labelled average, the spherical field. Step 3: in an octahedral field the two orange bars rise by three fifths of Delta o to form e g and the three blue bars fall by two fifths of Delta o to form t 2 g, with a double arrow marking Delta o. Step 4: in a tetrahedral field the pattern flips and shrinks: the three blue bars rise by two fifths of Delta t to form t 2, the two orange bars fall by three fifths of Delta t to form e, and a note says Delta t equals four ninths of Delta o.

The size of $Δ_{o}$ depends on the metal ion and on the ligands. A higher charge on the metal pulls the ligands closer and increases the splitting, and so does moving down a group to 4d and 5d metals. For a given metal ion, ligands can be ranked by the splitting they cause, measured from the light their complexes absorb. The ranking is the spectrochemical series:

spectrochemical series (weak field → strong field)

\begin{aligned} I^{-} < B r^{-} < S C N^{-} \\ < C l^{-} < S^{2 -} < F^{-} \\ < O H^{-} < C_{2} O_{4}^{2 -} \\ < H_{2} O < N C S^{-} \\ < E D T A^{4 -} < N H_{3} \\ < e n < C N^{-} < C O \end{aligned}

Definition

Spectrochemical series

Ligands arranged in order of increasing crystal field splitting, found experimentally from the absorption spectra of their complexes.

ExampleWater causes a smaller splitting than ammonia, which causes a smaller splitting than cyanide.

Now feed in the d electrons. In an octahedral complex the first three go singly into the three $t_{2 g}$ orbitals, as Hund's rule requires. The fourth electron faces a choice. It can enter an $e_{g}$ orbital, which costs $Δ_{o}$ , or pair up in a $t_{2 g}$ orbital, which costs the pairing energy $P$ , the extra repulsion of two electrons in one orbital.

Compare

Small gap or large gap?

$Δ_{o} < P$ : weak field

Climbing to $e_{g}$ costs less than pairing.
Fourth electron goes up: $t_{2 g}^{3} e_{g}^{1}$ .
Weak-field ligands give high-spin complexes.

$Δ_{o} > P$ : strong field

Pairing costs less than climbing.
Fourth electron pairs: $t_{2 g}^{4} e_{g}^{0}$ .
Strong-field ligands give low-spin complexes.

2/3Weak field: the fourth electron enters $e_{g}$ . Strong field: it pairs in $t_{2 g}$ .

Fig. 5.13From

d^{4}

onwards the size of

Δ_{o}

compared with

P

decides between high spin and low spin.NCERT p. 132

Two energy diagrams side by side. On the left, weak field: the two e g bars sit only a little above the three t 2 g bars, marked Delta o less than P. On the right, strong field: the e g bars sit far above, marked Delta o greater than P. Step 1 (d3): both diagrams have one up arrow in each t 2 g bar, 3 unpaired. Step 2 (d4): on the left the fourth electron goes up into e g, 4 unpaired, high spin, configuration t2g3 eg1; on the right it pairs in t 2 g, 2 unpaired, low spin, t2g4. Step 3 (d6): the left has t2g4 eg2 with 4 unpaired; the right has all six electrons paired in t 2 g, 0 unpaired.

The same contest decides $d^{5}$ , $d^{6}$ and $d^{7}$ ions. For $d^{8}$ , $d^{9}$ and $d^{10}$ there is only one way to fill the levels, so ligand strength cannot change the number of unpaired electrons. We can score each arrangement by its crystal field stabilisation energy (CFSE), counting $x$ electrons in $t_{2 g}$ and $y$ in $e_{g}$ :

E_{C F} = (\frac{3}{5} y - \frac{2}{5} x) Δ_{o}

(5.3)

where

$x, y$: numbers of electrons in $t_{2 g}$ and $e_{g}$
$E_{C F}$: add $P$ for each electron pair beyond those already present in the free ion

Run these numbers for any ion from $d^{4}$ to $d^{7}$ and the verdict is always the same: a strong field favours the low-spin arrangement and a weak field the high-spin one.

Example

High spin or low spin?

For $F e^{2 +}$ the pairing energy is about $210 k J m o l^{- 1}$ . The splitting is about $124 k J m o l^{- 1}$ in $[F e (H_{2} O)_{6}]^{2 +}$ and about $393 k J m o l^{- 1}$ in $[F e (C N)_{6}]^{4 -}$ . Predict the configuration, the number of unpaired electrons and the spin-only moment of each ion.

Try it yourself first. Then reveal the solution one step at a time.

(b) Splitting in a tetrahedral field

In a tetrahedral complex the four ligands sit between the axes, at alternate corners of a cube around the metal. Now $d_{x y}$ , $d_{x z}$ and $d_{y z}$ point closer to the ligands than $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ do. The pattern is therefore turned upside down: a lower pair $e$ and an upper trio $t_{2}$ (Fig. 5.12, last step). With only four ligands, none aimed straight at any orbital, the splitting is much smaller. Keep the metal, the ligands and the bond length unchanged, and theory gives

Δ_{t} = \frac{4}{9} Δ_{o}

(5.4)

Pairing energies are usually larger than such a small gap, so low-spin tetrahedral complexes are rare. The labels also lose their subscript g. That letter, from the German gerade (even), describes how an orbital behaves when it is inverted through the metal, a symmetry the octahedron has and the tetrahedron lacks; so the tetrahedral sets are written simply $e$ and $t_{2}$ .

InteractiveCrystal field lab

An energy diagram with two sets of d-orbital levels drawn as bars, for an octahedral or a tetrahedral complex. Electrons appear as up and down arrows. Buttons change the number of d electrons from 1 to 10 and switch the geometry; a slider sets the splitting Delta as a multiple of the pairing energy P from 0.2 to 3. The lab fills the levels in the lowest-energy way and shows the configuration (for example t2g4 eg2), the number of unpaired electrons, whether the complex is high spin or low spin, the spin-only magnetic moment in Bohr magnetons and the crystal field stabilisation energy, and lists real ions with that many d electrons.

Fig. 5.14Choose the number of d electrons and the geometry, then slide the splitting relative to the pairing energy and watch the electrons rearrange.

5.5.5 Colour in Coordination Compounds

Sunlight or white lamplight carries every visible wavelength. Pass it through a solution of a complex and one band of wavelengths is soaked up while the rest come through. Our eyes read what is left as the complementary colour of the band removed, the colour opposite it on a colour wheel (Fig. 5.15). A complex that absorbs yellow light therefore looks violet, and one that absorbs blue looks yellow-orange.

2/3 $[C r (H_{2} O)_{6}]^{3 +}$ absorbs yellow light near 575 nm and looks violet.

Fig. 5.15A complex looks the colour opposite to the one it absorbs.NCERT Table 5.3 · p. 133

A ring divided into six coloured sectors, each labelled with its name: red at the top, then clockwise orange, yellow, green, blue and violet. Step 1 notes that opposite colours are complementary. Step 2 shows hexaaquachromium(III): the yellow sector is outlined as absorbed, near 575 nanometres, an arrow crosses the wheel to the violet sector, outlined with a dashed line, and the text says you see violet. Step 3 shows hexaamminechromium(III): blue, near 465 nanometres, is absorbed and the arrow points to orange; the text says you see yellow-orange.

In crystal field terms, colour comes from an electron jumping from a lower d level to a higher one, a d–d transition. The photon absorbed must carry exactly the energy of the gap, so the wavelength absorbed measures the splitting directly. The simplest case is a $d^{1}$ ion such as titanium(III) in water. There is only one possible jump, $t_{2 g}^{1} e_{g}^{0} \to t_{2 g}^{0} e_{g}^{1}$ , so there is one absorption band, near 500 nm in the green–blue, and $[T i (H_{2} O)_{6}]^{3 +}$ solutions look purple. In general,

Δ = \frac{N_{A} h c}{λ}

(5.5)

where

$λ$: wavelength absorbed
$N_{A} h c$: $1.196 \times 10^{5} k J m o l^{- 1} n m$ , so $Δ$ in $k J m o l^{- 1}$ is $119 600 / λ$ with $λ$ in nm

Example

Splitting from the colour

$[C r (H_{2} O)_{6}]^{3 +}$ absorbs most strongly near 575 nm and $[C r (N H_{3})_{6}]^{3 +}$ near 465 nm. For these $d^{3}$ ions the lowest-energy absorption equals $Δ_{o}$ . Find $Δ_{o}$ for each in $k J m o l^{- 1}$ and check the result against the spectrochemical series.

Try it yourself first. Then reveal the solution one step at a time.

Because the splitting comes from the ligands, changing the ligands changes the colour, as the two chromium(III) complexes in Fig. 5.15 showed. Some ions have no d–d colour whatever surrounds them: with no d electrons ( $S c^{3 +}$ ) or a full d set ( $Z n^{2 +}$ ) there is no possible d–d jump, so their complexes are colourless. Taking ligands away can also remove a colour. Blue copper(II) sulphate crystals turn almost white when heated: with the water gone, each $C u^{2 +}$ is surrounded by sulphate ions, whose much weaker field moves the absorption out of the visible range into the infrared. The two activities below let you see both effects.

Try this

Colour from water

You’ll need

copper(II) sulphate pentahydrate (blue vitriol)
dry hard-glass test tube and holder
spirit lamp or burner
dropper with water

Predict first

What colour will the white powder become when water is added back?

it stays whiteblue againgreenblack

Gently heat a few blue crystals of copper(II) sulphate pentahydrate in a dry test tube until they turn white. Let the tube cool, then add two or three drops of water and watch.

Try this

A deep-blue complex

You’ll need

dilute copper(II) sulphate solution
dilute aqueous ammonia in a dropper bottle
test tube

Predict first

Ammonia is a stronger-field ligand than water. How should the colour change as ammonia replaces water around

C u^{2 +}

it fades to colourlessit turns a deeper, more violet-blueit turns yellowit does not change

Put about 2 mL of dilute copper(II) sulphate solution in a test tube and note its colour.
Add dilute aqueous ammonia one drop at a time, shaking after each drop.
Keep adding until no further change occurs, and compare the final colour with the starting one.

5.5.6 Limitations of Crystal Field Theory

Here is a fact the point-charge picture cannot explain. Fluoride is a small ion with a full negative charge, while carbon monoxide is uncharged and almost non-polar, so electrostatics alone would make fluoride by far the stronger ligand. The spectrochemical series says the opposite: CO splits the d orbitals much more than any halide does. The missing ingredient is covalency. Real metal–ligand bonds share electrons, and CO in particular accepts metal d electrons into its empty $π^{*}$ orbitals (Section 5.6), which widens the gap. Crystal field theory remains a quick, reliable guide to magnetism and colour; ligand field theory and molecular orbital theory, which you may meet later, add the covalent part.

Practice · 6 questions

Practice: bonding, magnetism and colour: questions with answers

Short answerMediumNCERT Intext Question 5.5
Nickel(II) forms red, square planar $[N i (d m g)_{2}]$ , which a magnet does not attract, and tetrahedral $[N i B r_{4}]^{2 -}$ , which it does. Account for the difference with valence bond theory.
Show solution
In both, nickel is $N i^{2 +}$ ( $3 d^{8}$ , two unpaired electrons). In $[N i B r_{4}]^{2 -}$ the weak bromide ligands use 4s and 4p ( $s p^{3}$ , tetrahedral), leaving the two unpaired 3d electrons: paramagnetic. In $[N i (d m g)_{2}]$ the eight 3d electrons pair in four orbitals, emptying one 3d orbital for $d s p^{2}$ hybridisation: square planar with no unpaired electrons, so diamagnetic.
Short answerMediumNCERT Intext Question 5.6
A magnet attracts $[M n C l_{4}]^{2 -}$ strongly but does not attract $[N i (C O)_{4}]$ at all. Shape cannot be the reason, since both are tetrahedral. What is?
Show solution
Mn is +2 in $[M n C l_{4}]^{2 -}$ : $3 d^{5}$ , and chloride does not cause pairing, so five unpaired electrons remain ( $s p^{3}$ ). Ni is 0 in $[N i (C O)_{4}]$ : $3 d^{8} 4 s^{2}$ ; CO pushes the 4s electrons into 3d to give $3 d^{10}$ , all paired, and the empty 4s and 4p form $s p^{3}$ hybrids.
Short answerMediumNCERT Intext Question 5.7
The magnetic moment of $[C r (H_{2} O)_{6}]^{2 +}$ is close to 4.9 BM, but that of $[C r (C N)_{6}]^{4 -}$ is only about 3 BM. Account for the difference.
Show solution
Water is a weak ligand: no pairing, $s p^{3} d^{2}$ outer orbital complex with four unpaired electrons (4.90 BM). Cyanide is strong: the four 3d electrons pair as far as possible in three orbitals, leaving two unpaired electrons and two empty 3d orbitals for $d^{2} s p^{3}$ hybridisation (2.83 BM).
Short answerMediumNCERT Intext Question 5.8
Valence bond theory classes $[C o (N O_{2})_{6}]^{3 -}$ as an inner orbital complex but $[N i (N O_{2})_{6}]^{4 -}$ as an outer orbital one, although both metals hold six N-bonded nitrite ligands. Why?
Show solution
$C o^{3 +}$ is $d^{6}$ . Nitrite pairs the electrons into three 3d orbitals, freeing two for $d^{2} s p^{3}$ : inner orbital, diamagnetic. $N i^{2 +}$ is $d^{8}$ ; even fully paired its electrons need four 3d orbitals, so only one is free. Octahedral bonding must therefore use 4d orbitals: $s p^{3} d^{2}$ , outer orbital, two unpaired electrons.
Quick answerEasyNCERT Intext Question 5.9
Gold(III) chloride dissolves in hydrochloric acid as the square planar ion $[A u C l_{4}]^{-}$ . How many of its electrons are unpaired?
Show answer
Answer 0
Au is +3 ( $5 d^{8}$ ). Square planar $d s p^{2}$ bonding needs one empty 5d orbital, so the eight electrons pair in four orbitals: no unpaired electrons.
Short answerMediumNCERT Intext Question 5.10
Both $[C o (H_{2} O)_{6}]^{2 +}$ and $[C o (N O_{2})_{6}]^{4 -}$ contain cobalt(II), yet one has three unpaired electrons and the other only one. Which is which? Explain with crystal field theory.
Show solution
Water is a weak-field ligand, so $Δ_{o} < P$ : the seven electrons spread out as $t_{2 g}^{5} e_{g}^{2}$ , three unpaired, in $[C o (H_{2} O)_{6}]^{2 +}$ . N-bonded nitrite is a strong-field ligand, so $Δ_{o} > P$ : six electrons fill $t_{2 g}$ before the seventh enters $e_{g}$ , giving $t_{2 g}^{6} e_{g}^{1}$ with one unpaired electron in $[C o (N O_{2})_{6}]^{4 -}$ .

5.6 Bonding in Metal Carbonyls

Carbon monoxide combines directly with most transition metals to give homoleptic carbonyls, compounds in which CO is the only ligand. Most are volatile liquids or low-melting solids, and their shapes follow from the number of CO groups each metal holds (Fig. 5.16 and the table).

Table 5.8Some homoleptic metal carbonyls

Carbonyl	Name	Structure
$[N i (C O)_{4}]$	tetracarbonylnickel(0)	tetrahedral
$[F e (C O)_{5}]$	pentacarbonyliron(0)	trigonal bipyramidal
$[C r (C O)_{6}]$	hexacarbonylchromium(0)	octahedral
$[M n_{2} (C O)_{10}]$	decacarbonyldimanganese(0)	two $M n (C O)_{5}$ square pyramids joined by an Mn–Mn bond; no CO bridges
$[C o_{2} (C O)_{8}]$	octacarbonyldicobalt(0)	Co–Co bond with two CO groups bridging the two cobalt atoms

Fig. 5.16Each small ball stands for one CO ligand, bonded to the metal through carbon.NCERT Fig. 5.13 · p. 136

Four panels. Top left: a nickel atom with four small CO balls at the corners of a tetrahedron, labelled Ni(CO)4, tetrahedral. Top right: an iron atom with five CO balls, three around the middle and one above and below, labelled Fe(CO)5, trigonal bipyramidal. Bottom left: a chromium atom with six CO balls at the corners of an octahedron, labelled Cr(CO)6, octahedral. Bottom right: two manganese atoms joined by a bond, each carrying five CO balls arranged as a square pyramid, labelled Mn2(CO)10, two square pyramids.

Carbon monoxide is a feeble Lewis base with almost no dipole, yet it binds metals in low oxidation states very strongly. The reason is that the metal–carbon bond works in both directions at once.

Forward (σ) donation: carbon's lone pair, not oxygen's, slots into an empty metal orbital and forms a $σ$ bond from ligand to metal.
Back (π) donation: the metal, rich in d electrons at oxidation state zero, pushes some of that charge sideways into the empty antibonding $π^{*}$ orbitals of CO, which have the right shape to overlap with its d orbitals.

The two effects help each other, which is why the bonding is called synergic. Electrons gained by σ donation make the metal richer in electrons and better able to give some back; giving electrons back makes the metal a better acceptor again. The metal–carbon bond ends up stronger than either effect alone would make it. Because the back-donated electrons enter an antibonding orbital of CO, the C–O bond becomes slightly weaker: the C–O stretching vibration falls from $2143 c m^{- 1}$ in free CO to about $2000 c m^{- 1}$ in $[C r (C O)_{6}]$ .

2/3A filled metal d orbital feeds electrons into the empty $π^{*}$ orbital of CO.

Fig. 5.17σ donation and π back-donation reinforce each other: the M–C bond strengthens while the C–O bond weakens slightly.NCERT Fig. 5.14 · p. 136

A metal M on the left is bonded to the carbon of a C≡O group drawn with a triple line. Step 1: a blue lone-pair lobe on carbon points at the metal, and an arrow above, labelled sigma donation, runs from C to M. Step 2: four lobes of a filled metal d orbital appear around M, and lobes of the empty pi-star orbital of CO appear above and below the carbon and oxygen; a curved arrow below, labelled pi back-donation, runs from the metal lobe to the CO lobe. Step 3 adds two labels on the right: M–C stronger, and C–O weaker, with a lower stretching frequency.

Quick check

1 of 3

In the synergic bonding of a metal carbonyl, the π bond forms when

CO donates its carbon lone pair to the metaloxygen donates a lone pair to the metalthe metal donates electrons into the σ orbital of COa filled metal d orbital donates electrons into the empty

π^{*}

orbital of CO

Show answer

Answer (D) a filled metal d orbital donates electrons into the empty $π^{*}$ orbital of CO

The π component is back-donation from metal d orbitals into the antibonding

π^{*}

orbitals of CO. The carbon lone pair forms the σ bond, not the π bond.

5.7 Importance and Applications of Coordination Compounds

From the green of a leaf to the drug in a chemotherapy drip, complexes do jobs that simple salts cannot. The table collects some of them, grouped by where they work.

Table 5.9Coordination compounds at work

Area	Substance	Use
Living things	chlorophyll (Mg), haemoglobin (Fe), vitamin $B_{12}$ (Co)	photosynthesis, oxygen transport, healthy red blood cells (prevents pernicious anaemia)
Enzymes	carbonic anhydrase (Zn), cytochromes (Fe)	remove $C O_{2}$ from tissues quickly; carry electrons in respiration
Medicine	cisplatin; EDTA, D-penicillamine, desferrioxamine B	slows the growth of tumours; removes excess lead, copper or iron
Extracting metals	$[A u (C N)_{2}]^{-}$ , $[A g (C N)_{2}]^{-}$	dissolve gold and silver out of crushed ore; zinc then frees the metal
Purifying metals	$[N i (C O)_{4}]$	volatile carbonyl forms from impure nickel and decomposes to pure nickel on heating
Industrial catalysis	Wilkinson's catalyst, a rhodium complex	hydrogenation of alkenes
Electroplating	$[A g (C N)_{2}]^{-}$ , $[A u (C N)_{2}]^{-}$ baths	smooth, even coatings because only a few free metal ions are present at a time
Detecting ions	dimethylglyoxime; thiocyanate; Nessler's reagent, $K_{2} [H g I_{4}]$	a rosy-red solid shows $N i^{2 +}$ ; blood-red colour shows $F e^{3 +}$ ; brown colour shows ammonia
Hardness of water	$N a_{2} H_{2} E D T A$ titration	binds $C a^{2 +}$ and $M g^{2 +}$ ; their different stabilities let each be estimated
Photography	$[A g (S_{2} O_{3})_{2}]^{3 -}$	hypo dissolves unexposed silver bromide when film is fixed

Gold extraction shows complex formation doing real work. Crushed ore is stirred with dilute sodium cyanide while air is blown through. Oxygen oxidises the gold, and cyanide immediately captures each $A u^{+}$ as the very stable linear ion $[A u (C N)_{2}]^{-}$ :

\begin{aligned} 4 A u + 8 C N^{-} + O_{2} \\ + 2 H_{2} O \\ \overset{}{\to} 4 [A u (C N)_{2}]^{-} \\ + 4 O H^{-} \end{aligned}

Zinc dust added to the clear solution then displaces the gold, because zinc forms the even more favoured $[Z n (C N)_{4}]^{2 -}$ :

\begin{aligned} 2 [A u (C N)_{2}]^{-} + Z n \\ \overset{}{\to} [Z n (C N)_{4}]^{2 -} \\ + 2 A u \end{aligned}

Nickel is purified in a similar spirit by the Mond process. Carbon monoxide passed over impure nickel at about 330 K forms gaseous $[N i (C O)_{4}]$ and leaves the impurities behind. Heating the gas to about 450 K deposits pure nickel and releases the CO for reuse.

In a hardness titration, EDTA wraps around each $C a^{2 +}$ or $M g^{2 +}$ ion through six donor atoms (Fig. 5.18). The five chelate rings make the complexes very stable, so the end point is sharp. The same grip makes calcium disodium EDTA a treatment for lead poisoning: it carries the lead ions out of the body in the urine. D-penicillamine removes excess copper, desferrioxamine B removes excess iron, and cisplatin, the cis isomer from Section 5.4.1, binds to DNA in dividing cancer cells and stops them copying it.

Fig. 5.18One EDTA ion supplies all six donor atoms of an octahedral complex, closing five chelate rings.NCERT p. 137

A calcium ion Ca sits at the centre, bonded to six donor atoms arranged around it: two nitrogen atoms on the left and four oxygen atoms on the top, right and bottom. Thick green bands show the EDTA chain: one band joins the two nitrogens on the far left, and from each nitrogen two bands curve round to two of the oxygens, one above and one below. The labels say six donor atoms, two N and four O, and five chelate rings, with the formula of the calcium–EDTA complex ion.

Quick check

1 of 3

In which form is gold held in solution when it is leached from its ore?

[A g (S_{2} O_{3})_{2}]^{3 -}

[N i (C O)_{4}]

[Z n (C N)_{4}]^{2 -}

[A u (C N)_{2}]^{-}

Show answer

Answer (D) $[A u (C N)_{2}]^{-}$

Gold dissolves as dicyanidoaurate(I),

[A u (C N)_{2}]^{-}

[Z n (C N)_{4}]^{2 -}

forms later, when zinc displaces the gold.

Recap

Coordination Compounds: key points and summary

In a coordination compound a metal atom or ion holds a definite set of ligands, each of which donates an electron pair (Section 5.2).
Werner's primary valence (ionisable, the oxidation state) and secondary valence (the coordination number, fixed in direction) explain silver nitrate and conductivity results (Section 5.1).
Ligands may be uni-, di- or polydentate; a ligand that binds through two or more atoms forms a chelate, and an ambidentate ligand can use either of two donor atoms ((c) Ligands).
Coordination number counts donor atoms; the oxidation number follows from charge balance (coordination number, oxidation number).
IUPAC formulas list the metal first, then ligands alphabetically; names give the cation first, ligands alphabetically, the metal (with -ate in anions) and its oxidation state (Section 5.3).
Geometrical isomers (cis/trans, fac/mer) arise in square planar and octahedral complexes, never in tetrahedral ones (Section 5.4.1).
Chiral complexes such as $[M (A A)_{3}]$ and cis- $[M (A A)_{2} B_{2}]$ exist as enantiomers that rotate polarised light in opposite directions (Section 5.4.2).
Linkage, coordination, ionisation and solvate isomers differ in which atoms are bonded (Section 5.4.3 to Section 5.4.6).
Valence bond theory links hybridisation to shape and explains inner orbital (low spin) and outer orbital (high spin) complexes; the spin-only moment is $\sqrt{n (n + 2)}$ BM (Section 5.5.1, Section 5.5.2).
Crystal field theory splits the d orbitals into $t_{2 g}$ and $e_{g}$ by $Δ_{o}$ in an octahedral field and into $e$ and $t_{2}$ by $Δ_{t} = \frac{4}{9} Δ_{o}$ in a tetrahedral one; $Δ_{o}$ versus $P$ decides high or low spin (Section 5.5.4).
Complexes are coloured because d–d transitions absorb visible light; the colour seen is the complement of that absorbed (Section 5.5.5).
Metal carbonyls are held by synergic σ donation and π back-donation; complexes matter in analysis, metallurgy, biology, catalysis and medicine (Section 5.6, Section 5.7).

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Intext Question 5.1	Practice: formulas and names · Q1
NCERT Intext Question 5.2	Practice: formulas and names · Q2
NCERT Intext Question 5.3	Practice: isomerism · Q1
NCERT Intext Question 5.4	Practice: isomerism · Q2
NCERT Intext Question 5.5	Practice: bonding, magnetism and colour · Q1
NCERT Intext Question 5.6	Practice: bonding, magnetism and colour · Q2
NCERT Intext Question 5.7	Practice: bonding, magnetism and colour · Q3
NCERT Intext Question 5.8	Practice: bonding, magnetism and colour · Q4
NCERT Intext Question 5.9	Practice: bonding, magnetism and colour · Q5
NCERT Intext Question 5.10	Practice: bonding, magnetism and colour · Q6
NCERT Exercises 5.1	Exercises · Q1
NCERT Exercises 5.2	Exercises · Q2
NCERT Exercises 5.3	Exercises · Q3
NCERT Exercises 5.4	Exercises · Q4
NCERT Exercises 5.5	Exercises · Q5
NCERT Exercises 5.6	Exercises · Q6
NCERT Exercises 5.7	Exercises · Q7
NCERT Exercises 5.8	Exercises · Q8
NCERT Exercises 5.9	Exercises · Q9
NCERT Exercises 5.10	Exercises · Q10
NCERT Exercises 5.11	Exercises · Q11
NCERT Exercises 5.12	Exercises · Q12
NCERT Exercises 5.13	Exercises · Q13
NCERT Exercises 5.14	Exercises · Q14
NCERT Exercises 5.15	Exercises · Q15
NCERT Exercises 5.16	Exercises · Q16
NCERT Exercises 5.17	Exercises · Q17
NCERT Exercises 5.18	Exercises · Q18
NCERT Exercises 5.19	Exercises · Q19
NCERT Exercises 5.20	Exercises · Q20
NCERT Exercises 5.21	Exercises · Q21
NCERT Exercises 5.22	Exercises · Q22
NCERT Exercises 5.23	Exercises · Q23
NCERT Exercises 5.24	Exercises · Q24
NCERT Exercises 5.25	Exercises · Q25
NCERT Exercises 5.26	Exercises · Q26
NCERT Exercises 5.27	Exercises · Q27
NCERT Exercises 5.28	Exercises · Q28
NCERT Exercises 5.29	Exercises · Q29
NCERT Exercises 5.30	Exercises · Q30
NCERT Exercises 5.31	Exercises · Q31

Quick check: MCQs on Coordination Compounds

5.1 Werner's Theory of Coordination Compounds4 questions
(g) Oxidation number of the central atom and (h) homoleptic and heteroleptic complexes4 questions
5.3.2 Naming of Mononuclear Coordination Compounds3 questions
5.4.2 Optical Isomerism4 questions
5.4.6 Solvate Isomerism4 questions
5.5.3 Limitations of Valence Bond Theory3 questions
(b) Splitting in a tetrahedral field4 questions
5.5.6 Limitations of Crystal Field Theory4 questions
5.6 Bonding in Metal Carbonyls3 questions
5.7 Importance and Applications of Coordination Compounds3 questions

Glossary · 15

Key terms (English–Hindi)

ambidentate ligand: A ligand with two different possible donor atoms that binds through either one, such as $N O_{2}^{-}$ (N or O) or $S C N^{-}$ (S or N).First seen in the chapter →

central atom or ionकेंद्रीय परमाणुkendriya paramanu: The metal atom or ion at the centre of a coordination entity; it accepts electron pairs from the ligands, so it acts as a Lewis acid.First seen in the chapter →
chelateकीलेटkeelet: A ring formed when one ligand binds the same metal ion through two or more donor atoms; chelate complexes are usually more stable than similar non-chelated ones.
coordination compoundउपसहसंयोजन यौगिकupsahsanyojan yaugik: A compound in which a metal atom or ion holds a definite number of neighbouring ions or molecules, each donating an electron pair, such as $[C u (N H_{3})_{4}] S O_{4}$ .First seen in the chapter →
coordination entity: The central atom together with the ligands bonded to it, written inside square brackets; it may be a cation, an anion or a neutral molecule.First seen in the chapter →
coordination numberउपसहसंयोजन संख्याupsahsanyojan sankhya: The number of ligand donor atoms joined to the central atom by $σ$ bonds.First seen in the chapter →
coordination polyhedron: The shape traced by the donor atoms around the central atom, for example octahedral, tetrahedral or square planar.First seen in the chapter →
coordination sphere: The central atom and its ligands, enclosed in square brackets; ions written outside the brackets are counter ions.First seen in the chapter →
crystal field splittingक्रिस्टल क्षेत्र विपाटनcrystal kshetra vipatan: The energy gap that opens between groups of d orbitals when ligands surround a metal ion; $Δ_{o}$ in an octahedral field and $Δ_{t}$ in a tetrahedral one.First seen in the chapter →

enantiomer: One of a pair of molecules or ions that are mirror images but cannot be superimposed (they are chiral); the two rotate plane-polarised light in opposite directions.First seen in the chapter →

homoleptic complex: A complex with only one kind of ligand, such as $[F e (C O)_{5}]$ . A heteroleptic complex has two or more kinds.First seen in the chapter →

inner orbital complex: A complex whose hybrid orbitals use the inner $(n - 1) d$ orbitals; for $d^{4}$ – $d^{6}$ ions this forces the d electrons to pair (low spin). An outer orbital complex uses $n d$ orbitals and keeps its electrons unpaired (high spin).First seen in the chapter →

ligandलिगन्डligand: An ion or molecule that donates at least one electron pair to the central atom. The number of donor atoms it uses is its denticity.

spectrochemical series: Ligands listed in order of the splitting they cause, found from the light their complexes absorb: halide ions at the weak end, $C N^{-}$ and $C O$ at the strong end.First seen in the chapter →
synergic bonding: Two-way bonding in metal carbonyls: CO donates a lone pair to the metal ( $σ$ ) and the metal donates d electrons back into CO's empty $π^{*}$ orbitals ( $π$ ); each strengthens the other.First seen in the chapter →

Coordination Compounds: Class 12 Chemistry · Chapter 5

What you will learn in Coordination Compounds

5.1 Werner's Theory of Coordination Compounds

5.2 Definitions of Some Important Terms Pertaining to Coordination Compounds

(a) Coordination entity and (b) central atom

(c) Ligands

(d) Coordination number and (e) coordination sphere

(f) Coordination polyhedron

(g) Oxidation number of the central atom and (h) homoleptic and heteroleptic complexes

5.3 Nomenclature of Coordination Compounds

5.3.1 Formulas of Mononuclear Coordination Entities

5.3.2 Naming of Mononuclear Coordination Compounds

Practice: formulas and names: questions with answers

5.4 Isomerism in Coordination Compounds

5.4.1 Geometric Isomerism

5.4.2 Optical Isomerism

5.4.3 Linkage Isomerism

5.4.4 Coordination Isomerism

5.4.5 Ionisation Isomerism

5.4.6 Solvate Isomerism

Practice: isomerism: questions with answers

5.5 Bonding in Coordination Compounds

5.5.1 Valence Bond Theory

5.5.2 Magnetic Properties of Coordination Compounds

5.5.3 Limitations of Valence Bond Theory

5.5.4 Crystal Field Theory

(a) Splitting in an octahedral field

(b) Splitting in a tetrahedral field

5.5.5 Colour in Coordination Compounds

5.5.6 Limitations of Crystal Field Theory

Practice: bonding, magnetism and colour: questions with answers

5.6 Bonding in Metal Carbonyls

5.7 Importance and Applications of Coordination Compounds

Coordination Compounds: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Coordination Compounds

Key terms (English–Hindi)

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