Coordination Compounds — JEE Mains Chemistry Formulas and Traps

Werner's Theory and Ionisable Ligands

Counting the ions outside the coordination sphere

Precipitate from the ionisable ions

n_{\mathrm{AgCl}} = n_{\text{complex}} \times (\text{number of } \mathrm{Cl^-} \text{ outside } [\;]) \qquad \Delta T_f = i\,K_f\,m

Primary valency, secondary valency and double salts

Oxidation state of the central metal

x + \sum q_{\text{ligands}} = q_{\text{complex ion}}

Common traps

Chloride inside the bracket never precipitates

In

\mathrm{[Cr(H_2O)_4Cl_2]Cl}

only one of the three chlorides reaches the silver ion. Counting all three triples the answer, and that value is always among the options.

Use the portion that was actually tested

When a mixed solution is split in two and each half meets a different reagent, each half carries only half the moles. Working from the whole volume doubles every answer.

Read which ratio the question wants

Moles of complex per mole of AgCl and moles of AgCl per mole of complex are reciprocals. For

\mathrm{[Cr(H_2O)_6]Cl_3}

the first is 1/3 and the second is 3.

Primary valency is the oxidation state, not the ionisable count

\mathrm{[Co(NH_3)_4Cl_2]Cl}

has only one ionisable chloride, but cobalt is still +3, so its primary valency is 3. The number of ions outside the bracket tells you the formula, not the valency.

A complex counter-ion changes the metal's charge

In

\mathrm{Hg[Co(SCN)_4]}

the mercury ion is +2, so the complex anion is 2− and cobalt is +2, with coordination number 4. A 2025 key gave this complex primary valency 3; the charge balance gives 2.

Ligands, Denticity and Nomenclature

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IUPAC names, oxidation state and d-electron count of complexes

Oxidation state and d count

x + \sum q_{\text{ligands}} = q_{\text{complex}} \qquad n_d = \text{group number} - x

Denticity and types of ligands

Ligand	Donor atom(s)	Denticity	Type
$\mathrm{NH_3}$ , $\mathrm{H_2O}$	N; O	1	Monodentate, neutral
$\mathrm{NO_2^-}$	N (nitro) or O (nitrito)	1	Ambidentate
$\mathrm{SCN^-}$	S (thiocyanato) or N (isothiocyanato)	1	Ambidentate
$\mathrm{CN^-}$	C (cyanido) or N (isocyanido)	1	Ambidentate
en, $\mathrm{H_2NCH_2CH_2NH_2}$	Two N	2	Chelating, neutral
Oxalate, $\mathrm{C_2O_4^{2-}}$	Two O	2	Chelating, not ambidentate Oxalate uses both oxygens at once; it has no choice of donor atom, so it is not ambidentate.
Biuret, $\mathrm{H_2NCONHCONH_2}$	Two carbonyl O, or two deprotonated amide N in alkali	2	Chelating
EDTA⁴⁻	Two N and four O	6	Chelating; octahedral even around Ca²⁺
$\mathrm{PPh_3}$ (in Wilkinson's catalyst)	P	1	σ-donor and π-acceptor

Chelating and ambidentate are different ideas: a chelate uses two atoms together, an ambidentate ligand uses one of two.

Nickel dimethylglyoximate and copper sulphate pentahydrate

Feature	Value	Reason
Colour of $\mathrm{[Ni(dmgH)_2]}$	Red (rosy red precipitate)	Used to detect Ni²⁺ in ammoniacal solution
Geometry and magnetism of $\mathrm{[Ni(dmgH)_2]}$	Square planar, diamagnetic	d⁸ Ni²⁺ with dsp² hybridisation; N–Ni–N angles close to 90°
H atoms in $\mathrm{[Ni(dmgH)_2]}$	14	Two ligands of $\mathrm{C_4H_7N_2O_2^-}$
H atoms in hydrogen bonds	2	One O–H···O bridge on each side; the other 12 H are in methyl groups
Five-membered rings in $\mathrm{[Ni(dmgH)_2]}$	2	One chelate ring per dmgH⁻; the other two rings are six-membered
Charge of the ligand as bound	−1 (dmgH⁻)	One oxime proton is lost; dmgH₂ itself is neutral
Waters bonded to Cu in $\mathrm{CuSO_4\cdot 5H_2O}$	4	Secondary valency 4 in the textbook formula $\mathrm{[Cu(H_2O)_4]SO_4\cdot H_2O}$
Hydrogen-bonded water in $\mathrm{CuSO_4\cdot 5H_2O}$	1	Held between sulphate and coordinated water, not bonded to Cu

Nickel dimethylglyoximate and blue vitriol are asked as facts, so the numbers in this table are worth memorising.

Common traps

A chelating ligand is not ambidentate

Oxalate and ethane-1,2-diamine bind through two atoms together, so they chelate. Ambidentate means one bond through either of two different atoms, as in

\mathrm{NO_2^-}

,

\mathrm{SCN^-}

and

\mathrm{CN^-}

.

Nitrogen and phosphorus ligands bond differently

\mathrm{N(CH_3)_3}

and

\mathrm{P(CH_3)_3}

both donate a lone pair, but only phosphorus can also accept π electrons back from the metal. A statement that their bonding is always the same is false.

Wilkinson's catalyst carries triphenylphosphine

Wilkinson's catalyst is

\mathrm{[RhCl(PPh_3)_3]}

. Its ligand is triphenylphosphine, a phosphine, not

\mathrm{PH_3}

itself. A 2024 key treated 'phosphine acts as a ligand in Wilkinson catalyst' as false, reading phosphine as

\mathrm{PH_3}

.

dmgH⁻ is an anion when it binds

Dimethylglyoxime loses one proton before it binds nickel, so the ligand is dmgH⁻ and the complex is neutral. Calling it a neutral bidentate ligand is the error in a common statement question.

Blue vitriol has a longer story than four waters

JEE keys follow the textbook formula

\mathrm{[Cu(H_2O)_4]SO_4\cdot H_2O}

: four waters on copper, secondary valency 4. In the real crystal two sulphate oxygens also sit on copper at longer distances, so copper is six-coordinate. Answer with 4 unless the question describes the crystal.

Oxido, not oxo; manganate, not permanganate

Current IUPAC names call the oxide ligand oxido, and an anionic manganese complex is a manganate. Permanganate is the common name of

\mathrm{MnO_4^-}

only, and the charge fixes the oxidation state:

\mathrm{K_2MnO_4}

is manganese(VI).

Bis and tris for ligands that already carry a number

Two trimethylphosphine ligands are bis(trimethylphosphine), not di(trimethylphosphine). Both trioxalatocobaltate(III) and tris(oxalato)cobaltate(III) are accepted for

\mathrm{[Co(C_2O_4)_3]^{3-}}

; pick the one the options offer.

Isomerism in Coordination Compounds

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Optical isomers and total stereoisomers of octahedral complexes

Counting stereoisomers

N_{\text{stereo}} = N_{\text{achiral geometrical}} + 2\,N_{\text{chiral geometrical}}

Types of structural isomerism in complexes

Type	What changes	Example pair or member	How to tell
Linkage	Donor atom of an ambidentate ligand	$\mathrm{[Co(NH_3)_5(NO_2)]Cl_2}$ and $\mathrm{[Co(NH_3)_5(ONO)]Cl_2}$	Colour and infrared spectrum differ
Ionisation	Which anion is inside the bracket	$\mathrm{[Co(NH_3)_5SO_4]Br}$ and $\mathrm{[Co(NH_3)_5Br]SO_4}$	$\mathrm{AgNO_3}$ test for the halide, $\mathrm{BaCl_2}$ test for sulphate
Coordination	Which metal holds which ligands	$\mathrm{[Co(NH_3)_6][Cr(CN)_6]}$ and $\mathrm{[Cr(NH_3)_6][Co(CN)_6]}$	Needs a complex cation AND a complex anion of different metals
Solvate (hydrate)	Water inside or outside the sphere	$\mathrm{[Cr(H_2O)_6]Cl_3}$ and $\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}$	Number of chlorides precipitated by $\mathrm{AgNO_3}$ (3 and 2)

Ionisation and hydrate isomers are told apart by what precipitates; linkage isomers need an ambidentate ligand.

Geometrical isomers of square planar and octahedral complexes

Type	Example	Geometrical isomers	Names
Tetrahedral $\mathrm{MABCD}$	An sp³ complex MABXL	0	All corners equivalent
Square planar $\mathrm{MA_2B_2}$	$\mathrm{[Pt(NH_3)_2Cl_2]}$	2	cis and trans (cisplatin is the cis form)
Square planar $\mathrm{MABCD}$	$\mathrm{[Pt(py)(NH_3)BrCl]}$	3	Each of B, C, D trans to A in turn
Octahedral $\mathrm{MA_4B_2}$	$\mathrm{[Co(NH_3)_4Cl_2]^+}$	2	cis and trans
Octahedral $\mathrm{MA_3B_3}$	$\mathrm{[Co(NH_3)_3Cl_3]}$	2	fac and mer
Octahedral $\mathrm{M(AA)_2B_2}$	$\mathrm{[Co(en)_2Cl_2]^+}$	2	cis and trans (cis is also chiral)
Octahedral $\mathrm{M(AA)_3}$	$\mathrm{[Co(en)_3]^{3+}}$	0	Only optical isomers
Octahedral $\mathrm{MA_5B}$	$\mathrm{[Co(CN)_5(NC)]^{3-}}$	0	One odd ligand has only one kind of position

Count geometrical isomers from the formula type; the metal and the ligands' names do not matter.

Common traps

Coordination isomerism needs two different metals

NCERT defines coordination isomerism as an exchange of ligands between the cationic and anionic complexes of DIFFERENT metal ions. By that definition

\mathrm{[Co(NH_3)_6][Co(CN)_6]}

, with cobalt in both ions, is not counted. A 2026 key follows this definition.

Ionisation isomers give different ions, not different amounts

\mathrm{[Co(NH_3)_5SO_4]Cl}

gives chloride and

\mathrm{[Co(NH_3)_5Cl]SO_4}

gives sulphate. If one isomer precipitates with

\mathrm{AgNO_3}

and the other with

\mathrm{BaCl_2}

, the isomerism is ionisation, not linkage or coordination.

Tetrahedral complexes have no cis–trans isomers

Every pair of corners of a tetrahedron is adjacent, so no ligand can be trans to another. An sp³ complex MABXL has 0 geometrical isomers, even with four different ligands.

Do not count the cis enantiomers as geometrical isomers

\mathrm{[Co(en)_2Cl_2]^+}

has 2 geometrical isomers, cis and trans. The cis form also has a mirror image, which makes 3 stereoisomers, but the extra one is an optical isomer.

Changing Ni²⁺ to Pt²⁺ changes the geometry

\mathrm{[NiCl_2Br_2]^{2-}}

is tetrahedral and paramagnetic with no geometrical isomers.

\mathrm{[PtCl_2Br_2]^{2-}}

is square planar, diamagnetic and has cis and trans forms.

A stereoisomer count includes the optical isomers

\mathrm{[CrCl_2(ox)_2]^{3-}}

has 2 geometrical isomers but 3 stereoisomers, because its cis form is chiral.

\mathrm{[CrCl_3(py)_3]}

has 2 stereoisomers, since fac and mer are both achiral.

Look for the mirror plane, not for four different groups

The carbon rule of four different groups does not apply to complexes. cis-

\mathrm{[PtCl_2(en)_2]^{2+}}

, with only two kinds of ligand, is chiral; trans-

\mathrm{[Co(NH_3)_4Cl_2]^+}

is not.

Hybridisation and Magnetism: Valence Bond Theory

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Inner-orbital and outer-orbital octahedral complexes

Octahedral hybridisation in valence bond theory

\text{strong field: } d^2sp^3 = (n-1)d^2\,ns\,np^3 \;(\text{inner, low spin}) \qquad \text{weak field: } sp^3d^2 = ns\,np^3\,nd^2 \;(\text{outer, high spin})

Four-coordinate complexes: tetrahedral or square planar

Complex	Metal and d count	Hybridisation and shape	Unpaired electrons
$\mathrm{Ni(CO)_4}$	Ni(0), 3d¹⁰ after 4s → 3d	sp³, tetrahedral	0 (diamagnetic) Ni(CO)₄ is diamagnetic; a statement calling it paramagnetic is false.
$\mathrm{[Ni(CN)_4]^{2-}}$	Ni²⁺, d⁸	dsp², square planar	0 (diamagnetic)
$\mathrm{[NiCl_4]^{2-}}$	Ni²⁺, d⁸	sp³, tetrahedral	2 (paramagnetic)
$\mathrm{[PtCl_4]^{2-}}$ , $\mathrm{[Pt(NH_3)_2Cl_2]}$	Pt²⁺, 5d⁸	dsp², square planar	0 (diamagnetic)
$\mathrm{[Cu(NH_3)_4]^{2+}}$	Cu²⁺, d⁹	Square planar	1 (paramagnetic)
$\mathrm{[Cu(CN)_4]^{3-}}$	Cu⁺, d¹⁰	sp³, tetrahedral	0 (diamagnetic)
$\mathrm{[Zn(NH_3)_4]^{2+}}$	Zn²⁺, d¹⁰	sp³, tetrahedral	0 (diamagnetic)
$\mathrm{[CoCl_4]^{2-}}$	Co²⁺, d⁷	sp³, tetrahedral	3 (paramagnetic)
$\mathrm{[MnBr_4]^{2-}}$	Mn²⁺, d⁵	sp³, tetrahedral	5 (paramagnetic)

For a four-coordinate complex decide the shape first; the magnetism follows from it.

Hybridisation, geometry and the limits of valence bond theory

Hybridisation	Coordination number and shape	d orbital used	Example
sp	2, linear	None	$\mathrm{[Ag(NH_3)_2]^+}$ , $\mathrm{[Ag(CN)_2]^-}$
sp³	4, tetrahedral	None	$\mathrm{[MnBr_4]^{2-}}$ , $\mathrm{Ni(CO)_4}$
dsp²	4, square planar	Inner $3d_{x^2-y^2}$	$\mathrm{[Ni(CN)_4]^{2-}}$
dsp³	5, trigonal bipyramidal	Inner $3d_{z^2}$	$\mathrm{Fe(CO)_5}$
d²sp³	6, octahedral (inner orbital)	Inner $3d_{x^2-y^2}$ and $3d_{z^2}$	$\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Co(C_2O_4)_3]^{3-}}$
sp³d²	6, octahedral (outer orbital)	Outer $4d_{x^2-y^2}$ and $4d_{z^2}$	$\mathrm{[CoF_6]^{3-}}$ , $\mathrm{[FeF_6]^{3-}}$

The d orbitals used are the ones that point at the ligands.

Common traps

Spin paired is low spin; spin free is high spin

\mathrm{[Co(NH_3)_6]^{3+}}

is a spin-paired (inner-orbital) complex and

\mathrm{[CoF_6]^{3-}}

is spin free (outer orbital). Swapping the two terms is the whole of one recurring question.

Octahedral nickel(II) is always outer orbital

Ni²⁺ is d⁸. Even with ammonia or en, pairing can empty only one 3d orbital, so octahedral nickel(II) complexes are sp³d² with 2 unpaired electrons.

Two textbook shortcuts that JEE keys have used

Tris(carbonato)cobaltate(III) is, like the oxalato complex, low spin and diamagnetic; a 2026 key treated

\mathrm{K_3[Co(CO_3)_3]}

as high spin sp³d² with 4.90 BM. And a 2023 key gave

\mathrm{[Fe(NH_3)_6]^{2+}}

as d²sp³, although measured it is high spin. Answer with the key's rule only where the options force it.

Ni(CO)₄ and [NiCl₄]²⁻ are both tetrahedral but differ in d count

Ni(CO)₄ is nickel(0), 3d¹⁰, diamagnetic.

\mathrm{[NiCl_4]^{2-}}

is nickel(II), d⁸, with 2 unpaired electrons. They share a shape, not a configuration.

[Ni(CN)₄]²⁻ is dsp², not sp³

Cyanide pairs the eight d electrons of Ni²⁺ into four orbitals, which empties one 3d orbital for dsp² bonding. The complex is square planar and diamagnetic.

sp³ can be diamagnetic or paramagnetic

Hybridisation alone does not fix the magnetism.

\mathrm{Ni(CO)_4}

is sp³ and diamagnetic;

\mathrm{[MnBr_4]^{2-}}

is sp³ with 5 unpaired electrons. Count the d electrons for each row of a match list.

Anionic ligands are not the strongest

A pure point-charge picture predicts that anions split the d orbitals most, yet halides sit at the weak end of the series and neutral CO at the strong end. So the statement that crystal field theory explains the strength of anionic ligands is false.

Crystal Field Splitting, the Spectrochemical Series and Colour

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Octahedral and tetrahedral splitting of the d orbitals

Crystal field splitting

E(e_g) = +0.6\Delta_o,\; E(t_{2g}) = -0.4\Delta_o \qquad E(t_2) = +0.4\Delta_t,\; E(e) = -0.6\Delta_t \qquad \Delta_t = \tfrac{4}{9}\Delta_o

Colour, absorbed wavelength and the splitting energy

Energy of the light absorbed

\Delta = \frac{hc}{\lambda} \qquad \Delta_{\text{molar}} = \frac{N_A\,hc}{\lambda}

Spectrochemical series and the size of the splitting

Ligand	Donor atom	Place in the series	Field
$\mathrm{I^-}$ , $\mathrm{Br^-}$	I, Br	Weakest	Weak
$\mathrm{SCN^-}$	S	Between Br⁻ and Cl⁻	Weak
$\mathrm{Cl^-}$ , $\mathrm{S^{2-}}$ , $\mathrm{F^-}$	Cl, S, F	Below OH⁻	Weak
$\mathrm{OH^-}$ , $\mathrm{C_2O_4^{2-}}$	O	Just below water	Weak
$\mathrm{H_2O}$	O	Middle of the series	Weak for most M²⁺; strong enough to pair Co³⁺
$\mathrm{NCS^-}$ , $\mathrm{EDTA^{4-}}$	N; N and O	Just above water	Intermediate
$\mathrm{NH_3}$ , en	N	Above EDTA⁴⁻; en above NH₃	Strong for M³⁺
$\mathrm{CN^-}$ , CO	C	Strongest	Strong CO is neutral yet the strongest ligand: its π back-bonding, not its charge, widens the gap.

Weak to strong: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO.

Common traps

The tetrahedral pattern is upside down

In a tetrahedron

d_{xy}

,

d_{xz}

,

d_{yz}

lie ABOVE

d_{x^2-y^2}

and

d_{z^2}

. Writing the octahedral order for a tetrahedral complex such as

\mathrm{[NiCl_4]^{2-}}

reverses every comparison.

Convert Δt to Δo with 9/4

If a tetrahedral complex absorbs light of energy

\Delta_t

, the octahedral splitting for the same metal and ligand is

\tfrac{9}{4}\Delta_t

, not

\Delta_t

. Forgetting the factor gives an answer 4/9 of the right one.

S-bonded thiocyanate is weak, N-bonded is not

\mathrm{SCN^-}

bonded through sulphur sits between Br⁻ and Cl⁻. Bonded through nitrogen,

\mathrm{NCS^-}

sits above water. Read which atom the question binds.

Splitting energy and CFSE are different quantities

\Delta_o

is the gap; the CFSE is that gap times a factor set by the d count. Among the hexaaqua ions of Ti³⁺, Cr³⁺, Mn³⁺ and Fe³⁺, a 2023 key picked Cr³⁺ for the 'highest

\Delta_o

'. That is true of the CFSE in

\Delta_o

units (

-1.2\Delta_o

for d³), not of the measured gap.

A stronger field absorbs a SHORTER wavelength

Energy and wavelength are inverse.

\mathrm{[CoCl(NH_3)_5]^{2+}}

, with the weaker Cl⁻, absorbs at a LONGER wavelength than

\mathrm{[Co(NH_3)_5(H_2O)]^{3+}}

. Wavenumber, by contrast, rises with field strength.

Absorbed colour is not the colour seen

A complex that absorbs orange-red light looks blue-green. Match the ORDER of absorbed wavelengths to field strength first; convert to a seen colour only if the question asks for it.

Energy absorbed is not intensity

A 2021 key ranked 'intensity of colour' of

\mathrm{[Ni(CN)_4]^{2-}}

,

\mathrm{[Ni(H_2O)_4]^{2+}}

and

\mathrm{[NiCl_4]^{2-}}

by field strength. That orders the ENERGY absorbed; tetrahedral complexes actually absorb more intensely. Another key calls

\mathrm{[Ni(CN)_4]^{2-}}

colourless, though its solutions are pale yellow.

High and Low Spin Configurations and CFSE

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High-spin and low-spin octahedral configurations

Spin state criterion

\Delta_o > P \Rightarrow \text{low spin } (t_{2g} \text{ filled first}) \qquad \Delta_o < P \Rightarrow \text{high spin}

Crystal field stabilisation energy of octahedral complexes

Octahedral CFSE

\text{CFSE} = \left(-0.4\,n_{t_{2g}} + 0.6\,n_{e_g}\right)\Delta_o

Tetrahedral configurations and their CFSE

d count	Configuration	Unpaired electrons	CFSE
d⁰	$e^0t_2^0$	0	0
d¹	$e^1t_2^0$	1	$-0.6\Delta_t$
d²	$e^2t_2^0$	2	$-1.2\Delta_t$
d³	$e^2t_2^1$	3	$-0.8\Delta_t$
d⁴	$e^2t_2^2$	4	$-0.4\Delta_t$
d⁵	$e^2t_2^3$	5	0
d⁶	$e^3t_2^3$	4	$-0.6\Delta_t$
d⁷	$e^4t_2^3$	3	$-1.2\Delta_t$
d⁸	$e^4t_2^4$	2	$-0.8\Delta_t$
d¹⁰	$e^4t_2^6$	0	0

Tetrahedral complexes are always high spin, so each d count has exactly one row.

Common traps

t₂g³eg¹ is the weak-field configuration

Putting the fourth electron in eg means the pairing energy was larger than

\Delta_o

: a weak-field ligand and a high-spin complex. The strong-field d⁴ configuration is

t_{2g}^4

.

Pairs and unpaired electrons are different counts

Low-spin

\mathrm{[Fe(CN)_6]^{4-}}

has six t₂g electrons: 3 PAIRS and 0 unpaired. Read whether the question wants electrons, pairs or unpaired electrons.

In a tetrahedron, e is filled first

The lower set is e, not t₂. Writing a tetrahedral d⁷ ion as

t_2^4e^3

copies the octahedral order and gets the configuration and the CFSE wrong. It is

e^4t_2^3

, CFSE

-1.2\Delta_t

.

A tetrahedral CFSE is measured in Δt

Paramagnetic

\mathrm{[Ni(PPh_3)_2Cl_2]}

is tetrahedral, so its CFSE is

-0.8\Delta_t

, not

-0.8\Delta_o

. A statement giving it in

\Delta_o

is incorrect.

CFSE is not the splitting energy

For

\mathrm{[Ti(H_2O)_6]^{3+}}

(d¹) the CFSE is

-0.4\Delta_o

, so

\Delta_o

is 2.5 times the CFSE magnitude. The light absorbed matches

\Delta_o

, not the CFSE.

Zero CFSE means high-spin d⁵ or d¹⁰

A complex with 'CFSE = 0' and a moment near 5.9 BM has five unpaired electrons: high-spin d⁵, such as Mn²⁺ or Fe³⁺ with a weak ligand like SCN⁻. The splitting itself is not zero.

Rules of thumb can clash with the numbers

A 2021 key ordered CFSE as

\mathrm{[Co(H_2O)_6]^{2+} < [CoF_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(en)_3]^{3+}}

, putting the +2 ion lowest by charge. In

\Delta_o

units the Co²⁺ aqua ion is

-0.8\Delta_o

and

\mathrm{[CoF_6]^{3-}}

only

-0.4\Delta_o

. Use the charge rule when comparing the SAME configuration.

Spin-Only Magnetic Moment

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Spin-only magnetic moment from unpaired electrons

Spin-only magnetic moment

\mu_{\text{spin-only}} = \sqrt{n(n+2)}\ \text{BM}

Counting paramagnetic species in a list

Paramagnetic test

n \geq 1 \Rightarrow \text{paramagnetic} \qquad n = 0 \Rightarrow \text{diamagnetic}

Unpaired electrons and moments of high-spin aqua ions

Aqua ion (high spin)	d count	Unpaired electrons	Spin-only moment (BM)
$\mathrm{Ti^{3+}}$	d¹	1	1.73
$\mathrm{V^{3+}}$	d²	2	2.83
$\mathrm{V^{2+}}$ , $\mathrm{Cr^{3+}}$	d³	3	3.87
$\mathrm{Cr^{2+}}$ , $\mathrm{Mn^{3+}}$	d⁴	4	4.90
$\mathrm{Mn^{2+}}$ , $\mathrm{Fe^{3+}}$	d⁵	5	5.92 The maximum: a d⁵ ion with a weak-field ligand.
$\mathrm{Fe^{2+}}$ , $\mathrm{Co^{3+}}$ (with F⁻)	d⁶	4	4.90
$\mathrm{Co^{2+}}$	d⁷	3	3.87
$\mathrm{Ni^{2+}}$	d⁸	2	2.83
$\mathrm{Cu^{2+}}$	d⁹	1	1.73
$\mathrm{Zn^{2+}}$	d¹⁰	0	0

Cr³⁺ is 3.87 BM with every ligand, because d³ has only one octahedral configuration.

Common traps

Copper(I) is diamagnetic

CuI and

\mathrm{K_3[Cu(CN)_4]}

contain Cu⁺, which is d¹⁰, so their moment is 0, not 1.73 BM. Only copper(II) has one unpaired electron.

Watch the units the answer is asked in

2.83 BM written in units of

10^{-1}

BM is 28. Writing 3 (rounded BM) or 283 in that blank loses the mark.

The same oxidation state does not explain different moments

Fe is +3 in both

\mathrm{[Fe(CN)_6]^{3-}}

(1.73 BM) and

\mathrm{[Fe(H_2O)_6]^{3+}}

(5.92 BM). The difference comes from the ligand's field strength, so a reason that cites the common oxidation state is true but not the explanation.

Fe²⁺ and Fe³⁺ aqua ions differ by one unpaired electron

High-spin Fe³⁺ (d⁵) has 5 unpaired electrons and Fe²⁺ (d⁶) has 4. The extra electron in d⁶ pairs up, so adding an electron here LOWERS the moment.

Change of ligand can reverse an order

\mathrm{[FeF_6]^{3-}}

(5 unpaired) is above

\mathrm{[CoF_6]^{3-}}

(4), but

\mathrm{[Fe(CN)_6]^{3-}}

(1) is below

\mathrm{[Mn(CN)_6]^{3-}}

(2). Decide the spin state for each complex before ordering.

Low-spin d⁶ is diamagnetic

\mathrm{[Fe(CN)_6]^{4-}}

,

\mathrm{[Co(NH_3)_6]^{3+}}

and

\mathrm{[Co(C_2O_4)_3]^{3-}}

have all six electrons paired in t₂g. Counting them as paramagnetic because 'iron and cobalt are magnetic' is the commonest slip in these lists.

Ferricyanide has one unpaired electron

\mathrm{[Fe(CN)_6]^{3-}}

is low-spin d⁵,

t_{2g}^5

: one unpaired electron, so it is paramagnetic, while ferrocyanide is not.

Metal Carbonyls, Stability and Applications

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Stability constants and the chelate effect

Overall stability constant

\beta_n = \frac{[\mathrm{ML}_n]}{[\mathrm{M}][\mathrm{L}]^n} = K_1K_2\cdots K_n \qquad K_{\text{diss}} = \frac{1}{\beta_n}

Synergic bonding and structures of metal carbonyls

Carbonyl	Shape at each metal	Bridging CO	Metal–metal bonds
$\mathrm{Ni(CO)_4}$	Tetrahedral	0	0
$\mathrm{Fe(CO)_5}$	Trigonal bipyramidal	0	0
$\mathrm{Cr(CO)_6}$ , $\mathrm{W(CO)_6}$	Octahedral	0	0
$\mathrm{Mn_2(CO)_{10}}$	Octahedral (five CO and one Mn–Mn bond)	0	1 Mn–Mn Decacarbonyldimanganese(0) has ten terminal CO groups and no bridge.
$\mathrm{Co_2(CO)_8}$	Two Co(CO)₃ units joined by two CO bridges	2 (with 6 terminal)	1 Co–Co

Only the dicobalt carbonyl here has bridging CO groups.

Complexes in biology, medicine, industry and analysis

Substance	Metal	Role
Chlorophyll	Mg	Photosynthetic pigment
Haemoglobin	Fe	Oxygen carrier in blood
Vitamin B₁₂ (cyanocobalamin)	Co	Anti-pernicious-anaemia factor
Cisplatin	Pt	Anticancer drug
Wilkinson's catalyst	Rh	Hydrogenation of alkenes
Ziegler–Natta catalyst	Ti (with Al)	Polymerisation of alkenes
Grubbs catalyst	Ru	Alkene metathesis
$\mathrm{[Ag(S_2O_3)_2]^{3-}}$ (from hypo)	Ag	Fixing in black-and-white photography Photography uses the thiosulphate complex, not $\mathrm{[Ag(CN)_2]^-}$ .
$\mathrm{[Ag(CN)_2]^-}$ , $\mathrm{[Au(CN)_2]^-}$	Ag, Au	Extraction by cyanide leaching; electroplating
EDTA	Ca, Mg (and Pb)	Water-hardness titration; treatment of lead poisoning
D-Penicillamine	Cu	Chelating drug for excess copper

Most match lists pair a biological or catalytic name with its metal.

Common traps

Synergic bonding strengthens the metal–carbon bond

Back-donation adds a π bond to the σ bond, so the M–C bond becomes STRONGER. It is the C–O bond that weakens, because electrons enter the antibonding π* orbital of CO.

π-acceptors, not π-donors, stabilise low oxidation states

A metal in the zero oxidation state is electron rich and needs ligands that take electrons away. A reason saying low oxidation states need π-DONOR ligands is false.

Use the FREE ligand, not the total added

The ligand bound in the complex is not free. If 1.0 mol of metal ion takes 2 mol of ligand into the complex, subtract those 2 mol from the total before putting [L] into

\beta

.

Chelation raises stability at the same metal and charge

Replacing two

\mathrm{NH_3}

by one en keeps the donor atoms the same but adds a ring, and each step raises the stability:

\mathrm{[Co(en)_3]^{2+}}

is the most stable of the ammine–en series.

Chlorophyll is magnesium, vitamin B₁₂ is cobalt

The two porphyrin-like biological complexes are easily swapped. Chlorophyll holds Mg²⁺; vitamin B₁₂ holds cobalt; haemoglobin holds iron.

EDTA is not an anticancer drug

EDTA and D-penicillamine are chelating agents that remove unwanted metals (lead, copper). Tumour growth is inhibited by platinum complexes such as cisplatin.

Coordination Compounds formulas

Werner's Theory and Ionisable Ligands

Counting the ions outside the coordination sphere

Primary valency, secondary valency and double salts

Common traps

Ligands, Denticity and Nomenclature

IUPAC names, oxidation state and d-electron count of complexes

Denticity and types of ligands

Nickel dimethylglyoximate and copper sulphate pentahydrate

Common traps

Isomerism in Coordination Compounds

Optical isomers and total stereoisomers of octahedral complexes

Types of structural isomerism in complexes

Geometrical isomers of square planar and octahedral complexes

Common traps

Hybridisation and Magnetism: Valence Bond Theory

Inner-orbital and outer-orbital octahedral complexes

Four-coordinate complexes: tetrahedral or square planar

Hybridisation, geometry and the limits of valence bond theory

Common traps

Crystal Field Splitting, the Spectrochemical Series and Colour

Octahedral and tetrahedral splitting of the d orbitals

Colour, absorbed wavelength and the splitting energy

Spectrochemical series and the size of the splitting

Common traps

High and Low Spin Configurations and CFSE

High-spin and low-spin octahedral configurations

Crystal field stabilisation energy of octahedral complexes

Tetrahedral configurations and their CFSE

Common traps

Spin-Only Magnetic Moment

Spin-only magnetic moment from unpaired electrons

Counting paramagnetic species in a list

Unpaired electrons and moments of high-spin aqua ions

Common traps

Metal Carbonyls, Stability and Applications

Stability constants and the chelate effect

Synergic bonding and structures of metal carbonyls

Complexes in biology, medicine, industry and analysis

Common traps

More JEE Mains Chemistry formula sheets