Coordination Compounds Class 12 Notes for JEE and NEET

ChemistryClass 12 Chemistry, Unit 5For JEE and NEET

The whole chapter on one page for JEE and NEET: Werner's theory, ligands and naming, isomerism, valence bond and crystal field theory, magnetism, colour, stability and uses, each with the condition under which it holds. A table shows how often each idea was needed in past JEE and NEET papers, and ten past-paper questions from both exams are solved at the end. For the full explanation of a topic, follow the links to our concept pages below.

1. Ligands, Werner's theory and naming

A coordination compound has a central metal atom or ion bonded to ligands, ions or molecules that each give it at least one lone pair. The metal and its ligands form the coordination entity, written in square brackets; ions outside the brackets are counter ions.

Term	Meaning	Example
Denticity	How many donor atoms one ligand uses	unidentate $N H_{3}$ , $C l^{-}$ ; didentate en, $C_{2} O_{4}^{2 -}$ (ox); hexadentate $EDT A^{4 -}$
Chelate	A ligand bound through two or more atoms, so it closes a ring with the metal	en, ox, EDTA, dimethylglyoxime
Ambidentate	Two possible donor atoms, but only one binds at a time	$N O_{2}^{-}$ (through N) or $ON O^{-}$ (through O); $SC N^{-}$ or $NC S^{-}$
Coordination number	The number of donor atoms bonded to the metal	6 in $[Co (en)_{3}]^{3 +}$ (3 ligands × 2)
Oxidation state	The charge left on the metal when the ligands leave with their electron pairs	+3 in $[Co (N H_{3})_{6}] C l_{3}$
Homoleptic	Only one kind of ligand (heteroleptic: more than one)	$[Co (N H_{3})_{6}]^{3 +}$ (homoleptic), $[Co (N H_{3})_{4} C l_{2}]^{+}$ (heteroleptic)

Werner's theory. A metal shows two kinds of valency. The primary valency is its oxidation state: it is ionisable and satisfied by anions. The secondary valency is its coordination number: it is not ionisable and fixes the shape. Only the ions outside the brackets are free in solution, so only they precipitate with $AgN O_{3}$ or $BaC l_{2}$ . The ions in solution, which set the conductivity, are these counter ions plus the complex ion itself.

Old formula	Werner's formula	AgCl per mole	Ions in solution
$CoC l_{3} \cdot 6N H_{3}$	$[Co (N H_{3})_{6}] C l_{3}$	3	4
$CoC l_{3} \cdot 5N H_{3}$	$[Co (N H_{3})_{5} Cl] C l_{2}$	2	3
$CoC l_{3} \cdot 4N H_{3}$	$[Co (N H_{3})_{4} C l_{2}] Cl$	1	2

Naming (IUPAC).

Name the cation first, then the anion, as in any salt.
Inside the entity, list the ligands in alphabetical order, then the metal. Prefixes di, tri, tetra count simple ligands; bis, tris, tetrakis count ligands whose names already hold a number (bis(ethane-1,2-diamine)).
Anionic ligands end in -o: chlorido, cyanido, hydroxido, oxalato, nitrito-N (bound through N), nitrito-O. Neutral ligands keep their names, except aqua ( $H_{2} O$ ), ammine ( $N H_{3}$ ), carbonyl (CO) and nitrosyl (NO).
Give the metal's oxidation state in Roman numerals. In an anionic complex the metal ends in -ate: ferrate, cuprate, argentate, aurate, plumbate, stannate.
Examples: $K_{3} [Fe (CN)_{6}]$ is potassium hexacyanidoferrate(III); $[Co (N H_{3})_{4} C l_{2}] Cl$ is tetraamminedichloridocobalt(III) chloride; $[Ag (H_{2} O)_{2}] [Ag (CN)_{2}]$ is diaquasilver(I) dicyanidoargentate(I).

2. Isomerism

Structural isomers have different bonds.

Type	What changes	Example pair
Linkage	An ambidentate ligand binds through its other atom	$[Co (N H_{3})_{5} (N O_{2})] C l_{2}$ and $[Co (N H_{3})_{5} (ONO)] C l_{2}$
Ionisation	An ion inside the brackets swaps with one outside	$[Co (N H_{3})_{5} (S O_{4})] Br$ and $[Co (N H_{3})_{5} Br] S O_{4}$
Coordination	Ligands swap between a complex cation and a complex anion	$[Co (N H_{3})_{6}] [Cr (CN)_{6}]$ and $[Cr (N H_{3})_{6}] [Co (CN)_{6}]$
Solvate (hydrate)	Water inside the brackets swaps with an ion outside	$[Cr (H_{2} O)_{6}] C l_{3}$ (violet) and $[Cr (H_{2} O)_{5} Cl] C l_{2} \cdot H_{2} O$ (grey-green)

Stereoisomers have the same bonds arranged differently in space: geometrical (cis and trans, fac and mer) and optical (non-superimposable mirror images). In fac the three like ligands sit on one face of the octahedron; in mer they lie around a meridian, in one plane with the metal (two of them trans to each other). A complex is chiral when it has no plane of symmetry: cis- $[Co (en)_{2} C l_{2}]^{+}$ and $[Co (en)_{3}]^{3 +}$ are chiral; trans- $[Co (en)_{2} C l_{2}]^{+}$ is not, and no square planar complex with simple ligands is chiral.

Type	Example	Geometrical isomers	Stereoisomers in all
square planar $M A_{2} B_{2}$	$[Pt (N H_{3})_{2} C l_{2}]$	2	2
square planar $MABCD$	$[Pt (N H_{3}) (H_{2} O) BrCl]$	3	3
tetrahedral $M A_{2} B_{2}$	$[Zn (N H_{3})_{2} C l_{2}]$	1	1
octahedral $M A_{4} B_{2}$	$[Co (N H_{3})_{4} C l_{2}]^{+}$	2	2
octahedral $M A_{3} B_{3}$	$[Co (N H_{3})_{3} (N O_{2})_{3}]$	2	2
octahedral $M A_{2} B_{2} C_{2}$	$[Co (N H_{3})_{2} (H_{2} O)_{2} C l_{2}]^{+}$	5	6
octahedral $M (AA)_{2} B_{2}$	$[Co (en)_{2} C l_{2}]^{+}$	2	3
octahedral $M (AA)_{3}$	$[Co (en)_{3}]^{3 +}$	1	2

The counts above were checked by placing the ligands on the corners in every possible way and comparing the arrangements by rotation (and by reflection, for geometrical isomers).

Syllabus: JEE Main and NEET list "isomerism". The JEE Advanced 2026 text names only cis-trans and ionisation isomerism, yet JEE Advanced 2026 asked linkage isomers and a mer isomer; learn them all.

3. Valence bond theory and magnetism

Find the metal's oxidation state and its number of $d$ electrons. A strong-field ligand ( $C N^{-}$ , CO, en, and $N H_{3}$ with Co(III)) pairs the $d$ electrons, so inner $d$ orbitals hybridise (inner orbital, low spin). A weak-field ligand ( $F^{-}$ , $C l^{-}$ , $H_{2} O$ ) leaves them unpaired, so the outer $d$ orbitals are used (outer orbital, high spin). For six-coordinate (octahedral) 3d ions this choice arises only for $d^{4}$ to $d^{7}$ : $d^{1}$ to $d^{3}$ ions are always inner orbital ( $d^{2} s p^{3}$ ), an octahedral $N i^{2 +}$ complex ( $d^{8}$ ) is always outer orbital ( $s p^{3} d^{2}$ , 2 unpaired), $C o^{3 +}$ pairs its electrons with every common ligand except $F^{-}$ , and 4d and 5d ions are low spin even with $C l^{-}$ ( $[PtC l_{4}]^{2 -}$ is $d s p^{2}$ ). Four-coordinate $N i^{2 +}$ ( $d^{8}$ ) also has a choice: $d s p^{2}$ square planar with $C N^{-}$ , $s p^{3}$ tetrahedral with $C l^{-}$ .

Coordination number	Hybridisation	Shape	Examples (unpaired electrons)
2	$s p$	linear	$[Ag (N H_{3})_{2}]^{+}$ (0)
4	$s p^{3}$	tetrahedral	$[NiC l_{4}]^{2 -}$ (2), $[Ni (CO)_{4}]$ (0)
4	$d s p^{2}$	square planar	$[Ni (CN)_{4}]^{2 -}$ (0), $[PtC l_{4}]^{2 -}$ (0)
6	$d^{2} s p^{3}$ (inner)	octahedral	$[Co (N H_{3})_{6}]^{3 +}$ (0), $[Fe (CN)_{6}]^{3 -}$ (1)
6	$s p^{3} d^{2}$ (outer)	octahedral	$[Co F_{6}]^{3 -}$ (4), $[Fe F_{6}]^{3 -}$ (5)

Spin-only magnetic moment

μ = n (n + 2) BM

$n$ = number of unpaired electrons. $n$ = 1, 2, 3, 4, 5 gives 1.73, 2.83, 3.87, 4.90, 5.92 BM. A complex with $n = 0$ is diamagnetic. Spin only: the orbital contribution is ignored, which works well for 3d ions.

4. Crystal field theory, CFSE and colour

Crystal field theory treats the ligands as point charges (anions) or point dipoles (neutral molecules). They raise the energy of the $d$ orbitals that point at them, so the five $d$ orbitals split.

Figure 1: Splitting of the five

d

orbitals, drawn to scale from their average energy (left). In an octahedral field

e_{g}

rises by

0.6 Δ_{o}

and

t_{2 g}

falls by

0.4 Δ_{o}

; in a tetrahedral field the order flips and the gap is only about

\frac{4}{9}

as large, so tetrahedral complexes are almost always high spin.

High spin or low spin. The fourth electron of an octahedral ion either pairs in $t_{2 g}$ (cost: the pairing energy $P$ ) or enters $e_{g}$ (cost: $Δ_{o}$ ). If $Δ_{o} < P$ (weak field) the complex is high spin; if $Δ_{o} > P$ (strong field) it is low spin. Only $d^{4}$ to $d^{7}$ ions differ:

Ion	High spin (weak field)	Low spin (strong field)
$d^{4}$	$t_{2 g}^{3} e_{g}^{1}$ (4 unpaired)	$t_{2 g}^{4}$ (2 unpaired)
$d^{5}$	$t_{2 g}^{3} e_{g}^{2}$ (5 unpaired)	$t_{2 g}^{5}$ (1 unpaired)
$d^{6}$	$t_{2 g}^{4} e_{g}^{2}$ (4 unpaired)	$t_{2 g}^{6}$ (0 unpaired)
$d^{7}$	$t_{2 g}^{5} e_{g}^{2}$ (3 unpaired)	$t_{2 g}^{6} e_{g}^{1}$ (1 unpaired)

Spectrochemical series (ligands in order of field strength, weak to strong):

$I^{-}$ < $B r^{-}$ < $SC N^{-}$ < $C l^{-}$ < $S^{2 -}$ < $F^{-}$ < $O H^{-}$ < $C_{2} O_{4}^{2 -}$ < $H_{2} O$ < $NC S^{-}$ < $EDT A^{4 -}$ < $N H_{3}$ < en < $C N^{-}$ < CO

Crystal field stabilisation energy (CFSE)

CFSE = (- 0.4 x + 0.6 y) Δ_{o}

$x$ electrons in $t_{2 g}$ and $y$ in $e_{g}$ . Tetrahedral: $(- 0.6 x + 0.4 y) Δ_{t}$ with $x$ in $e$ , $y$ in $t_{2}$ and $Δ_{t} \approx \frac{4}{9} Δ_{o}$ . Examples: low-spin $d^{6}$ , $- 2.4 Δ_{o}$ ; high-spin $d^{5}$ , 0. The pairing energy $P$ is left out; some questions add $P$ for each extra pair of electrons.

Colour. A complex absorbs light whose energy matches $Δ$ and shows the complementary colour. A larger $Δ$ means a shorter wavelength absorbed: $Δ$ (kJ/mol) $= 119627/ λ$ (nm). $[Ti (H_{2} O)_{6}]^{3 +}$ absorbs near 498 nm (240 kJ/mol), blue-green light, and looks violet. Ions with $d^{0}$ or $d^{10}$ have no $d$ - $d$ transition, so $[Sc (H_{2} O)_{6}]^{3 +}$ and $[Zn (H_{2} O)_{6}]^{2 +}$ are colourless.

Syllabus: JEE Main and NEET say "basic ideas of Crystal field theory"; JEE Advanced names octahedral and tetrahedral fields, the spectrochemical series, and colour of 3d-series compounds.

5. Stability, metal carbonyls and uses

Stability. For $M + 4L ⇌ M L_{4}$ the overall stability constant is $β_{4}$ $= [M L_{4}] / ([M] [L]^{4})$ ; a larger $β$ means a more stable complex. Chelating ligands give more stable complexes than the same number of unidentate ligands (the chelate effect): $[Ni (en)_{3}]^{2 +}$ is far more stable than $[Ni (N H_{3})_{6}]^{2 +}$ .

Metal carbonyls

In $[Ni (CO)_{4}]$ (tetrahedral), $[Fe (CO)_{5}]$ (trigonal bipyramidal) and $[Cr (CO)_{6}]$ (octahedral) the metal is in the zero oxidation state. Carbon gives its lone pair to an empty metal orbital (a $σ$ bond) and a filled metal $d$ orbital gives electrons back into the empty $π^{*}$ orbital of CO (a $π$ bond). This synergic bonding makes the metal-carbon bond stronger and the C-O bond weaker.

Syllabus: metal carbonyls and stability are named only in JEE Advanced 2026; JEE Main 2026 still asked about synergic bonding, and NEET 2018 asked about $[Fe (CO)_{5}]$ .

Uses

Where	Example
Tests in qualitative analysis	$N i^{2 +}$ with dimethylglyoxime gives red $[Ni (dmg)_{2}]$ (square planar, diamagnetic); $F e^{3 +}$ with $[Fe (CN)_{6}]^{4 -}$ gives Prussian blue; $C u^{2 +}$ gives chocolate-brown $C u_{2} [Fe (CN)_{6}]$ ; $C o^{2 +}$ with $KN O_{2}$ gives yellow $K_{3} [Co (N O_{2})_{6}]$ ; $S^{2 -}$ turns sodium nitroprusside purple
Hardness of water	$C a^{2 +}$ and $M g^{2 +}$ are titrated with EDTA
Extraction of metals	Silver and gold dissolve as $[Ag (CN)_{2}]^{-}$ and $[Au (CN)_{2}]^{-}$ (cyanide process); zinc then displaces the metal
Biology	chlorophyll (Mg), haemoglobin (Fe), vitamin $B_{12}$ (Co)
Medicine and industry	cisplatin, cis- $[Pt (N H_{3})_{2} C l_{2}]$ , treats some cancers; EDTA removes lead from the body; Wilkinson's catalyst $[(P h_{3} P)_{3} RhCl]$ hydrogenates alkenes

N a_{2} [Fe (CN)_{5} NO] + N a_{2} S N a_{4} [Fe (CN)_{5} NOS] (purple)

4Au + 8C N^{-} + 2 H_{2} O + O_{2} 4 [Au (CN)_{2}]^{-} + 4O H^{-}

2 [Au (CN)_{2}]^{-} + Zn 2Au + [Zn (CN)_{4}]^{2 -}

6. Which ideas JEE and NEET use most

We read all 116 questions on coordination compounds in our 55 past papers (JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026) and noted the ideas each solution needs; one question can need several. The percentages are shares of each exam's questions on this chapter. The ideas used most:

Idea or formula	JEE (48 papers)	NEET (7 papers)
Crystal field theory: splitting $Δ_{o}$ and $Δ_{t} \approx \frac{4}{9} Δ_{o}$ , high or low spin, the spectrochemical series	3737%	424%
Spin-only magnetic moment $μ = n (n + 2)$ BM; paramagnetic or diamagnetic	3333%	318%
Valence bond theory: $s p^{3}$ , $d s p^{2}$ , $d^{2} s p^{3}$ , $s p^{3} d^{2}$ ; inner and outer orbital complexes	1717%	318%
Werner's theory: ions outside the brackets, AgCl precipitated, conductivity	1111%	16%
Geometrical isomerism: cis-trans, fac-mer, counting isomers	1010%	212%
CFSE: $- 0.4 Δ_{o}$ per $t_{2 g}$ electron, $+ 0.6 Δ_{o}$ per $e_{g}$ electron	1111%	00%
Uses: tests in qualitative analysis, extraction of metals, biology, catalysts	1010%	16%
Ligands, denticity, chelates, ambidentate ligands, oxidation state, homoleptic	88%	318%

Show the other 6

Idea or formula	JEE (48 papers)	NEET (7 papers)
Colour: absorbed wavelength falls as $Δ$ rises; the complementary colour is seen	77%	212%
Structural isomerism: linkage, ionisation, coordination, solvate	66%	212%
Stability and the chelate effect	33%	16%
Optical isomerism	33%	16%
Metal carbonyls and synergic bonding	22%	16%
IUPAC names and formulas	11%	212%

Questions per paper: JEE 99 in 48 papers (about 2.1 a paper), NEET 17 in 7 papers (about 2.4 a paper). The NEET column is smaller only because we have fewer NEET papers. Our bank has no NEET 2020 or 2021 paper.

7. Exam traps

Watch out

Only the ions outside the square brackets precipitate with $AgN O_{3}$ or $BaC l_{2}$ , but the complex ion counts for conductivity too: $[Co (N H_{3})_{6}] C l_{3}$ gives 4 ions in solution.
Find the oxidation state first: CO and $N H_{3}$ are neutral, so Ni is 0 in $[Ni (CO)_{4}]$ ( $d^{10}$ ).
$[NiC l_{4}]^{2 -}$ is tetrahedral and paramagnetic; $[Ni (CN)_{4}]^{2 -}$ is square planar and diamagnetic; $[Ni (CO)_{4}]$ is tetrahedral and diamagnetic.
Co(III) pairs its electrons even with $N H_{3}$ , so $[Co (N H_{3})_{6}]^{3 +}$ is low spin and diamagnetic; $[Co F_{6}]^{3 -}$ is high spin.
Tetrahedral complexes are almost always high spin, and a tetrahedral $M A_{2} B_{2}$ has no geometrical isomers.
For $d^{1}$ to $d^{3}$ and $d^{8}$ to $d^{10}$ the octahedral configuration is the same with strong and weak ligands: an octahedral $N i^{2 +}$ complex always has 2 unpaired electrons.
A larger $Δ$ means a shorter wavelength absorbed; the colour seen is the complementary colour.
4d and 5d metals have a larger $Δ$ , so their complexes are low spin even with $C l^{-}$ ; the $d^{8}$ ions $P d^{2 +}$ and $P t^{2 +}$ are square planar, as in $[PtC l_{4}]^{2 -}$ .
Synergic bonding makes the metal-carbon bond stronger and the C-O bond weaker.
cis- $[Co (en)_{2} C l_{2}]^{+}$ is chiral and trans is not: count both enantiomers when asked for stereoisomers.

8. What the 2026 syllabi say

JEE Main 2026 (NTA syllabus, Unit 12) and NEET (UG) 2026 (NMC syllabus, Unit 12) use the same words: "Introduction to coordination compounds. Werner's theory, ligands, coordination number, denticity, chelation, IUPAC nomenclature of mononuclear co-ordination compounds, isomerism, Bonding: Valence bond approach and basic ideas of Crystal field theory, colour and magnetic properties, importance of coordination compounds (in qualitative analysis, extraction of metals and in biological systems)."

JEE Advanced 2026 (Information Brochure, Annexure-I, Chemistry): "Werner's theory; Nomenclature, cis-trans and ionization isomerism, hybridization and geometries (linear, tetrahedral, square planar and octahedral) of mononuclear coordination compounds; Bonding [VBT and CFT (octahedral and tetrahedral fields)]; Magnetic properties (spin-only) and colour of 3d-series coordination compounds; Ligands and spectrochemical series; Stability; Importance and applications; Metal carbonyls."

So metal carbonyls and stability are named only in the JEE Advanced text, yet JEE Main 2026 asked synergic bonding and NEET 2023 asked the chelate effect (PYQ 9 below). Linkage, coordination, solvate and optical isomerism are not named in the JEE Advanced text, yet JEE Advanced 2026 asked linkage isomers and a mer isomer. Learn the whole chapter for every exam.

9. Solved past-paper questions

Ten questions from JEE Main, JEE Advanced and NEET, chosen to cover the ideas asked most. The question and the answer key are as set in the paper; the working is ours.

PYQ 1: Werner's theory: counting ions

JEE Main 2026 Apr 6 Shift 2, Chemistry Q21

An excess of AgNO $_{3}$ is added to 100 mL of a 0.05 M solution of tetraaquadichloridochromium(III) chloride. The number of moles of AgCl precipitated will be _____ × 10 $^{- 3}$ . (Nearest integer)

Solution:

The name gives the formula. Tetraaqua = four $H_{2} O$ , dichlorido = two $C l^{-}$ inside the brackets, chromium(III) = Cr at +3. The entity's charge is $+ 3 - 2 = + 1$ , so one $C l^{-}$ sits outside: $[Cr (H_{2} O)_{4} C l_{2}] Cl$ .
Moles of complex $= 0.05 \times 0.100$ $= 5 \times 1 0^{- 3}$ . Only the outside chloride precipitates, one per formula unit, so $AgCl$ $= 5 \times 1 0^{- 3}$ mol.

Answer key 5

PYQ 2: Ambidentate ligands

NEET 2026, Chemistry Q31

Which one of the following is an ambidentate ligand?

(A) Oxalate
(B) Ethane–1, 2–diamine
(C) Thiocyanate
(D) Ethylenediaminetetraacetate ion

Solution:

An ambidentate ligand has two possible donor atoms but binds through only one at a time.
Thiocyanate binds through S ( $SC N^{-}$ , thiocyanato-S) or through N ( $NC S^{-}$ , thiocyanato-N): ambidentate.
Oxalate and ethane-1,2-diamine bind through two atoms at once (didentate chelates); EDTA through six. So (C).

Answer key (C) Thiocyanate

PYQ 3: Counting stereoisomers

JEE Main 2025 Apr 4 Shift 1, Chemistry Q9

Number of stereoisomers possible for the complexes, $[C r C l_{3} (p y)_{3}]$ and $[C r C l_{2} (o x)_{2}]^{3 -}$ are respectively

(py = pyridine, ox = oxalate)

(A) 3 & 3
(B) 2 & 2
(C) 2 & 3
(D) 1 & 2

Solution:

$[CrC l_{3} (py)_{3}]$ is octahedral $M A_{3} B_{3}$ : the three Cl can lie on one face (fac) or on a meridian (mer). Each has a mirror plane, so there are 2 stereoisomers.
$[CrC l_{2} (ox)_{2}]^{3 -}$ is $M (AA)_{2} B_{2}$ . Trans (the two Cl opposite) has a mirror plane. Cis is chiral and exists as two mirror images. That makes 3 stereoisomers, so the answer is 2 and 3 (C).

Answer key (C) 2 & 3

PYQ 4: Types of structural isomerism

NEET 2024, Chemistry Q2

Match List - I with List - II:

List - I (Complex)	List-II (Type of isomerism)
A. $[Co(NH_{3})_{5} (NO_{2})] Cl_{2}$	I. Solvate isomerism
B. $[Co(NH_{3})_{5} (SO_{4})] Br$	II. Linkage isomerism
C. $[Co(NH_{3})_{6}] [Cr (CN)_{6}]$	III. Ionization isomerism
D. $[Co(H_{2} O)_{6}] Cl_{3}$	IV. Coordination isomerism

Choose the correct answer from the options given below:

(A) A-I, B-IV, C-III, D-II
(B) A-II, B-IV, C-III, D-I
(C) A-II, B-III, C-IV, D-I
(D) A-I, B-III, C-IV, D-II

Solution:

A: $N O_{2}^{-}$ can bind through N or O, so linkage isomerism (II).
B: $S O_{4}^{2 -}$ inside and $B r^{-}$ outside can swap places, so ionisation isomerism (III).
C: both ions are complexes, so ligands can swap between Co and Cr: coordination isomerism (IV).
D: a water inside can swap with a $C l^{-}$ outside: solvate (hydrate) isomerism (I). A-II, B-III, C-IV, D-I (C).

Answer key (C) A-II, B-III, C-IV, D-I

PYQ 5: Nickel dimethylglyoxime

JEE Main 2026 Jan 23 Shift 1, Chemistry Q13

The statements that are incorrect about the nickel (II) complex of dimethylglyoxime are :

A. It is red in colour

B. It has a high solubility in water at pH = 9

C. The Ni ion has two unpaired d-electrons

D. The N–Ni–N bond angle is almost close to 90°

E. The complex contains four five-membered metallacycles (metal containing rings)

Choose the correct answer from the options given below :

(A) C and E only
(B) A, D and B only
(C) B, C and E only
(D) C and D only

Solution:

$N i^{2 +}$ is $d^{8}$ ; with the strong-field dimethylglyoximate it is square planar ( $d s p^{2}$ ): all eight $d$ electrons are paired, so C is false.
It is the red precipitate of the nickel test: it comes down from ammoniacal (basic, about pH 9) solution because it is insoluble in water, so B is false; A is true. Square planar means N-Ni-N angles close to $9 0^{\circ}$ , so D is true.
Each dimethylglyoximate closes one five-membered Ni-N-C-C-N ring (two in all); the two O-H…O hydrogen bonds close six-membered rings. So E is false. The incorrect statements are B, C and E (C).

Answer key (C) B, C and E only

PYQ 6: Paramagnetic complexes

JEE Main 2026 Apr 8 Shift 2, Chemistry Q21

Number of paramagnetic complexes among the following is ________.

$[MnBr_{4}]^{2 -}$ , $[NiCl_{4}]^{2 -}$ , $[Ni (CN)_{4}]^{2 -}$ , $[Ni (CO)_{4}]$ , $[CoF_{6}]^{3 -}$ , $[Fe (CN)_{6}]^{4 -}$ , $[Mn (CN)_{6}]^{3 -}$ , $[Ti (CN)_{6}]^{3 -}$ , $[Cu (H_{2} O)_{6}]^{2 +}$ , $[Co (C_{2} O_{4})_{3}]^{3 -}$

Solution:

Count the unpaired electrons ( $n$ ) from the oxidation state, the $d$ count and the field strength:

Complex	Metal ion	Field, shape	$n$
$[MnB r_{4}]^{2 -}$	$M n^{2 +}$ , $d^{5}$	weak, tetrahedral	5
$[NiC l_{4}]^{2 -}$	$N i^{2 +}$ , $d^{8}$	weak, tetrahedral	2
$[Ni (CN)_{4}]^{2 -}$	$N i^{2 +}$ , $d^{8}$	strong, square planar	0
$[Ni (CO)_{4}]$	Ni(0), $d^{10}$	tetrahedral	0
$[Co F_{6}]^{3 -}$	$C o^{3 +}$ , $d^{6}$	weak, octahedral	4
$[Fe (CN)_{6}]^{4 -}$	$F e^{2 +}$ , $d^{6}$	strong, octahedral	0
$[Mn (CN)_{6}]^{3 -}$	$M n^{3 +}$ , $d^{4}$	strong, octahedral	2
$[Ti (CN)_{6}]^{3 -}$	$T i^{3 +}$ , $d^{1}$	octahedral	1
$[Cu (H_{2} O)_{6}]^{2 +}$	$C u^{2 +}$ , $d^{9}$	octahedral	1
$[Co (C_{2} O_{4})_{3}]^{3 -}$	$C o^{3 +}$ , $d^{6}$	low spin ( $C o^{3 +}$ ), octahedral	0

Six complexes have unpaired electrons, so the answer is 6.

Answer key 6

PYQ 7: CFSE in a tetrahedral field

JEE Main 2026 Apr 6 Shift 1, Chemistry Q11

Match the List-I with List-II:

List-I (Electronic configuration of tetrahedral metal ion)		List-II (CFSE in units of $Δ_{t}$ )
A.	$d^{2}$	I.	$- 0.6$
B.	$d^{4}$	II.	$- 0.8$
C.	$d^{6}$	III.	$- 1.2$
D.	$d^{8}$	IV.	$- 0.4$

Choose the correct answer from the options given below:

(A) A-III, B-IV, C-II, D-I
(B) A-III, B-I, C-IV, D-II
(C) A-III, B-IV, C-I, D-II
(D) A-II, B-I, C-IV, D-III

Solution:

A tetrahedral ion is high spin: electrons fill $e$ (each $- 0.6 Δ_{t}$ ) and $t_{2}$ (each $+ 0.4 Δ_{t}$ ) singly before pairing.
$d^{2}$ : $e^{2}$ , $- 1.2 Δ_{t}$ (III). $d^{4}$ : $e^{2} t_{2}^{2}$ , $- 1.2 + 0.8 = - 0.4 Δ_{t}$ (IV).
$d^{6}$ : $e^{3} t_{2}^{3}$ , $- 1.8 + 1.2 = - 0.6 Δ_{t}$ (I). $d^{8}$ : $e^{4} t_{2}^{4}$ , $- 2.4 + 1.6 = - 0.8 Δ_{t}$ (II). So (C).

Answer key (C) A-III, B-IV, C-I, D-II

PYQ 8: Field strength and the energy absorbed

NEET 2022, Chemistry Q41

The order of energy absorbed which is responsible for the color of complexes

(A) [Ni(H $_{2}$ O) $_{2}$ (en) $_{2}$ ] $^{2 +}$

(B) [Ni(H $_{2}$ O) $_{4}$ (en)] $^{2 +}$ and

(A) (C) $>$ (B) $>$ (A)
(B) (C) $>$ (A) $>$ (B)
(C) (B) $>$ (A) $>$ (C)
(D) (A) $>$ (B) $>$ (C)

Solution:

en is a stronger-field ligand than $H_{2} O$ , so each en that replaces two waters raises $Δ_{o}$ .
(C) has three en, (A) two and (B) one, so the energy absorbed is (C) > (A) > (B), option (B).

Answer key (B) (C) $>$ (A) $>$ (B)

PYQ 9: The chelate effect

NEET 2023, Chemistry Q44

Which complex compound is most stable?

(A) $[Co (NH_{3})_{3} (NO_{3})_{3}]$
(B) $[CoCl_{2} (en)_{2}] NO_{3}$
(C) $[Co (NH_{3})_{6}]_{2} (SO_{4})_{3}$
(D) $[Co (NH_{3})_{4} (H_{2} O) Br] (NO_{3})_{2}$

Solution:

All four are Co(III) complexes. Only $[CoC l_{2} (en)_{2}] N O_{3}$ has chelating ligands: two en rings.
Chelation makes a complex more stable than one with unidentate ligands only, so (B) is the most stable.

Answer key (B) $[CoCl_{2} (en)_{2}] NO_{3}$

PYQ 10: High and low spin d4

JEE Advanced 2025 Paper 2, Chemistry Section 3 Q7

The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br) $_{6}$ ] $^{3 -}$ and [Mn(CN) $_{6}$ ] $^{3 -}$ is ______.

Solution:

In both ions Mn is $+ 3$ (six ligands with charge $- 1$ each and an overall charge of $- 3$ ), so $d^{4}$ .
$B r^{-}$ is a weak-field ligand: high spin $t_{2 g}^{3} e_{g}^{1}$ , $n = 4$ , $μ = 24 = 4.90$ BM.
$C N^{-}$ is a strong-field ligand: low spin $t_{2 g}^{4}$ , $n = 2$ , $μ = 8 = 2.83$ BM. The sum is 7.73 BM.

Answer key 7.73

Every number on this page is checked by a script, and every worked answer is recomputed and matched with the answer key. Past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026; the questions were read by hand (lists checked 2026-10-19).

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