JEE Mains Chemistry · Coordination Compounds

Hybridisation and Magnetism: Valence Bond Theory

Valence bond theory reads a complex's hybridisation, shape and magnetism from the metal's d-electron count and the ligand: a strong-field ligand pairs the d electrons and frees inner d orbitals (d²sp³, dsp²), a weak one leaves them unpaired and uses outer d orbitals (sp³d²).

All Coordination Compounds notes JEE Mains Chemistry strategy

Why this matters

Thirty-two PYQs, twenty-six of them multiple choice, and six from 2026. Thirteen decide between inner-orbital d²sp³ and outer-orbital sp³d² for an octahedral complex; thirteen sort the four-coordinate nickel, platinum and copper complexes by shape and magnetism; six match complexes to their hybridisation or test what valence bond theory cannot explain.

Concept 1 of 3: Inner-orbital and outer-orbital octahedral complexes

Six ligands need six empty orbitals on the metal. If two of the 3d orbitals can be emptied, the metal uses them: 3d + 4s + 4p gives d²sp³, an inner-orbital, low-spin complex. A strong-field ligand such as

\mathrm{CN^-}

, NH₃ with Co³⁺, en or oxalate with Co³⁺ forces the d electrons to pair and so empties them. A weak-field ligand such as F⁻ or Cl⁻ leaves the electrons spread out, and the metal must use the outer 4d orbitals: sp³d², outer-orbital, high spin.

Definition

d¹, d², d³: two 3d orbitals are always empty, so the complex is d²sp³ whatever the ligand ( $\mathrm{[Cr(NH_3)_6]^{3+}}$ , 3 unpaired).
d⁴ to d⁷: the ligand decides. Strong field: d²sp³, electrons paired ( $\mathrm{[Co(NH_3)_6]^{3+}}$ 0 unpaired, $\mathrm{[Fe(CN)_6]^{3-}}$ 1, $\mathrm{[Mn(CN)_6]^{3-}}$ 2). Weak field: sp³d², as many unpaired as possible ( $\mathrm{[CoF_6]^{3-}}$ 4, $\mathrm{[FeF_6]^{3-}}$ 5, $\mathrm{[Fe(H_2O)_6]^{2+}}$ 4).
d⁸ octahedral: only one 3d orbital could ever be emptied, so it is always sp³d² with 2 unpaired ( $\mathrm{[Ni(NH_3)_6]^{2+}}$ ).
Low spin = spin paired = inner orbital. High spin = spin free = outer orbital.
Cobalt(III) is low spin with NH₃, en, oxalate and CN⁻; F⁻ leaves it high spin.

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})

Worked example

Give the hybridisation, the number of unpaired electrons and the magnetic nature of

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q108Moderate

Example 1 · Coordination Compounds · Hybridization and Magnetism

The type of hybridization and the magnetic property of

\left\lbrack MnCl_{6} \right\rbrack^{3 -}

are :

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.

Drill 12 more on inner-orbital and outer-orbital octahedral complexes

Concept 2 of 3: Four-coordinate complexes: tetrahedral or square planar

Four ligands can sit at the corners of a tetrahedron (sp³) or a square (dsp²). Square planar needs one empty 3d orbital, so it belongs to d⁸ ions with a strong-field ligand, and to all d⁸ Pd(II) and Pt(II) complexes. Nickel(0) in Ni(CO)₄ is special: CO pushes the two 4s electrons into 3d, giving 3d¹⁰ and a tetrahedral, diamagnetic complex.

Definition

d⁸ + strong field (CN⁻): dsp², square planar, 0 unpaired: $\mathrm{[Ni(CN)_4]^{2-}}$ .
d⁸ + weak field (Cl⁻, Br⁻, PPh₃ with Cl⁻): sp³, tetrahedral, 2 unpaired: $\mathrm{[NiCl_4]^{2-}}$ , $\mathrm{[Ni(PPh_3)_2Cl_2]}$ .
4d⁸ and 5d⁸ (Pd²⁺, Pt²⁺): always square planar and diamagnetic, even with Cl⁻.
d¹⁰ (Ni(0) in Ni(CO)₄, Cu⁺, Zn²⁺): sp³, tetrahedral, diamagnetic.
$\mathrm{[Cu(NH_3)_4]^{2+}}$ : Cu²⁺ is d⁹, square planar, 1 unpaired electron, paramagnetic.

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.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q45Moderate

Example 2 · Coordination Compounds · Hybridization and Magnetism

The correct statement among the following is :

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.

Drill 12 more on four-coordinate complexes: tetrahedral or square planar

Concept 3 of 3: Hybridisation, geometry and the limits of valence bond theory

Match-list questions pair a complex with a hybridisation and a magnetic character. Fix the coordination number, then the d count, then the ligand strength, and the row is decided. Valence bond theory gives these answers but cannot give numbers: it does not explain colour, it cannot say how strongly paramagnetic a complex is, and it cannot tell strong- from weak-field ligands on its own.

Definition

Coordination number 2: sp, linear. 4: sp³ (tetrahedral) or dsp² (square planar). 5: dsp³, trigonal bipyramidal. 6: d²sp³ (inner) or sp³d² (outer), octahedral.
Limitations of valence bond theory: it makes several assumptions; it gives no quantitative account of magnetic data; it does not explain colour; it gives no quantitative account of the thermodynamic or kinetic stability of complexes; it does not predict whether a four-coordinate complex is tetrahedral or square planar; and it does not distinguish weak- from strong-field ligands.
Crystal field theory explains colour and magnetism, but it cannot explain the ORDER of the spectrochemical series (why neutral CO is stronger than the anions).

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.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q33Moderate

Example 3 · Coordination Compounds · Hybridization and Magnetism

Match List-I with List-II.

List-I (Complex)	List-II (Hybridisation and magnetic character)
(A) $[MnBr_{4}]^{2-}$	(I) $d^{2}sp^{3}$ , diamagnetic
(B) $[FeF_{6}]^{3-}$	(II) $sp^{3}d^{2}$ , paramagnetic
(C) $[Co(C_{2}O_{4})_{3}]^{3-}$	(III) $sp^{3}$ , diamagnetic
(D) $[Ni(CO)_{4}]$	(IV) $sp^{3}$ , paramagnetic

Choose the correct answer from the options given below :

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.

Drill 5 more on hybridisation, geometry and the limits of valence bond theory

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Inner-orbital and outer-orbital octahedral complexes
Octahedral hybridisation in valence bond theory
$\text{strong field: } d^2sp^3 = (n-1)d^2\,ns\,np^3 \;(\text{inner, low spin}) \qquad \text{weak field: } sp^3d^2 = ns\,np^3\,nd^2 \;(\text{outer, high spin})$

Reference tables (2)

Four-coordinate complexes: tetrahedral or square planar9 rows

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 theory6 rows

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.

Watch out for (7)

Spin paired is low spin; spin free is high spin
→ Inner-orbital and outer-orbital octahedral complexes
Octahedral nickel(II) is always outer orbital
→ Inner-orbital and outer-orbital octahedral complexes
Two textbook shortcuts that JEE keys have used
→ Inner-orbital and outer-orbital octahedral complexes
Ni(CO)₄ and [NiCl₄]²⁻ are both tetrahedral but differ in d count
→ Four-coordinate complexes: tetrahedral or square planar
[Ni(CN)₄]²⁻ is dsp², not sp³
→ Four-coordinate complexes: tetrahedral or square planar
sp³ can be diamagnetic or paramagnetic
→ Hybridisation, geometry and the limits of valence bond theory
Anionic ligands are not the strongest
→ Hybridisation, geometry and the limits of valence bond theory

Test yourself on Coordination Compounds

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Coordination Compounds chapter test Drill all 32 questions