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MHT-CET Chemistry · Coordination Compounds

Bonding in Complexes: Hybridisation, Magnetism, EAN and Stability

Valence bond theory reads a complex's geometry and hybridisation from the metal's d-electron count and the ligand's field strength — a strong-field ligand pairs the d electrons and gives an inner-orbital, low-spin complex — the unpaired electrons left over fix the spin-only magnetic moment, the effective atomic number counts the electrons around the metal after the ligands donate, and the stability of complexes with one ligand follows the metal's charge and the Irving–Williams order.

All Coordination Compounds notesMHT-CET Chemistry strategy

Why this matters

48 PYQs, one HARD, and now the chapter's largest page. Of the twenty-one set from 2021 to 2025, six are hybridisation and geometry — dsp² square planar [Ni(CN)₄]²⁻, asked four times, and d²sp³ [Co(NH₃)₆]³⁺, twice; five are unpaired electrons and the magnetic moment — zero in [Co(NH₃)₆]³⁺, four in [CoF₆]³⁻, 1.73 BM for one electron; five compute an EAN; five are stability orders. Four cards.

Concept 1 of 4: Hybridisation and Geometry from d Electrons and Field Strength

Four steps. (1) Find the metal's oxidation state and its d-electron count: Ni²⁺ is 3d⁸, Co³⁺ is 3d⁶, Fe²⁺ is 3d⁶, Zn²⁺ is 3d¹⁰. (2) Decide whether the ligand is strong field (CN⁻, CO, en, NH₃ with Co³⁺) or weak (halides, H₂O). A strong-field ligand pairs up the d electrons and frees inner 3d orbitals. (3) Count the orbitals needed — one per ligand. (4) Name them. [Ni(CN)₄]²⁻: pairing the eight d electrons frees one 3d orbital, so 3d + 4s + two 4p gives dsp², square planar. [Co(NH₃)₆]³⁺: pairing the six electrons into three orbitals frees two 3d, giving d²sp³, an inner-orbital octahedron. [CoF₆]³⁻: F⁻ does not pair, so the outer 4d is used — sp³d², outer-orbital.

Definition

  • Coordination number 4: dsp2dsp^2 square planar (d⁸ with strong field: [Ni(CN)4]2−[\text{Ni(CN)}_4]^{2-}, all Pt(II) complexes such as cisplatin); sp3sp^3 tetrahedral ([NiCl4]2−[\text{NiCl}_4]^{2-}, Ni(CO)4\text{Ni(CO)}_4, [Zn(NH3)4]2+[\text{Zn(NH}_3)_4]^{2+}); [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+} is square planar.
  • Coordination number 6: d2sp3d^2sp^3 inner-orbital, low spin ([Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+}, [Fe(CN)6]4−[\text{Fe(CN)}_6]^{4-}, [Fe(CN)6]3−[\text{Fe(CN)}_6]^{3-}); sp3d2sp^3d^2 outer-orbital, high spin ([CoF6]3−[\text{CoF}_6]^{3-}, [FeF6]3−[\text{FeF}_6]^{3-}).
  • Coordination number 2: spsp linear ([Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+, [Ag(CN)2]−[\text{Ag(CN)}_2]^-).
  • Crystal field view of the same fact: in an octahedron the d orbitals split into t2gt_{2g} (lower) and ege_g (upper); a strong-field ligand makes the gap larger than the pairing energy, so [Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+} is t2g6eg0t_{2g}^6e_g^0.

Hybridisation by coordination number

CN 2:sp;CN 4:sp3 (tetrahedral), dsp2 (square planar);CN 6:d2sp3 (inner), sp3d2 (outer)\text{CN } 2: sp;\quad \text{CN } 4: sp^3 \text{ (tetrahedral)},\ dsp^2 \text{ (square planar)};\quad \text{CN } 6: d^2sp^3 \text{ (inner)},\ sp^3d^2 \text{ (outer)}

Worked example

Predict the hybridisation and geometry of [Fe(CN)₆]³⁻.
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The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q71Moderate

Example 1 · Coordination Compounds · Bonding Theories, EAN, Crystal Field, Hybridization and Magnetism

Hybridisation of [Ni(CN)4]2−[\text{Ni(CN)}_4]^{2-} is

Calling every four-coordinate complex tetrahedral

Four ligands can be tetrahedral (sp³) OR square planar (dsp²). A d⁸ ion with a strong-field ligand — [Ni(CN)₄]²⁻, every Pt(II) complex — is square planar.
Drill 5 more on hybridisation and geometry from d electrons and field strength

Concept 2 of 4: Unpaired Electrons and the Spin-Only Magnetic Moment

The same Co³⁺ ion (3d⁶, four unpaired as a free ion) is diamagnetic in [Co(NH₃)₆]³⁺ and strongly paramagnetic in [CoF₆]³⁻. NH₃ pairs all six electrons into t₂g — zero unpaired, low spin, diamagnetic. F⁻ is weak, so the electrons spread over all five d orbitals as in the free ion — four unpaired, high spin. Once n is known, the spin-only moment is √(n(n+2)) Bohr magnetons: one unpaired electron gives √3 = 1.73 BM.

Definition

  • Low spin (strong field): [Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+} 0 unpaired, diamagnetic; [Fe(CN)6]4−[\text{Fe(CN)}_6]^{4-} 0; [Fe(CN)6]3−[\text{Fe(CN)}_6]^{3-} 1; [Ni(CN)4]2−[\text{Ni(CN)}_4]^{2-} 0.
  • High spin (weak field): [CoF6]3−[\text{CoF}_6]^{3-} 4 unpaired; [FeF6]3−[\text{FeF}_6]^{3-} 5; [NiCl4]2−[\text{NiCl}_4]^{2-} 2.
  • Free Co3+\text{Co}^{3+} (before hybridisation) has 4 unpaired electrons; in [Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+} all are paired, so it is diamagnetic and not high spin.
  • Spin-only moments: n = 1 → 1.73 BM; 2 → 2.83; 3 → 3.87; 4 → 4.90; 5 → 5.92.

Spin-only magnetic moment

μ=n(n+2) BM\mu = \sqrt{n(n+2)}\ \text{BM}

Worked example

Find the spin-only magnetic moment of [FeF₆]³⁻.
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The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q95Moderate

Example 2 · Coordination Compounds · Bonding Theories, EAN, Crystal Field, Hybridization and Magnetism

What is the number of unpaired electrons present in CoF63−CoF_6^{3-}?

Using the free-ion count for a strong-field complex

Co³⁺ has four unpaired electrons only before the ligands arrive. With NH₃ or CN⁻ they pair — [Co(NH₃)₆]³⁺ has zero. Decide the ligand's field strength before counting.
Drill 4 more on unpaired electrons and the spin-only magnetic moment

Concept 3 of 4: Effective Atomic Number (EAN)

Count the electrons around the metal in the complex: start from its atomic number, take away the electrons it lost to reach its oxidation state, and add two for every donor atom. Many stable complexes land on 36, krypton's count: [Fe(CN)₆]⁴⁻ is 26 − 2 + 12, [Co(NH₃)₆]³⁺ is 27 − 3 + 12, [Zn(NH₃)₄]²⁺ is 30 − 2 + 8. Not every complex does — [Cu(NH₃)₄]²⁺ is 29 − 2 + 8 = 35 — so compute, do not assume 36.

Definition

  • EAN=Z−oxidation state+2×coordination number\text{EAN} = Z - \text{oxidation state} + 2 \times \text{coordination number}.
  • 36: [Fe(CN)6]4−[\text{Fe(CN)}_6]^{4-} (Z 26), [Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+} (Z 27), [Zn(NH3)4]2+[\text{Zn(NH}_3)_4]^{2+} (Z 30), Ni(CO)4\text{Ni(CO)}_4 (Z 28), Cr(CO)6\text{Cr(CO)}_6 (Z 24).
  • Not 36: [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+} 35; [Fe(CN)6]3−[\text{Fe(CN)}_6]^{3-} 35; [Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+ 50.

EAN

EAN=Z−x+2 CN\text{EAN} = Z - x + 2\,\text{CN}

Worked example

Calculate the EAN of Ni in Ni(CO)₄ and of Pt in [PtCl₆]²⁻ (Z: Ni 28, Pt 78).
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The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q89Moderate

Example 3 · Coordination Compounds · Bonding Theories, EAN, Crystal Field, Hybridization and Magnetism

What is EAN of Cu in [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+}?

Adding one electron per ligand, or per bidentate ligand

Each donor atom gives a PAIR, so add 2 × coordination number. With en or oxalate count donor atoms, not ligands: [Co(en)₃]³⁺ is 27 − 3 + 12 = 36.
Drill 4 more on effective atomic number (ean)

Concept 4 of 4: Stability of Complexes: Metal Charge and the Irving–Williams Order

Stability is about how tightly the metal holds its ligands. Two rules cover the paper. First, with the same ligand, the divalent first-row ions follow the Irving–Williams order Mn²⁺ < Fe²⁺ < Co²⁺ < Ni²⁺ < Cu²⁺, and the textbook adds Cd²⁺ at the bottom: Cu²⁺ > Ni²⁺ > Co²⁺ > Fe²⁺ > Mn²⁺ > Cd²⁺. Second, a higher charge on the metal holds ligands more firmly, so a Co³⁺ complex outranks a Cu²⁺ one, which outranks an Ag⁺ one. Chelating ligands (en, oxalate, EDTA) also give more stable complexes than the same number of monodentate ones.

Definition

  • Same ligand, divalent ions: Cu2+>Ni2+>Co2+>Fe2+>Mn2+>Cd2+\text{Cu}^{2+} > \text{Ni}^{2+} > \text{Co}^{2+} > \text{Fe}^{2+} > \text{Mn}^{2+} > \text{Cd}^{2+}. Most stable Cu2+\text{Cu}^{2+}; least stable Cd2+\text{Cd}^{2+}.
  • Higher metal charge → more stable: [Co(NH3)6]3+>[Cu(CN)4]2−>[Ag(CN)2]−[\text{Co(NH}_3)_6]^{3+} > [\text{Cu(CN)}_4]^{2-} > [\text{Ag(CN)}_2]^- (Co³⁺ > Cu²⁺ > Ag⁺), the paper's keyed order.
  • Chelate effect: [Ni(en)3]2+[\text{Ni(en)}_3]^{2+} is more stable than [Ni(NH3)6]2+[\text{Ni(NH}_3)_6]^{2+}.
  • Stability is measured by the overall formation constant β\beta; a larger β\beta means a more stable complex.

Irving–Williams order (same ligand)

Cu2+>Ni2+>Co2+>Fe2+>Mn2+>Cd2+\text{Cu}^{2+} > \text{Ni}^{2+} > \text{Co}^{2+} > \text{Fe}^{2+} > \text{Mn}^{2+} > \text{Cd}^{2+}

Worked example

Arrange the complexes of Ni²⁺, Mn²⁺ and Cu²⁺ with ethylenediamine in decreasing stability.
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The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q63Moderate

Example 4 · Coordination Compounds · Bonding Theories, EAN, Crystal Field, Hybridization and Magnetism

Which from following is a correct stability order of complex formed by metal ions if the ligand remains same?

Ranking by atomic number

Stability rises from Mn²⁺ to Cu²⁺ but it is not 'heavier is more stable': Cd²⁺ is heavier than all of them and forms the least stable complexes.
Drill 4 more on stability of complexes: metal charge and the irving–williams order

Summary — formulas & gotchas at a glance

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Formulas (4)

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