CUET UG Chemistry Inorganic Chemistry d-Block Elements Properties Oxidation States Complex Compounds

Must-Know (15–20 detailed bullets)

• Transition elements are d-block elements where d-orbitals are progressively filled; general electronic configuration is (n–1)d¹⁻¹⁰ ns⁰⁻². Example: Fe (Z = 26) has [Ar] 3d⁶ 4s².

• Zn, Cd, and Hg are not considered transition metals because they have completely filled d-orbitals in both atomic and common +2 oxidation states. Example: Zn²⁺ is 3d¹⁰.

• Transition metals exhibit variable oxidation states due to comparable energies of (n–1)d and ns orbitals. Example: Mn shows +2 to +7; Mn²⁺, MnO₂ (+4), MnO₄⁻ (+7).

• The highest oxidation state of a transition metal equals the sum of (n–1)d and ns electrons. Example: Cr (3d⁵4s¹) shows +6 in CrO₃.

• Scandium (Sc³⁺) does not show variable oxidation states; its only stable state is +3. Sc²⁺ is unstable in aqueous solution.

• Cr²⁺ is a strong reducing agent; it gets oxidized to Cr³⁺. E°(Cr³⁺/Cr²⁺) = –0.41 V (verify from NCERT).

• Mn³⁺ is a strong oxidizing agent; it gets reduced to Mn²⁺. E°(Mn³⁺/Mn²⁺) = +1.51 V (verify from NCERT).

• Transition metals form coloured ions due to d-d transitions. Example: Cu²⁺ (blue), Cr³⁺ (green/violet), Fe³⁺ (yellow/brown).

• The colour of a complex depends on the metal, its oxidation state, and the ligand. Example: [CoF₆]³⁻ is blue, [Co(H₂O)₆]³⁺ is pink.

• Transition metals form paramagnetic substances due to unpaired electrons. Magnetic moment μ = √[n(n+2)] BM, where n = number of unpaired electrons. Example: Fe³⁺ (d⁵) has μ ≈ 5.92 BM.

• Transition metals form interstitial compounds with small atoms like H, C, N, B. These are hard, high melting, and retain metallic conductivity. Example: TiC, Fe₃H.

• Transition metals are good catalysts. Example: Fe in Haber process, V₂O₅ in Contact process.

• Complex compounds consist of a central metal ion bonded to ligands via coordinate bonds. Example: [Co(NH₃)₆]³⁺.

• Ligands are classified by denticity: monodentate (NH₃), bidentate (en = ethane-1,2-diamine), hexadentate (EDTA⁴⁻).

• Coordination number is the number of coordinate bonds with the metal. Example: [Fe(CN)₆]⁴⁻ has CN = 6.

• IUPAC name of [Co(NH₃)₆]Cl₃ is hexaamminecobalt(III) chloride.

• Werner’s theory proposed primary and secondary valences. In [Co(NH₃)₆]Cl₃, primary valence = 3, secondary valence = 6.

• Crystal field splitting in octahedral field: d-orbitals split into t₂g (dxy, dyz, dzx) and eg (dx²–y², dz²). Δ₀ = 10 Dq.

• In tetrahedral complexes, splitting is smaller: Δₜ = (4/9)Δ₀; configuration often high spin.

• Strong field ligands (e.g., CN⁻, CO) cause large splitting and favour pairing; weak field ligands (e.g., I⁻, Cl⁻) cause small splitting.

Difficulty Level

Intermediate — requires understanding of electronic configurations, oxidation states, and coordination chemistry basics, but most facts are directly from NCERT Class 12 Chapter 8 and 9.

Common CUET Traps (3 bullets)

Trap: Assuming all d-block elements are transition metals.
Avoid: Remember Zn, Cd, Hg have full d¹⁰ configuration and are not transition metals.

Trap: Thinking higher oxidation states are always stable.
Avoid: Higher oxidation states become less stable down the group for 3d metals; e.g., Mn⁺⁷ is stable but Fe⁺⁶ is rare and strong oxidant.

Trap: Confusing coordination number with oxidation state.
Avoid: In [Co(NH₃)₅Cl]Cl₂, Co oxidation state is +3, coordination number is 6.

Practice MCQs (5 questions)

Q1. Which of the following ions shows the highest magnetic moment?
A. Ti³⁺
B. Cr³⁺
C. Mn²⁺
D. Fe³⁺
Answer: C
Explanation: Mn²⁺ has 5 unpaired electrons (d⁵), highest among options.
Why others fail: Fe³⁺ also has d⁵, but Mn²⁺ and Fe³⁺ have same μ; if both present, either correct — but here only Mn²⁺ listed.

Q2. Which element does not exhibit variable oxidation states?
A. Mn
B. Cr
C. Fe
D. Sc
Answer: D
Explanation: Scandium exhibits only +3 oxidation state in compounds.
Why others fail: Mn, Cr, Fe all show multiple oxidation states (e.g., Mn from +2 to +7).

Q3. The correct IUPAC name of [Co(NH₃)₅(NO₂)]Cl₂ is:
A. Pentaamminenitrocobalt(III) chloride
B. Pentaamminenitritocobalt(III) chloride
C. Pentaamminenitro-O-cobalt(III) chloride
D. Pentaamminenitro-N-cobalt(III) chloride
Answer: D
Explanation: NO₂ binds through nitrogen in nitro form; correct name uses "nitro-N".
Why others fail: Option B assumes O-binding (nitrito), which is incorrect for this linkage isomer.

Q4. Which of the following is a bidentate ligand?
A. H₂O
B. NH₃
C. en
D. Cl⁻
Answer: C
Explanation: Ethane-1,2-diamine (en) has two donor N atoms, making it bidentate.
Why others fail: H₂O, NH₃, Cl⁻ are monodentate.

Q5. In an octahedral complex, the crystal field splitting energy (Δ₀) causes d-orbitals to split into:
A. Two orbitals of higher energy and three of lower energy
B. Three orbitals of higher energy and two of lower energy
C. Three t₂g and two eg orbitals, with eg higher in energy
D. Three t₂g and two eg orbitals, with t₂g higher in energy
Answer: C
Explanation: In octahedral field, eg orbitals (dx²–y², dz²) are higher in energy than t₂g.
Why others fail: Option A describes tetrahedral splitting, not octahedral.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ Zn²⁺ is colourless — no d-d transition (d¹⁰).
• ⚠️ CrO₄²⁻ is yellow, Cr₂O₇²⁻ is orange — pH dependent interconversion.
• ⚠️ KMnO₄ is purple due to charge transfer, not d-d transition.
• ⚠️ Co³⁺ is unstable in water unless stabilized by ligands like NH₃.
• ⚠️ [Ni(CO)₄] is tetrahedral and diamagnetic — Ni in zero oxidation state.
• ⚠️ Fe³⁺ more stable than Fe²⁺ in air — Fe²⁺ oxidizes to Fe³⁺.
• ⚠️ Cu⁺ disproportionates in aqueous solution: 2Cu⁺ → Cu + Cu²⁺.
• ⚠️ Lanthanoid contraction causes similar atomic radii of Zr and Hf.
• ⚠️ IUPAC: ligands named alphabetically, not by denticity.
• ⚠️ EDTA forms 6 coordinate bonds — hexadentate ligand.
• ⚠️ Inner orbital complexes use (n–1)d orbitals (e.g., d²sp³).
• ⚠️ Outer orbital complexes use nd orbitals (e.g., sp³d²).
• ⚠️ Strong field ligands: CO > CN⁻ > en > NH₃ > H₂O > F⁻ > Cl⁻ > Br⁻ > I⁻ (remember: "CoCa Ena FHiClBrI").
• ⚠️ [FeF₆]³⁻ is high spin, [Fe(CN)₆]³⁻ is low spin — due to ligand strength.
• ⚠️ Paramagnetism increases with number of unpaired electrons.
• ⚠️ K₃[Fe(CN)₆] is red crystalline — used to test Fe²⁺ (gives blue ppt).
• ⚠️ K₄[Fe(CN)₆] + Fe³⁺ → Prussian blue (Fe₄[Fe(CN)₆]₃).
• ⚠️ Linkage isomerism occurs in ambidentate ligands like NO₂⁻ and SCN⁻.
• ⚠️ Coordination isomerism occurs in complexes with both cation and anion as complexes.
• ⚠️ Hydrate isomerism example: [Cr(H₂O)₆]Cl₃ (violet), [Cr(H₂O)₅Cl]Cl₂·H₂O (green).