Class 12 Chemistry Chapter 5: Coordination Compounds worksheet with answer key
A 20-mark worksheet on Coordination Compounds for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (c) SCN⁻
SCN⁻ is ambidentate: it binds through S or through N. Oxalate and ethane-1,2-diamine are didentate, using both donor atoms at once; NH₃ has only N.
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Answer: (c) [Co(NH₃)₂(en)₂]Cl₃
Diammine is two NH₃, bis(ethane-1,2-diamine) two en; all four ligands are neutral, so the cation is [Co(NH₃)₂(en)₂]³⁺ and three Cl⁻ balance it. No chlorido ligand is named, so Cl is not inside the bracket.
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Answer: (b) 3 ions
Four NH₃ fill four of the six positions, so two Cl⁻ are inside: [Pt(NH₃)₄Cl₂]Cl₂. The two outer Cl⁻ and the complex cation give 3 ions.
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Answer: (c) two N atoms and four O atoms
EDTA⁴⁻ is (⁻OOCCH₂)₂N–CH₂–CH₂–N(CH₂COO⁻)₂: each of the two N atoms and each of the four carboxylate O atoms donates a pair, six in all.
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Answer: (c) [Co(NH₃)₄(H₂O)Cl]²⁺
A heteroleptic complex has more than one kind of donor group; [Co(NH₃)₄(H₂O)Cl]²⁺ has three (NH₃, H₂O, Cl⁻). The other three carry one kind each and are homoleptic.
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Zn²⁺ is 3d¹⁰: the 3d orbitals are full, so the vacant 4s and three 4p orbitals hybridise to sp³, giving a tetrahedron that accepts four NH₃ pairs. All electrons are paired, so the ion is diamagnetic.
- Zn²⁺ 3d¹⁰; 4s and 4p used: sp³ (1)
- no unpaired electron: diamagnetic (1)
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(i) Chlorophyll is a magnesium complex; (ii) vitamin B12, cyanocobalamine, is a cobalt complex. Cis-platin inhibits tumour growth.
- magnesium in chlorophyll and cobalt in vitamin B12 (1)
- cis-platin (1)
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(i) Li⁺ is the cation; [AlH₄]⁻ is an anion with four hydrido ligands and Al in +3 (x − 4 = −1): lithium tetrahydridoaluminate(III). (ii) A neutral molecule: ligands alphabetically, aqua before chlorido, and Cu is +2 (x − 2 = 0): diaquadichloridocopper(II).
- lithium tetrahydridoaluminate(III) (1)
- diaquadichloridocopper(II) (1)
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(i) Six NO₂⁻ bound through N to Co³⁺ give [Co(NO₂)₆]³⁻, balanced by three Na⁺: Na₃[Co(NO₂)₆]. (ii) Four neutral NH₃ on Cu²⁺ give [Cu(NH₃)₄]²⁺ with one SO₄²⁻: [Cu(NH₃)₄]SO₄. (iii) [Fe(CO)₅].
- Na₃[Co(NO₂)₆] (1)
- [Cu(NH₃)₄]SO₄ (1)
- [Fe(CO)₅] (1)
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(a) Only one Cl⁻ is ionisable, so two Cl⁻ join four NH₃ inside the sphere: [Ir(NH₃)₄Cl₂]Cl. (b) Primary valence 3, the ionisable valence, read today as the oxidation state +3; secondary valence 6, the non-ionisable valence, the coordination number. (c) [Ir(NH₃)₄Cl₂]⁺ is octahedral [MX₂L₄]: two isomers, cis (Cl at 90°) and trans (Cl at 180°).
- [Ir(NH₃)₄Cl₂]Cl (1)
- primary 3 = oxidation state; secondary 6 = coordination number (1)
- two: cis and trans (1)
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Co²⁺ is 3d⁷. Δo < P (high spin): t₂g⁵eg², three unpaired electrons, μ = √(3 × 5) = √15 = 3.87 BM. Δo > P (low spin): t₂g⁶eg¹, one unpaired electron, μ = √(1 × 3) = 1.73 BM.
- Co²⁺ 3d⁷; Δo < P: t₂g⁵eg², 3 unpaired (1)
- Δo > P: t₂g⁶eg¹, 1 unpaired (1)
- μ = 3.87 BM and 1.73 BM (1)
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