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Chapter-5 Coordination Compounds — Online MCQ Test

CHEMISTRY · CLASS 12 SECOND PUC · Karnataka State Board

Practice Chapter-5 Coordination Compounds with a free chapter-wise online MCQ test for Karnataka State Board CLASS 12 SECOND PUC CHEMISTRY. This chapter covers: ligands - IUPAC nomenclature - isomerism - Valence Bond Theory (VBT) - Crystal Field Theory (CFT). AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-5 Coordination Compounds — Important Questions & Answers (FAQ)

Frequently asked questions from Karnataka State Board CLASS 12 SECOND PUC CHEMISTRY — Chapter-5 Coordination Compounds, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is a ligand in coordination chemistry?
  • A. A molecule or ion that donates electron pairs to the central metal atom ✓
  • B. A positively charged metal ion
  • C. A negatively charged counterion
  • D. A complex salt that is insoluble in water
Answer: A. A molecule or ion that donates electron pairs to the central metal atom
A ligand is a Lewis base that donates electron pairs to the central metal atom (Lewis acid) to form a coordinate covalent bond.
Which of the following is a monodentate ligand?
  • A. Ethylenediamine (en)
  • B. EDTA (ethylenediaminetetraacetate)
  • C. Ammonia (NH₃) ✓
  • D. Oxalate (C₂O₄²⁻)
Answer: C. Ammonia (NH₃)
Ammonia has only one lone pair on nitrogen that can coordinate to the metal, making it monodentate. Ethylenediamine and EDTA are polydentate, while oxalate is bidentate.
The IUPAC name of [Ni(CO)₄] is:
  • A. Nickel tetracarbonyl
  • B. Tetracarbonylnickel(0) ✓
  • C. Tetracarbonyl nickel
  • D. Nickel(II) tetracarbonyl
Answer: B. Tetracarbonylnickel(0)
In IUPAC nomenclature for neutral complexes, the central metal's oxidation state must be mentioned in parentheses. Here Ni is 0, so the name is tetracarbonylnickel(0).
Which complex would show optical isomerism?
  • A. [Ni(NH₃)₄]²⁺ (square planar)
  • B. [Co(en)₃]³⁺ (octahedral) ✓
  • C. [PtCl₄]²⁻ (square planar)
  • D. [Zn(NH₃)₄]²⁺ (tetrahedral)
Answer: B. [Co(en)₃]³⁺ (octahedral)
[Co(en)₃]³⁺ has three bidentate ethylenediamine ligands in octahedral geometry, which creates a chiral structure with no plane of symmetry, showing optical isomerism (enantiomers).
Consider the spectrochemical series: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < H₂O < NCS⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO. Which complex would have the maximum d-orbital splitting (Δ)?
  • A. [Fe(H₂O)₆]²⁺
  • B. [Fe(NH₃)₆]²⁺
  • C. [Fe(CN)₆]⁴⁻ ✓
  • D. [Fe(Cl)₆]⁴⁻
Answer: C. [Fe(CN)₆]⁴⁻
CN⁻ is the strongest field ligand in the series (except CO), producing the maximum d-orbital splitting. [Fe(CN)₆]⁴⁻ would have the largest Δ value among the given options.

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