Chapter Exam

Chapter-5 Coordination Compounds — Online MCQ Test

Q: Which of the following is a monodentate ligand?

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.

Q: The IUPAC name of [Ni(CO)₄] is:

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).

Q: Which complex would show optical isomerism?

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).

Q: 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)?

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.

CHEMISTRY · CLASS 12 INTER II YEAR · Andhra State Board

Practice Chapter-5 Coordination Compounds with a free chapter-wise online MCQ test for Andhra State Board CLASS 12 INTER II YEAR 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 Andhra State Board CLASS 12 INTER II YEAR 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.