Unit 12: Coordination Compounds — Online MCQ Test
CHEMISTRY · NEET (All) · NEET
Practice Unit 12: Coordination Compounds with a free chapter-wise online MCQ test.
This chapter covers: Introduction to coordination compounds. Werner's theory; ligands coordination number denticity. chelation; IUPAC nomenclature of mononuclear co-ordination compounds isomerism; Bond....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Unit 12: Coordination Compounds — Important Questions & Answers
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
- A. SCN- < F- < CN- < C2O4 2-
- B. F- < SCN- < C2O4 2- < CN-
- C. CN- < C2O4 2- < SCN- < F-
- D. SCN- < F- < C2O4 2- < CN-
Answer: D. SCN- < F- < C2O4 2- < CN-
According to the spectrochemical series, the order of increasing field strength for these ligands is SCN- < F- < C2O4 2- < CN-.
According to the spectrochemical series, the order of increasing field strength for these ligands is SCN- < F- < C2O4 2- < CN-.
Which of the following are paramagnetic? A. [NiCl4]²⁻ B. Ni(CO)₄ C. [Ni(CN)₄]²⁻ D. [Ni(H₂O)₆]²⁺ E. Ni(PPh₃)₄
- A. A and C only
- B. B and E only
- C. A and D only
- D. A, D and E only
Answer: C. A and D only
[NiCl₄]²⁻ (sp³) is paramagnetic (2 unpaired e⁻). Ni(CO)₄, [Ni(CN)₄]²⁻, and Ni(PPh₃)₄ are diamagnetic. [Ni(H₂O)₆]²⁺ (sp³d²) is paramagnetic (2 unpaired e⁻).
[NiCl₄]²⁻ (sp³) is paramagnetic (2 unpaired e⁻). Ni(CO)₄, [Ni(CN)₄]²⁻, and Ni(PPh₃)₄ are diamagnetic. [Ni(H₂O)₆]²⁺ (sp³d²) is paramagnetic (2 unpaired e⁻).
The correct order of the wavelength of light absorbed by the following complexes is: A. [Co(NH₃)₆]³⁺ B. [Co(CN)₆]³⁻ C. [Cu(H₂O)₄]²⁺ D. [Ti(H₂O)₆]³⁺
- A. B < D < A < C
- B. B < A < D < C
- C. C < D < A < B
- D. C < A < D < B
Answer: B. B < A < D < C
Wavelength absorbed is inversely proportional to crystal field splitting (Δ). Stronger ligands cause higher splitting, hence lower wavelength absorbed. Order of field strength: CN⁻ > NH₃ > H₂O.
Wavelength absorbed is inversely proportional to crystal field splitting (Δ). Stronger ligands cause higher splitting, hence lower wavelength absorbed. Order of field strength: CN⁻ > NH₃ > H₂O.
Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
- A. [Co(NH₃)₃Cl₃]
- B. [Co(NH₃)₄Cl₂]Cl
- C. [Co(NH₃)₆]Cl₃
- D. [Co(NH₃)₅Cl]Cl₂
Answer: A. [Co(NH₃)₃Cl₃]
Conductance in solution depends on the number of ions produced. [Co(NH₃)₃Cl₃] is a non-electrolyte (does not dissociate), therefore it has the minimum conductance.
Conductance in solution depends on the number of ions produced. [Co(NH₃)₃Cl₃] is a non-electrolyte (does not dissociate), therefore it has the minimum conductance.
Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
- A. [Co(NH₃)₃Cl₃]
- B. [Co(NH₃)₄Cl₂]Cl
- C. [Co(NH₃)₆]Cl₃
- D. [Co(NH₃)₅Cl]Cl₂
Answer: A. [Co(NH₃)₃Cl₃]
Conductance depends on the number of ions produced in solution. [Co(NH₃)₃Cl₃] is a non-electrolyte and produces no ions, hence minimum conductance.
Conductance depends on the number of ions produced in solution. [Co(NH₃)₃Cl₃] is a non-electrolyte and produces no ions, hence minimum conductance.
Match List-I with List-II. List-I: A. Haber process, B. Wacker oxidation, C. Wilkinson catalyst, D. Ziegler catalyst List-II: I. Fe catalyst, II. PdCl2, III. [(PPh3)3RhCl], IV. TiCl4 with Al(CH3)3
- A. A-I, B-II, C-IV, D-III
- B. A-II, B-III, C-I, D-IV
- C. A-I, B-II, C-III, D-IV
- D. A-I, B-IV, C-III, D-II
Answer: C. A-I, B-II, C-III, D-IV
Haber (Fe), Wacker (PdCl2), Wilkinson ([(PPh3)3RhCl]), Ziegler (TiCl4+Al(CH3)3).
Haber (Fe), Wacker (PdCl2), Wilkinson ([(PPh3)3RhCl]), Ziegler (TiCl4+Al(CH3)3).
Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?
- A. The solution is decolourized
- B. SO2 is reduced
- C. Green Cr2(SO4)3 is formed
- D. The solution turns blue
Answer: C. Green Cr2(SO4)3 is formed
Acidified K2Cr2O7 is an oxidizing agent that oxidizes SO2 to sulfate, reducing Cr(VI) to Cr(III), which forms green colored Cr2(SO4)3.
Acidified K2Cr2O7 is an oxidizing agent that oxidizes SO2 to sulfate, reducing Cr(VI) to Cr(III), which forms green colored Cr2(SO4)3.
Iron carbonyl, Fe(CO)5 is
- A. Tetranuclear
- B. Dinuclear
- C. Trinuclear
- D. Mononuclear
Answer: D. Mononuclear
Fe(CO)5 contains only one iron atom, therefore it is classified as mononuclear.
Fe(CO)5 contains only one iron atom, therefore it is classified as mononuclear.
Iron carbonyl, Fe(CO)5 is
- A. trinuclear
- B. tetranuclear
- C. dinuclear
- D. mononuclear
Answer: D. mononuclear
Fe(CO)5 contains only one iron atom, thus it is mononuclear.
Fe(CO)5 contains only one iron atom, thus it is mononuclear.
Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code: Column I: a. Co3+, b. Cr3+, c. Fe3+, d. Ni2+ Column II: i. 8 B.M., ii. 35 B.M., iii. 3 B.M., iv. 24 B.M., v. 15 B.M.
- A. (1) iv v ii i
- B. (2) i ii iii iv
- C. (3) iv i ii iii
- D. (4) iii v i ii
Answer: D. (4) iii v i ii
Spin magnetic moment = sqrt(n(n+2)) BM. Co3+ (3d6, n=4) -> sqrt(24)=4.9, Cr3+ (3d3, n=3) -> sqrt(15)=3.87, Fe3+ (3d5, n=5) -> sqrt(35)=5.9, Ni2+ (3d8, n=2) -> sqrt(8)=2.82. Matches: a-iii, b-v, c-i, d-ii.
Spin magnetic moment = sqrt(n(n+2)) BM. Co3+ (3d6, n=4) -> sqrt(24)=4.9, Cr3+ (3d3, n=3) -> sqrt(15)=3.87, Fe3+ (3d5, n=5) -> sqrt(35)=5.9, Ni2+ (3d8, n=2) -> sqrt(8)=2.82. Matches: a-iii, b-v, c-i, d-ii.