Practice free Co-ordination Compounds (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Homoleptic complex from the following complexes is:
Homoleptic = all ligands identical. Only (A) contains a single ligand type (oxalate).
Which complex compound is most stable?
$[\text{Co(NH}_3)_6]^{3+}$ is exceptionally inert and stable with six identical strong-field ligands.
Given below are two statements:
Statement I: Both $[\text{Co(NH}_3)_6]^{3+}$ and $[\text{CoF}_6]^{3-}$ complexes are octahedral but differ in their magnetic behaviour.
Statement II: $[\text{Co(NH}_3)_6]^{3+}$ is diamagnetic whereas $[\text{CoF}_6]^{3-}$ is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
NH₃ is strong field → low-spin $d^6$ (no unpaired e⁻, diamagnetic). F⁻ is weak field → high-spin $d^6$ (4 unpaired e⁻, paramagnetic).
Match List I with List II.
| List I (Complex) | List II (Type of isomerism) |
|---|---|
| A. $[\text{Co(NH}_3)_5(\text{NO}_2)]\text{Cl}_2$ | I. Solvate isomerism |
| B. $[\text{Co(NH}_3)_5(\text{SO}_4)]\text{Br}$ | II. Linkage isomerism |
| C. $[\text{Co(NH}_3)_6][\text{Cr(CN)}_6]$ | III. Ionization isomerism |
| D. $[\text{Co(H}_2\text{O})_6]\text{Cl}_3$ | IV. Coordination isomerism |
Choose the correct answer from the options given below:
NO₂ binds via N or O (linkage); SO₄/Br outside vs. inside (ionization); both ions complex (coordination); H₂O exchange (solvate).
Given below are two statements:
Statement I: $[\text{Co(NH}_3)_6]^{3+}$ is a homoleptic complex whereas $[\text{Co(NH}_3)_4\text{Cl}_2]^+$ is a heteroleptic complex.
Statement II: Complex $[\text{Co(NH}_3)_6]^{3+}$ has only one kind of ligands but $[\text{Co(NH}_3)_4\text{Cl}_2]^+$ has more than one kind of ligands.
In the light of the above statements, choose the correct answer from the options given below:
Homoleptic = one kind of ligand; heteroleptic = more than one. Both statements hold.
Which of the following are paramagnetic? A. $[\text{NiCl}_4]^{2-}$, B. $\text{Ni(CO)}_4$, C. $[\text{Ni(CN)}_4]^{2-}$, D. $[\text{Ni(H}_2\text{O)}_6]^{2+}$, E. $\text{Ni(PPh}_3)_4$. Choose the correct answer from the options given below:
$[\text{NiCl}_4]^{2-}$ (tetrahedral) and $[\text{Ni(H}_2\text{O)}_6]^{2+}$ (octahedral) have 2 unpaired electrons → paramagnetic. $\text{Ni(CO)}_4$, $[\text{Ni(CN)}_4]^{2-}$ (square planar) and $\text{Ni(PPh}_3)_4$ are diamagnetic.
The correct order of the wavelength of light absorbed by the following complexes is: A. $[\text{Co(NH}_3)_6]^{3+}$, B. $[\text{Co(CN)}_6]^{3-}$, C. $[\text{Cu(H}_2\text{O)}_4]^{2+}$, D. $[\text{Ti(H}_2\text{O)}_6]^{3+}$. Choose the correct answer:
Larger crystal-field splitting $\Delta$ means shorter absorbed wavelength. $\Delta$: B (CN⁻) > A (NH₃) > D (Ti³⁺/H₂O) > C (Cu²⁺/H₂O). So wavelength increases as B < A < D < C.
Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
$[\text{Co(NH}_3)_3\text{Cl}_3]$ is a non-electrolyte (no ions in solution), so it has the 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
Choose the correct answer:
Haber → Fe (I); Wacker → PdCl₂ (II); Wilkinson → [(PPh₃)₃RhCl] (III); Ziegler → TiCl₄ with Al(CH₃)₃ (IV).
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