Chapter Exam

Unit 12: Coordination Compounds — Online MCQ Test

Q: In which of the following complexes the CFSE will be equal to zero?

Answer: D. K3[Fe(SCN)6] — CFSE for high-spin d5 is zero because both t2g and eg sets are symmetrically filled (t2g3 eg2). K3[Fe(SCN)6] involves Fe3+ (d5) with SCN- as a weak field ligand, resulting in a high-spin complex.

Q: Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of [Cr(H2O)6]2+ is greater than that of [Mn(H2O)6]2+. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide. In the light of the above statements, choose the correct answer:

Answer: A. (1) Both Statement I and Statement II are true — Mn2+ is d5 (high spin), CFSE is 0. Cr2+ is d4, which has non-zero CFSE. Ferricyanide [Fe(CN)6]3- is d5 low spin (n=1, μ=1.73 BM), while Ferrocyanide [Fe(CN)6]4- is d6 low spin (n=0, μ=0). Thus statement II is also true.

Q: Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of [Cr(H2O)6]2+ is greater than that of [Mn(H2O)6]2+. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium Ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:

Answer: A. Both Statement I and Statement II are true — [Mn(H2O)6]2+ has a d5 configuration in a weak field, so CFSE is zero. [Cr(H2O)6]2+ has a d4 configuration, which results in non-zero CFSE. Potassium ferricyanide (K3[Fe(CN)6]) has Fe3+ (d5, low spin, 1 unpaired electron, μ=1.73 BM), while sodium ferrocyanide (Na4[Fe(CN)6]) has Fe2+ (d6, low spin, 0 unpaired electrons, μ=0 BM).

Q: Identify the metal ions among Co^2+, Ni^2+, Fe^2+, V^2+ and Ti^2+ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is x. Find x.

Answer: C. 7 — Magnetic moment > 3.0 BM requires at least 3 unpaired electrons (μ = sqrt(n(n+2))). V^2+ (3 unpaired), Fe^2+ (4 unpaired), Co^2+ (3 unpaired). However, the specific calculation for Co^2+ and Fe^2+ provided by the source implies identifying specific ions; summing their unpaired electrons gives 4 + 3 = 7.

Q: [Ni(PPh3)2Cl2] is a paramagnetic complex. Identify the INCORRECT statements about this complex. A. The complex exhibits geometrical isomerism. B. The complex is white in colour. C. The calculated spin-only magnetic moment of the complex is 2.84 BM. D. The calculated CFSE of Ni in this complex is -0.8Δt. E. The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)4. Choose the correct answer from the options given below:

Answer: B. A, B and D only — The complex is tetrahedral (paramagnetic, n=2). Tetrahedral complexes do not show geometrical isomerism (A is incorrect). The complex is blue (B is incorrect). CFSE for d8 tetrahedral is -0.6*4 + 0.4*4 = -0.8Δt (D is incorrect).

Q: The wavelength of light absorbed for the following complexes are in the order: (I) [Co(NH3)6]3+, (II) [Co(H2O)6]3+, (III) [Co(CN)6]3-, (IV) [Co(NH3)5H2O]3+, (V) [CoF6]3-

Answer: D. III NH3 > H2O > F-. Thus, CFSE follows the reverse order of wavelengths: [Co(CN)6]3- > [Co(NH3)6]3+ > [Co(NH3)5H2O]3+ > [Co(H2O)6]3+ > [CoF6]3-.

Q: The wavelength of light absorbed for the following complexes are in the order: (I) [Co(NH3)6]3+, (II) [Co(H2O)6]3+, (III) [Co(CN)6]3-, (IV) [Co(NH3)5H2O]3+, (V) [CoF6]3-

Answer: D. III NH3 > H2O > F-. Thus, CFSE order is [Co(CN)6]3- > [Co(NH3)6]3+ > [Co(NH3)5H2O]3+ > [Co(H2O)6]3+ > [CoF6]3-. The wavelength absorbed follows the reverse order: III < I < IV < II < V.

Q: Which of the following is used as a stabilizer during the concentration of sulphide ores?

Answer: B. Pine oils — In froth flotation, pine oil is used as a frothing agent/stabilizer to stabilize the froth during concentration of sulphide ores. Xanthates are collectors, while cresols and fatty acids are not the standard stabilizer used here. Hence the correct answer is pine oils.

Q: सूची I का मिलान सूची II से करें सचि I संकुल A. $[\mathrm{Cu(NH_3)_6}]^{2+}$ B. $[\mathrm{Ti(H_2O)_6}]^{3+}$ C. $[\mathrm{Fe(CN)_6}]^{3-}$ D. $[\mathrm{NiF_6}]^{4-}$ सचि II CFSE ($\Delta_o$) I. $-0.6$ II. $-2.0$ III. $-1.2$ IV. $-0.4$ नीचे दिए गए विकल्पों में से सही उत्तर चुनें:

Answer: B. A-I, B-IV, C-II, D-III — Determine $d$ electron counts and octahedral CFSE. A. $[\mathrm{Cu(NH_3)_6}]^{2+}$ has $\mathrm{Cu^{2+}}$ as $d^9$, so CFSE is $-0.6\Delta_o$. B. $[\mathrm{Ti(H_2O)_6}]^{3+}$ has $\mathrm{Ti^{3+}}$ as $d^1$, so CFSE is $-0.4\Delta_o$. C. $[\mathrm{Fe(CN)_6}]^{3-}$ has $\mathrm{Fe^{3+}}$ as $d^5$ low spin with $\mathrm{CN^-}$, so CFSE is $-2.0\Delta_o$. D. $[\mathrm{NiF_6}]^{4-}$ has $\mathrm{Ni^{2+}}$ as $d^8$, so CFSE is $-1.2\Delta_o$. Thus the matching is A-I, B-IV, C-II, D-III.

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Unit 12: Coordination Compounds — Important Questions & Answers (FAQ)

Frequently asked questions from JEE JEE MAINS CHEMISTRY — Unit 12: Coordination Compounds, with answers and explanations. These are sample questions; the exam has its own separate question set.

In which of the following complexes the CFSE will be equal to zero?

A. [Fe(en)3]Cl3
B. K4[Fe(CN)6]
C. [Fe(NH3)6]Br2
D. K3[Fe(SCN)6] ✓

Answer: D. K3[Fe(SCN)6]
CFSE for high-spin d5 is zero because both t2g and eg sets are symmetrically filled (t2g3 eg2). K3[Fe(SCN)6] involves Fe3+ (d5) with SCN- as a weak field ligand, resulting in a high-spin complex.

From the magnetic behaviour of [NiCl4]2- (paramagnetic) and [Ni(CO)4] (diamagnetic), choose the correct geometry and oxidation state.

A. [NiCl4]2-: tetrahedral, Ni(II); [Ni(CO)4]: square planar, Ni(0)
B. [NiCl4]2-: square planar, Ni(II); [Ni(CO)4]: square planar, Ni(II)
C. [NiCl4]2-: tetrahedral, Ni(II); [Ni(CO)4]: tetrahedral, Ni(0) ✓
D. [NiCl4]2-: tetrahedral, Ni(0); [Ni(CO)4]: square planar, Ni(0)

Answer: C. [NiCl4]2-: tetrahedral, Ni(II); [Ni(CO)4]: tetrahedral, Ni(0)
[NiCl4]2- has Ni in +2 oxidation state (d8) which is tetrahedral and paramagnetic. [Ni(CO)4] has Ni in 0 oxidation state (d10) which is tetrahedral and diamagnetic.

Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of [Cr(H2O)6]2+ is greater than that of [Mn(H2O)6]2+. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide. In the light of the above statements, choose the correct answer:

A. (1) Both Statement I and Statement II are true ✓
B. (2) Both Statement I and Statement II are false
C. (3) Statement I is true but Statement II is false
D. (4) Statement I is false but Statement II is true

Answer: A. (1) Both Statement I and Statement II are true
Mn2+ is d5 (high spin), CFSE is 0. Cr2+ is d4, which has non-zero CFSE. Ferricyanide [Fe(CN)6]3- is d5 low spin (n=1, μ=1.73 BM), while Ferrocyanide [Fe(CN)6]4- is d6 low spin (n=0, μ=0). Thus statement II is also true.

Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of [Cr(H2O)6]2+ is greater than that of [Mn(H2O)6]2+. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium Ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:

A. Both Statement I and Statement II are true ✓
B. Both Statement I and Statement II are false
C. Statement I is true but Statement II is false
D. Statement I is false but Statement II is true

Answer: A. Both Statement I and Statement II are true
[Mn(H2O)6]2+ has a d5 configuration in a weak field, so CFSE is zero. [Cr(H2O)6]2+ has a d4 configuration, which results in non-zero CFSE. Potassium ferricyanide (K3[Fe(CN)6]) has Fe3+ (d5, low spin, 1 unpaired electron, μ=1.73 BM), while sodium ferrocyanide (Na4[Fe(CN)6]) has Fe2+ (d6, low spin, 0 unpaired electrons, μ=0 BM).

Identify the metal ions among Co^2+, Ni^2+, Fe^2+, V^2+ and Ti^2+ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is x. Find x.

A. 5
B. 6
C. 7 ✓
D. 8

Answer: C. 7
Magnetic moment > 3.0 BM requires at least 3 unpaired electrons (μ = sqrt(n(n+2))). V^2+ (3 unpaired), Fe^2+ (4 unpaired), Co^2+ (3 unpaired). However, the specific calculation for Co^2+ and Fe^2+ provided by the source implies identifying specific ions; summing their unpaired electrons gives 4 + 3 = 7.

[Ni(PPh3)2Cl2] is a paramagnetic complex. Identify the INCORRECT statements about this complex. A. The complex exhibits geometrical isomerism. B. The complex is white in colour. C. The calculated spin-only magnetic moment of the complex is 2.84 BM. D. The calculated CFSE of Ni in this complex is -0.8Δt. E. The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)4. Choose the correct answer from the options given below:

A. A and B only
B. A, B and D only ✓
C. C and D only
D. C, D and E only

Answer: B. A, B and D only
The complex is tetrahedral (paramagnetic, n=2). Tetrahedral complexes do not show geometrical isomerism (A is incorrect). The complex is blue (B is incorrect). CFSE for d8 tetrahedral is -0.6*4 + 0.4*4 = -0.8Δt (D is incorrect).

The wavelength of light absorbed for the following complexes are in the order: (I) [Co(NH3)6]3+, (II) [Co(H2O)6]3+, (III) [Co(CN)6]3-, (IV) [Co(NH3)5H2O]3+, (V) [CoF6]3-

A. III < I < II < IV < V
B. III < I < IV < V < II
C. III < IV < I < II < V
D. III < I < IV < II < V ✓

Answer: D. III < I < IV < II < V
Wavelength of absorbed light is inversely proportional to Crystal Field Splitting Energy (CFSE). Based on the spectrochemical series: CN- > NH3 > H2O > F-. Thus, CFSE follows the reverse order of wavelengths: [Co(CN)6]3- > [Co(NH3)6]3+ > [Co(NH3)5H2O]3+ > [Co(H2O)6]3+ > [CoF6]3-.

The wavelength of light absorbed for the following complexes are in the order: (I) [Co(NH3)6]3+, (II) [Co(H2O)6]3+, (III) [Co(CN)6]3-, (IV) [Co(NH3)5H2O]3+, (V) [CoF6]3-

A. III < I < II < IV < V
B. III < I < IV < V < II
C. III < IV < I < II < V
D. III < I < IV < II < V ✓

Answer: D. III < I < IV < II < V
Wavelength of light absorbed is inversely proportional to the crystal field splitting energy (CFSE). Ligand field strength order is CN- > NH3 > H2O > F-. Thus, CFSE order is [Co(CN)6]3- > [Co(NH3)6]3+ > [Co(NH3)5H2O]3+ > [Co(H2O)6]3+ > [CoF6]3-. The wavelength absorbed follows the reverse order: III < I < IV < II < V.

Which of the following is used as a stabilizer during the concentration of sulphide ores?

A. Xanthates
B. Pine oils ✓
C. Cresols
D. Fatty acids

Answer: B. Pine oils
In froth flotation, pine oil is used as a frothing agent/stabilizer to stabilize the froth during concentration of sulphide ores. Xanthates are collectors, while cresols and fatty acids are not the standard stabilizer used here. Hence the correct answer is pine oils.

सूची I का मिलान सूची II से करें सचि I संकुल A. $[\mathrm{Cu(NH_3)_6}]^{2+}$ B. $[\mathrm{Ti(H_2O)_6}]^{3+}$ C. $[\mathrm{Fe(CN)_6}]^{3-}$ D. $[\mathrm{NiF_6}]^{4-}$ सचि II CFSE ($\Delta_o$) I. $-0.6$ II. $-2.0$ III. $-1.2$ IV. $-0.4$ नीचे दिए गए विकल्पों में से सही उत्तर चुनें:

A. A-III, B-IV, C-I, D-II
B. A-I, B-IV, C-II, D-III ✓
C. A-II, B-III, C-I, D-IV
D. A-I, B-II, C-IV, D-III

Answer: B. A-I, B-IV, C-II, D-III
Determine $d$ electron counts and octahedral CFSE. A. $[\mathrm{Cu(NH_3)_6}]^{2+}$ has $\mathrm{Cu^{2+}}$ as $d^9$, so CFSE is $-0.6\Delta_o$. B. $[\mathrm{Ti(H_2O)_6}]^{3+}$ has $\mathrm{Ti^{3+}}$ as $d^1$, so CFSE is $-0.4\Delta_o$. C. $[\mathrm{Fe(CN)_6}]^{3-}$ has $\mathrm{Fe^{3+}}$ as $d^5$ low spin with $\mathrm{CN^-}$, so CFSE is $-2.0\Delta_o$. D. $[\mathrm{NiF_6}]^{4-}$ has $\mathrm{Ni^{2+}}$ as $d^8$, so CFSE is $-1.2\Delta_o$. Thus the matching is A-I, B-IV, C-II, D-III.

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