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Valence-bond approach and basic crystal-field theory: NEET MCQs with answers

90 questions on this topic of Coordination Compounds (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. What is the electronic configuration of a d⁴ transition metal ion in an octahedral crystal field when Δₒ > P?

  1. t₂g³ eg¹
  2. t₂g¹ eg³
  3. t₂g² eg²
  4. t₂g⁴ eg⁰
Show answer and explanation

Answer: (D) t₂g⁴ eg⁰

When the crystal field splitting energy (Δₒ) is greater than the pairing energy (P), pairing of electrons in the lower-energy t₂g orbitals is energetically more favorable than promoting an electron to the eg set. Therefore, for a d⁴ ion, all four electrons occupy the t₂g set, giving a low-spin configuration of t₂g⁴ eg⁰.

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2. Given below are two statements: Statement I: The complex [Co(NH₃)₆]³⁺ is diamagnetic whereas [Ni(NH₃)₆]²⁺ is paramagnetic. Statement II: Ammonia always acts as a strong field ligand causing complete pairing of d-electrons in all first-row transition metal complexes. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because Co³⁺ (d⁶) with NH₃ forms a low-spin d²sp³ hybridized diamagnetic complex, while Ni²⁺ (3d⁸) with NH₃ retains two unpaired electrons in eg orbitals and is paramagnetic (sp³d²). Statement II is incorrect because NH₃ is at the border of weak and strong field ligands; it acts as a strong field ligand for Co³⁺ but does not cause forced pairing in Ni²⁺ or Mn²⁺.

hard

3. An octahedral complex [Cr(NH₃)₆]³⁺ and a tetrahedral complex [CoCl₄]²⁻ have spin-only magnetic moments of μ₁ and μ₂ respectively. What is the value of (μ₁ - μ₂)?

  1. 1.73 BM
  2. 1.10 BM
  3. 0 BM
  4. 2.05 BM
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Answer: (C) 0 BM

In [Cr(NH₃)₆]³⁺, Cr³⁺ is a 3d³ system with 3 unpaired electrons (t₂g³), so μ₁ = √(3 × 5) = √15 ≈ 3.87 BM. In [CoCl₄]²⁻, Co²⁺ is a 3d⁷ system in a tetrahedral field with configuration e⁴ t₂³, which also has 3 unpaired electrons, so μ₂ = √(3 × 5) = √15 ≈ 3.87 BM. Therefore, (μ₁ - μ₂) = 0 BM.

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4. Given below are two statements: Statement I: Square planar complexes are generally formed by d⁸ metal ions with strong field ligands. Statement II: Tetrahedral complexes rarely exhibit low-spin configurations because crystal field splitting (Δₜ) is much smaller than the pairing energy (P). In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Statement I is correct because strong field ligands cause electron pairing in d⁸ systems, leaving one empty d-orbital (specifically d(x²-y²)) to form dsp² square planar complexes. Statement II is also correct because Δₜ is approximately (4/9)Δₒ, which is almost always smaller than pairing energy P, leading to high-spin complexes.

easy

5. According to Valence Bond Theory, what is the hybridization of Ni and the geometry of the complex ion [Ni(CN)₄]²⁻?

  1. sp³, tetrahedral
  2. d²sp, square planar
  3. dsp², square planar
  4. sp³d, trigonal bipyramidal
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Answer: (C) dsp², square planar

In [Ni(CN)₄]²⁻, nickel is in the +2 oxidation state with a 3d⁸ configuration. Cyanide (CN⁻) is a strong field ligand that forces pairing of the two unpaired 3d electrons, leaving one 3d orbital vacant. This vacant 3d orbital undergoes dsp² hybridization with one 4s and two 4p orbitals, yielding a square planar geometry.

hard

6. Given below are two statements: Statement I: [PtCl₄]²⁻ has a square planar geometry and is diamagnetic. Statement II: For 4d and 5d transition metal ions, the crystal field splitting energy is sufficiently large that their d⁸ complexes are square planar even with weak-field ligands. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
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Answer: (A) Both Statement I and Statement II are correct.

Due to greater spatial extent of 5d orbitals, Pt²⁺ experiences large crystal field splitting (Δ > P) even with halides. Hence, [PtCl₄]²⁻ undergoes pairing to adopt a dsp² hybridized square planar, diamagnetic configuration.

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7. According to Valence Bond Theory, which of the following complexes is an inner orbital, diamagnetic octahedral complex?

  1. [Fe(H₂O)₆]²⁺
  2. [Ni(NH₃)₆]²⁺
  3. [FeF₆]³⁻
  4. [Co(NH₃)₆]³⁺
Show answer and explanation

Answer: (D) [Co(NH₃)₆]³⁺

In [Co(NH₃)₆]³⁺, Co³⁺ has a 3d⁶ configuration. Under the influence of NH₃ (a strong field ligand), electrons pair up to leave two 3d orbitals vacant, resulting in d²sp³ hybridization (inner orbital) and zero unpaired electrons (diamagnetic).

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8. In a tetrahedral crystal field, the splitting Δt is related to Δo by:

  1. Δt = Δo
  2. Δt = 2Δo
  3. Δt ≈ (9/4)Δo
  4. Δt ≈ (4/9)Δo
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Answer: (D) Δt ≈ (4/9)Δo

Tetrahedral splitting is smaller (Δt ≈ 4/9 Δo), so tetrahedral complexes are generally high-spin and pairing is rare.

NCERT Class 12

medium

9. According to the Valence Bond Theory, the hybridization and number of unpaired electrons in the complex ion [Cu(NH₃)₄]²⁺ are respectively:

  1. sp³, 1
  2. sp³d², 0
  3. dsp², 1
  4. dsp², 0
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Answer: (C) dsp², 1

In [Cu(NH₃)₄]²⁺, copper is in the +2 oxidation state with a 3d⁹ configuration. According to Valence Bond Theory, one 3d electron is promoted to the 4p orbital (or hybridizes as dsp² with a vacancy in 3d), giving a square planar geometry via dsp² hybridization. Since one unpaired electron remains, the complex is paramagnetic with 1 unpaired electron.

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10. A high-spin octahedral complex of a transition metal ion with d⁶ configuration has a crystal field stabilization energy (excluding pairing energy) equal to:

  1. -0.6 Δₒ
  2. -0.4 Δₒ
  3. -0.8 Δₒ
  4. -1.2 Δₒ
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Answer: (B) -0.4 Δₒ

In a high-spin octahedral complex (weak field), d⁶ has the configuration t₂g⁴ eg². Its CFSE is calculated as [4 × (-0.4) + 2 × (+0.6)] Δₒ = (-1.6 + 1.2) Δₒ = -0.4 Δₒ.

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11. Why are tetrahedral complexes generally not observed to exhibit orbital degeneracy splitting with a subscript 'g' in their crystal field states?

  1. They have a center of inversion (centrosymmetric)
  2. The splitting energy is always negative
  3. They lack a center of inversion
  4. Only two ligands approach along Cartesian axes
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Answer: (C) They lack a center of inversion

The subscript 'g' stands for 'gerade', meaning symmetric with respect to a center of inversion. Tetrahedral complexes do not possess a center of inversion, so the representations are designated simply as 'e' and 't₂' rather than 'eg' and 't₂g'.

hard

12. Which of the following ions will have the maximum value of crystal field splitting energy (Δₒ) when forming an octahedral complex with the same neutral ligand L?

  1. Co²⁺
  2. Co³⁺
  3. Rh³⁺
  4. Ir³⁺
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Answer: (D) Ir³⁺

Crystal field splitting energy increases down a triad from 3d to 4d to 5d series (3d < 4d < 5d) by about 30% to 50% for each step. Since Ir belongs to the 5d transition series and has a +3 oxidation state, it produces the largest Δₒ among the given ions.

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