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JEE Chemistry 2023 Question with Solution

Match List I with List II:

A matching table with List I containing complexes A to D and List II containing hybridisation types I to IV for coordination compounds.
  • A

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

  • B

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

  • C

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

  • D

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

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Match the coordination complexes with their hybridisations.

Find: The correct correspondence between List I and List II.

For [Fe(NH3)6]2+[Fe(NH_3)_6]^{2+}, Δ0<P\Delta_0 < P, hence the pairing of electrons does not occur in t2gt_{2g}. Therefore, the complex is outer orbital and its hybridisation is sp3d2sp^3d^2.

(A) [Ni(CO)4] - sp3\text{(A) } [Ni(CO)_4] \text{ - } sp^3(B) [Cu(NH3)4]2+ - dsp2\text{(B) } [Cu(NH_3)_4]^{2+} \text{ - } dsp^2(C) [Fe(NH3)6]2+ - sp3d2\text{(C) } [Fe(NH_3)_6]^{2+} \text{ - } sp^3d^2(D) [Fe(H2O)6]2+ - sp3d2\text{(D) } [Fe(H_2O)_6]^{2+} \text{ - } sp^3d^2

Thus, the correct match is:

A - I, B - II, C - IV, D - III\text{A - I, B - II, C - IV, D - III}

Therefore, the correct option is C.

Common mistakes

  • Confusing inner-orbital and outer-orbital octahedral complexes. This is wrong because the ligand strength determines whether pairing occurs before hybridisation. Check whether Δ0\Delta_0 is greater or smaller than pairing energy PP first.

  • Assuming all octahedral complexes use d2sp3d^2sp^3 hybridisation. This is wrong because weak-field ligands can give outer-orbital complexes with sp3d2sp^3d^2 hybridisation. Determine the field strength of the ligand before assigning hybridisation.

  • Treating [Ni(CO)4][Ni(CO)_4] as a square planar complex. This is wrong because [Ni(CO)4][Ni(CO)_4] is tetrahedral and corresponds to sp3sp^3 hybridisation. Use coordination number and ligand nature together.

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