Among the statements(a)-(d) the incorrect ones are :
(a) Octahedral CO(III) complexes with strong fields ligands have very high magnetic moments.
(b) When Δ₀ < P, the d-electron configuration of Co(III) in an octahedral complex is t_(eg)⁴e_g²
(c) Wavelength of light absorbed by [Co(en)₃]³⁺ is lower than that of [CoF₆]³⁻
(d) If the Δ₀ for an octahedral complex of CO(III) is 18,000 cm⁻¹, the Δₜ for its tetrahedral complex with the same ligand be 16,000 cm⁻¹

Solution

Chemistry ma’am · 2 steps

Write Co(III) as d⁶, then test each statement on spin, field strength and the 4/9 rule.

1. Start with Co(III). Co is 3d⁷ 4s², so Co³⁺ is d⁶. With a strong field (Δₒ > P) all six electrons pair in t_(2g). With a weak field (Δₒ < P) they spread out as t_(2g)⁴ e_g².

   Δₜ = 4/9Δₒ = 4/9 × 18000 cm⁻¹ = 8000 cm⁻¹

   Statement	What crystal field theory says	Verdict
   (a) strong field, very high moment	t_(2g)⁶ e_g⁰ has 0 unpaired electrons, so it is diamagnetic with μ = 0	incorrect
   (b) Δₒ < P gives t_(2g)⁴ e_g²	Pairing costs more than promotion, so the complex is high spin with 4 unpaired electrons	correct
   (c) [Co(en)₃]³⁺ absorbs a shorter λ	en is a stronger field ligand than F⁻, so Δ is larger. Since Δ = hc/λ, λ is shorter	correct
   (d) Δₜ = 16000 cm⁻¹	Δₜ = 4/9Δₒ = 8000 cm⁻¹	incorrect

2. So. Statements (a) and (d) are the incorrect ones, which is option (d).

The idea to keep. Octahedral d⁶ with a strong field ligand is t_(2g)⁶ with no unpaired electrons. A stronger ligand means a bigger Δ and a shorter wavelength absorbed. Also Δₜ = 4/9Δₒ, always under half.

Answer from the NTA final answer key. Solution by PrepWiser.