Let X be the number of complexes with a larger Δₒ than [Co(NH₃)₆]³⁺ among [Co(CN)₆]³⁻, [Co(en)₃]³⁺, [Co(H₂O)₆]³⁺, [CoF₆]³⁻ and [Co(NO₂)₆]³⁻. Let Y be the number of linkage isomers of [Co(NO₂)₂(NH₃)₄]⁺

Question

Let X be the number of complexes with a larger Δₒ than [Co(NH₃)₆]³⁺ among [Co(CN)₆]³⁻, [Co(en)₃]³⁺, [Co(H₂O)₆]³⁺, [CoF₆]³⁻ and [Co(NO₂)₆]³⁻. Let Y be the number of linkage isomers of [Co(NO₂)₂(NH₃)₄]⁺ (count by donor atom choice only). Let Z be the number of species showing geometrical isomerism among [Pt(en)₂]²⁺, [Pt(en)Cl₂], [Pt(gly)₂], [Pt(NH₃)₂Cl₂] and [Pt(NH₃)(py)ClBr]. Find (X + Y) × Z.

Answer:

18

Solution:

Stronger than NH₃: CN⁻, en and NO₂⁻, so X = 3. Linkage forms of two NO₂⁻: both N, both O, or one of each, so Y = 3. Geometrical isomers occur for [Pt(gly)₂], [Pt(NH₃)₂Cl₂] and [Pt(NH₃)(py)ClBr], so Z = 3. (3 + 3) × 3 = 18.

TestHub · Accepted Answer

Stronger than NH₃: CN⁻, en and NO₂⁻, so X = 3. Linkage forms of two NO₂⁻: both N, both O, or one of each, so Y = 3. Geometrical isomers occur for [Pt(gly)₂], [Pt(NH₃)₂Cl₂] and [Pt(NH₃)(py)ClBr], so Z = 3. (3 + 3) × 3 = 18.

Chemistry - Coordination Chemistry Question with Solution | TestHub

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