The number of geometrical isomers of the complex [RhH(C ≡ CR)₂ (PMe₃)₃] is · ABC26IN1136
Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Coordination Chemistry – General · Exam: CSIR NET JUNE 2019 · Marks: 2 · Difficulty: Easy
The number of geometrical isomers of the complex $\left[\mathrm{RhH}(\mathrm{C} \equiv \mathrm{CR})_{2}\left(\mathrm{PMe}_{3}\right)_{3}\right]$ is
(a)2
(b)3
(c)4
(d)1
Answer
Answer: B ✓ checked by 4AB · confidence medium
Explanation
Octahedral MA₃B₂C (A = PMe₃, B = alkynyl, C = H): fac/mer arrangements of the three phosphines with the alkynyls cis or trans give three geometrical isomers (b).