The number of geometrical isomers of the complex [RhH(C≡CR)₂(PMe₃)₃] is [CSIR-2020] · ABC26IN0153

Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Coordination Chemistry – General · Exam: CSIR NET 2019 · Marks: 2 · Difficulty: Easy

The number of geometrical isomers of the complex $[\mathrm{RhH(C{\equiv}CR)_2(PMe_3)_3}]$ is [CSIR-2020]
(a)2
(b)3
(c)4
(d)1
Answer
Answer (as printed): B
Explanation
No explanation was printed for this question.

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