Read the passage carefully and answer the Questions.
In the formation of coordination complexes, if the inner d orbital (n-1d) is used in hybridisation, the complex, is called an inner orbital or low spin or spin paired complex. And if it uses outer d orbital (nd) in hybridisation (like \( sp^3d^2 \)), it is called outer orbital or high spin or spin free complex. The degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the complex. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. For coordination complexes, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula, \( μ = \sqrt{n (n+2)} \) where \( n \) is the number of unpaired electrons and \( μ \) is the magnetic moment in units of Bohr magneton (BM). The coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine.
Which of the following metals belongs to Wilkinson catalyst?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Rh\)
Wilkinson catalyst is the rhodium complex \( \ce{[(Ph3P)3RhCl]} \), chloridotris(triphenylphosphine)rhodium(I). It is a homogeneous catalyst for the hydrogenation of alkenes: it dissolves in the reaction mixture, and \( \ce{H2} \) adds across the C=C bond at room temperature.
The oxidation state of rhodium follows from the charges, since triphenylphosphine is neutral and chloride is −1:
\( \displaystyle x + 3(0) + (-1) = 0 \Rightarrow x = +1 \)
So the metal of Wilkinson catalyst is rhodium (Rh).
NCERT (2026-27 edition):
Importance and Applications of Coordination Compounds – …
