Total number of geometrical isomers for the complex $\left[\mathrm{RhCl}(\mathrm{CO})(\mathrm{PPh}_{3})(\mathrm{NH}_{3})\right]$ is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Coordination Compounds

Total number of geometrical isomers for the complex

[RhCl (CO) (PPh_{3}) (NH_{3})]

Answer: 3

Solution

This is a square planar complex with 4 different ligands (Cl, CO, PPh₃, NH₃). For a square planar MA₄ type complex with 4 different ligands, the number of geometrical isomers = 3. These correspond to different relative positions of the ligands around the central metal atom.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer

Practise more: all JEE Advanced PYQs on Coordination Compounds →

More JEE Advanced questions on Coordination Compounds

The total number of all possible isomers for the square planar complex with formula $K [M (NCS) (NO_{2}) ($ gly $)$ ] is ____
( $M =$ metal ion and gly $= NH_{2} CH_{2} COO^{-}$ )
JEE Advanced 2026View solution →
Consider that the coordinating atoms of the ligands in cis- $[Co (NH_{3})_{4} Cl_{2}] Cl$ and mer- $[Co (NH_{3})_{3} Cl_{3}]$ octahedral complexes are at the vertices of an octahedron. The sum of total number of the triangular faces in both the compl …
JEE Advanced 2026View solution →
The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is
JEE Advanced 2025View solution →
The sum of the spin only magnetic moment values (in B.M.) of $[Mn (Br)_{6}]^{3 -}$ and $[Mn (CN)_{6}]^{3 -}$ is ____.
JEE Advanced 2025View solution →
Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other.
[ $en = H_{2} NCH_{2} CH_{2} NH_{2}$ ]
JEE Advanced 2024View solution →
Among $V (CO)_{6}$ , $Cr (CO)_{5}$ , $Cu (CO)_{3}$ , $Mn (CO)_{5}$ , $Fe (CO)_{5}$ , $[Co (CO)_{3}]^{3 -}$ , $[Cr (CO)_{4}]^{4 -}$ , and $Ir (CO)_{3}$ , t …
JEE Advanced 2024View solution →