Q23 / 75 · JEE Main 2026 (28 January Shift 1) · Chemistry
X is the number of geometrical isomers exhibited by $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{BrCl}\right]$.
Y is the number of optically inactive isomer(s) exhibited by $\left[\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-}$
Z is the number of geometrical isomers exhibited by $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]$.
The value of $\mathrm{X}+\mathrm{Y}+\mathrm{Z}$ is $\_\_\_\_$.
Y is the number of optically inactive isomer(s) exhibited by $\left[\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-}$
Z is the number of geometrical isomers exhibited by $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]$.
The value of $\mathrm{X}+\mathrm{Y}+\mathrm{Z}$ is $\_\_\_\_$.
Correct answer: 6
Solution
For $[Pt(NH_3)(H_2O)BrCl]$ with square planar geometry and four different ligands, geometrical isomerism arises from different positional arrangements: X = 3.
For $[CrCl_2(ox)_2]^{3-}$, the trans isomer is optically inactive (meso compound) while cis is optically active, giving Y = 1.
For $[Co(NH_3)_3(NO_2)_3]$ in octahedral geometry, facial (fac) and meridional (mer) isomers are possible: Z = 2.
Therefore, X + Y + Z = 3 + 1 + 2 = 6
For $[CrCl_2(ox)_2]^{3-}$, the trans isomer is optically inactive (meso compound) while cis is optically active, giving Y = 1.
For $[Co(NH_3)_3(NO_2)_3]$ in octahedral geometry, facial (fac) and meridional (mer) isomers are possible: Z = 2.
Therefore, X + Y + Z = 3 + 1 + 2 = 6
Hence, the correct answer is 6.