JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

X is the number of geometrical isomers exhibited by $[Pt(NH_{3}) (H_{2} O) BrCl]$ .

Y is the number of optically inactive isomer(s) exhibited by $[CrCl_{2} (ox)_{2}]^{3 -}$ .

Z is the number of geometrical isomers exhibited by $[Co(NH_{3})_{3} (NO_{2})_{3}]$ .

The value of X + Y + Z is ______.

Your answer

Solution

Square planar $[Pt(NH_{3}) (H_{2} O) BrCl]$ of type [Ma b c d]: 3 geometrical isomers (each of the three ways of pairing trans-partners). $\Rightarrow X = 3$ .

$[CrCl_{2} (ox)_{2}]^{3 -}$ : octahedral with two monodentate Cl and two bidentate oxalate ligands. Cis and trans geometrical isomers exist; the cis isomer is optically active (two enantiomers) and the trans isomer is optically inactive. Optically inactive isomers $\Rightarrow$ only the trans isomer is optically inactive. $\Rightarrow Y = 1$ .

$[Co(NH_{3})_{3} (NO_{2})_{3}]$ : [Ma $_{3}$ b $_{3}$ ] octahedral, gives 2 geometrical isomers (facial and meridional). $\Rightarrow Z = 2$ .

$X + Y + Z = 3 + 1 + 2 = 6$ .

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →