X is the number of geometrical isomers of [Pt(NH3)(H2O)BrCl], Y is the number of optically inactive isomers of [CrCl2(ox)2]3− and Z is the number of geometrical isomers of [Co(NH3)3(NO2)3].
Q. X is the number of geometrical isomers exhibited by [Pt(NH3)(H2O)BrCl].

Y is the number of optically inactive isomer(s) exhibited by [CrCl2(ox)2]3−.

Z is the number of geometrical isomers exhibited by [Co(NH3)3(NO2)3].

The value of X + Y + Z is ____ .

Correct Answer: 6

Explanation

Part X: [Pt(NH3)(H2O)BrCl]

Platinum(II) complexes are square planar. Since all four ligands are different, the complex shows two geometrical isomers (cis and trans).

X = 2

Part Y: [CrCl2(ox)2]3−

This is an octahedral complex with two bidentate oxalate (ox) ligands.

It shows:

• One cis form → optically active (Δ and Λ forms)
• One trans form → optically inactive

Number of optically inactive isomers = 1

Y = 1

Part Z: [Co(NH3)3(NO2)3]

This is an octahedral MA3B3 type complex.

Such complexes show two geometrical isomers:

• fac
• mer

Z = 2

Final Calculation

X + Y + Z = 2 + 1 + 2 = 6

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