JEE Chemistry 2026 Question with Solution

Given:

• $X$ is the number of geometrical isomers of $[\mathrm{Pt(NH_3)(H_2O)BrCl}]$.
• $Y$ is the number of optically inactive isomer(s) of $[\mathrm{CrCl_2(ox)_2}]^{3-}$.
• $Z$ is the number of geometrical isomers of $[\mathrm{Co(NH_3)_3(NO_2)_3}]$.

Find: $X + Y + Z$

For $[\mathrm{Pt(NH_3)(H_2O)BrCl}]$, the complex is square planar with a $d^8$ metal center $\mathrm{Pt(II)}$. Square planar complexes can exhibit geometrical isomerism. The possible isomers are cis and trans.

Therefore, $X = 2$.

For $[\mathrm{CrCl_2(ox)_2}]^{3-}$, the complex is octahedral and $\mathrm{ox}$ is oxalate, a bidentate ligand. The solution states that all isomers formed are optically inactive and that the complex has 2 geometrical isomers, cis and trans.

Therefore, $Y = 2$.

For $[\mathrm{Co(NH_3)_3(NO_2)_3}]$, the complex is octahedral and shows fac and mer geometrical isomerism.

Therefore, $Z = 2$.

Now add the values:

$$X + Y + Z = 2 + 2 + 2 = 6$$

Therefore, the final value of $X + Y + Z$ is $6$.

Given: Three coordination compounds are to be analyzed for geometrical and optical inactivity counts.

Find: The value of $X + Y + Z$.

1. For $[\mathrm{Pt(NH_3)(H_2O)BrCl}]$:

• This is a square planar complex.
• With four different monodentate ligands in square planar geometry, geometrical isomerism is possible.
• The two arrangements are cis and trans.

So,

$$X = 2$$

2. For $[\mathrm{CrCl_2(ox)_2}]^{3-}$:

• This is an octahedral complex.
• $\mathrm{ox}$ is a bidentate oxalate ligand.
• The solution states that the complex has two geometrical isomers, cis and trans.
• It also states that all isomers formed are optically inactive.

So,

$$Y = 2$$

3. For $[\mathrm{Co(NH_3)_3(NO_2)_3}]$:

• This is an octahedral complex of the type $[\mathrm{MA_3B_3}]$.
• Such complexes show fac and mer geometrical isomerism.

So,

$$Z = 2$$

Hence,

$$X + Y + Z = 2 + 2 + 2 = 6$$

Thus, the required numerical value is $6$.