Question

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

Answer 1

Correct Answer: 6

To determine \(X+Y+Z\), we will analyze each complex and calculate the relevant isomer counts:

1. Analyzing \([\mathrm{Pt(NH_3)(H_2O)BrCl}]\): 
This coordination complex is square planar (\(d^8\) metal center Pt(II)). Square planar complexes can exhibit geometrical isomerism.
Possible isomers:

• Cis: Adjacent NH₃ and H₂O, Br and Cl.
• Trans: Opposite NH₃ and H₂O, Br and Cl.

Thus, there are 2 geometrical isomers. Therefore, \(X=2\).

 

2. Analyzing \([\mathrm{CrCl_2(ox)_2}]^{3-}\):
This is an octahedral complex where 'ox' represents oxalate ions (\(C_2O_4^{2-}\)) which are bidentate ligands. Octahedral complexes with bidentate ligands can show geometric isomerism but not optical isomerism since 'ox' is planar.
All isomers formed are optically inactive.
The complex has 2 geometrical isomers (cis and trans). Thus, \(Y=2\).

3. Analyzing \([\mathrm{Co(NH_3)_3(NO_2)_3}]\):
This complex is also octahedral. The ligands are monodentate, allowing for geometrical isomerism:

• Facial (fac): All three NO₂ groups adjacent.
• Meridional (mer): NO₂ groups are along a line.

There are 2 geometrical isomers. Thus, \(Z=2\).

 

Calculating \(X+Y+Z\):
\(X+Y+Z=2+2+2=6\)
The computed value falls within the expected range [6,6].

Therefore, the final value of \(X+Y+Z\) is 6.