\(X\) 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 \(X + Y + Z\). }
\(X\) 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 \(X + Y + Z\). }
Correct Answer: 6
Solution and Explanation
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.
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