X is the number of geometrical isomers exhibited by $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{BrCl}\right]$.
Y is the number of optically inactive isomer(s) exhibited by $\left[\mathrm{CrCl}_2(\mathrm{ox})_2\right]^{3-}$
Z is the number of geometrical isomers exhibited by $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3\right]$.
The value of $\mathrm{X}+\mathrm{Y}+\mathrm{Z}$ is $\_\_\_\_$。
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Solution
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X = 3 (Two cis + one trans isomers)
Y = 1 (trans isomer)
Z = 2 (Fac- mer isomer)
X+Y+Z = 3 + 1 + 2 = 6
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