$X$ is the number of geometrical isomers exhibited by $[Pt(NH_3)(H_2O)BrCl]$.
$Y$ is the number of optically inactive isomer$(s)$ exhibited by $[CrCl_2(ox)_2]^{3-}$.
$Z$ is the number of geometrical isomers exhibited by $[Co(NH_3)_3(NO_2)_3]$.
The value of $X+Y+Z$ is . . . . . . .
$Y$ is the number of optically inactive isomer$(s)$ exhibited by $[CrCl_2(ox)_2]^{3-}$.
$Z$ is the number of geometrical isomers exhibited by $[Co(NH_3)_3(NO_2)_3]$.
The value of $X+Y+Z$ is . . . . . . .
- A$4$
- B$5$
- C$6$
- D$7$
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Similar Questions
Draw all the isomers (geometrical and optical) of:
$(i)$ $[CoCl_2(en)_2]^+$
$(ii)$ $[Co(NH_3)Cl(en)_2]^{2+}$
$(iii)$ $[Co(NH_3)_2Cl_2(en)]^+$
$(i)$ $[CoCl_2(en)_2]^+$
$(ii)$ $[Co(NH_3)Cl(en)_2]^{2+}$
$(iii)$ $[Co(NH_3)_2Cl_2(en)]^+$
Medium
View Solution$[CoCl_2(NH_3)_4]^+ + Cl^- \to [CoCl_3(NH_3)_3] + NH_3$. If in this reaction two isomers of the product are obtained,which is true for the initial (reactant) complex?
Which one of the following complexes exhibits chirality?
Which one has the highest paramagnetism?
Difficult
View SolutionThe spin only magnetic moment value of $Cr(CO)_6$ is (in $BM$)
Easy
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