The total number of isomers shown by $[C o (N H_{3})_{4} (N O_{2})_{2}] (N O_{3})$ complex is:

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 10
The complex $[C o (N H_{3})_{4} (N O_{2})_{2}] (N O_{3})]$ has ten isomers. They include geometrical, ionization and linkage isomers.
$N O_{3}$ and $N O_{2}$ groups can be either inside or outside the coordination sphere. This gives ionization isomers. $N O_{2}$ group can coordinate through N or O atom. This gives linkage isomerism. cis and trans isomers give geometrical isomers.

Suggest Corrections

Similar questions

[C o (N H_{3})_{4} (N O_{2})_{2}] N O_{3}

[R h C l (C O) (P P h_{3}) (N H_{3})]

[C o (e n)_{2} (N O_{2}) (S C N)]^{+}

C_{4} H_{10} O

Join BYJU'S Learning Program