Question

The complex $[Cr{({H_2}O)_4}B{r_2}]Cl$ gives the test for:
A. $B{r^ - }$
B. $C{l^ - }$
C. $C{r^{3 + }}$
D. $C{r^{3 + }}$ and $C{l^ - }$

Answer 1

Hint: A coordination compound is a compound where the central metal atom is linked with several ions or neutral molecules by coordinate bonds. This type of compound is made up of the coordination sphere which is non-ionizable and the ionization sphere which is ionizable.

Complete step by step answer:
In a coordinate compound also known as a complex compound, there is a central metal atom that is surrounded by ions or neutral molecules linked through coordinate bonds.
These compounds are formed with two parts namely the coordination sphere which is represented inside a square bracket $[ ]$ and the ionization sphere which comprises the part outside the square bracket.
The ionization sphere contains the part that can be ionized from the compound during the reaction.
Since the bromine ion is present inside the coordination sphere, it will not be ionized.
The chloride ion which is present outside the coordination sphere in the ionization sphere will be ionized first from the compound.
Chromium ion will also not ionize because it’s also in the coordination sphere thus non-ionizable.
So when the compound will react the chloride ion which is present in the ionization sphere will ionize and thus the compound will give the test for the chloride ion.

So, the correct answer is Option B.

Note: The ions and molecules in the coordination sphere are linked to the central metal through the coordinate bonds and thus don’t break initially.
Different ions in the ionization sphere of the compound will give rise to ionization isomerism and such complex compounds will be called ionization isomers. In an ionization isomer, the ligands and the ions are all the same. Since the ions outside the coordination sphere is different in ionization isomers, the properties will also differ.