A metal complex having composition $\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2$Br has been isolated in two forms (X) and (Y). The form (X) reacts with ${\mathrm{AgNO}}_3$ to give a white precipitate readily soluble in dilute aqueous Ammonia, whereas (Y) gives a pale-yellow precipitate soluble in concentrate ammonia.

Solution:

A metal complex having composition.In this case, Cr(NH3)4Cl2Br has been isolated in two forms A and B. The form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia. This is a test for chloride ion. B gives a pale-yellow precipitate soluble in concentrated ammonia. This is a test for bromide ion.
Ais[Cr(NH3)4ClBr]Cl.
[Cr(NH3)4ClBr]Cl+AgNO3→AgClWhite↓+[Cr(NH3)4ClBr]+NO−3
AgCl+2NH4OH→[Ag(NH3)2Cl]+2H2O
B is [Cr(NH3)4CI2]Br.
[Cr(NH3)4Cl2]Br+AgNO3→AgBrYellow↓+[Cr(NH3)4Cl2]+NO−3
AgBr+2NH4OH→[Ag(NH3)2Br]+2H2O
In both A and B :
The oxidation number of Cr is +3.
Atomic number = 24
Cr=[Ar]3d54s1
Cr3+=[Ar]3d34s0
d2sp3-Hybridization (octahedral)
There are three unpaired electrons, so the complex is paramagnetic.
Spin magnetic moment
μ=√n(n+2)BM=√3(3+2)BM=√15BM