A metal complex having composition `Cr(NH_3)_4 Cl_2 Br` has been isolated in two forms `A` and `B`. The form `A` reacts with `AgNO_3` to give a white precipitate readily soluble in dilute aqueous ammonia, whereas `B` gives a pale yellow precipitate soluble in concentrațed ammonia. Write the formula,of `A` and `B`

Explanation

Ans: .Cr(NH_3)_4 C_2, ~B_ r . has two isomèrs. Coordination number of chromium is 6 Hence .[Cr(NH_3)_4 Cl_2] Br. and .[Cr(NH_3^circ)_4-ClBr] Cl.
may be the isomers. . ^prime A^prime. forms white .precipitate with .Ag NO_3., which implies one chlorine is present outside the coordination sphere. Hence .A-[Cr(NH_3)_4 CIBr] Cl.
.[Cr(NH_3)_4 Cl Br] Cl underset./AgNO_3/longrarr[Cr(NH_3)_4 C tildeB B r]^2+.


NO_3^-*+AgCl downarrow


Since B forms a pale yellow precipitate with AgNO, Br atom must be presènt outside the coordination sphere. Hence, .B. is
Hybridisation of .Cr. in (A) and (B) is d.sp" having three unpaired electrons (3d.)
...