A metal complex having composition `Cr(NH_(3))_(4)CI_(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 concentrated ammonia.
(i) Write the formulae of isomers A and B.
(ii) State the hybridisation of chromium in each of them.
(iii) Calculate the magnetic moment (spin only) of the isomer A.

Explanation

`(i) [Cr(NH_(3))_(4)ClBr]Cl overset(aq)iff [Cr(NH_(3))_(4)ClBr]^(+)+Cl^(-),Ag^(+)+Cl^(-) to AgCl darr`(white), soluble in dilute `NH_(3)`.
`[Cr(NH_(3))_(4)Cl_(2)]Br overset(aq)iff [Cr(NH_(3))_(4)Cl_(2)]^(+) +Br^(-),Ag^(+)+Br^(-) to AgBr darr`(yellow), soluble in conc. `NH_(3)`. So , `A=[Cr(NH_(3))_(4)ClBr]Cl and B=[Cr(NH_(3))_(4)Cl_(2)]Br`.
(ii) In both complexes chromium is in +3 oxidation state. Chromium with `3d^(3)` configuration has 3 unpaired electrons with weak field as well as strong field as well as strong field ligand. So, the hybridisation scheme is as follow :

(iii) `mu=sqrt(n(n+2))=sqrt(15)`
(iv) EAN=24-3+12=22
(v) Yes, both have two ions per formula unit.
`(vi) AgCl+2NH_(3) iff [Ag(NH_(3))_(2)]Cl,AgBr+2NH_(3)iff[Ag(NH_(3))_(2)]Br`