Question

A metal complex having composition 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 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 value) of the isomer A

Answer 1

complex [Cr(NH₃)₄BrCl]Cl reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia .

Hence,  isomer A is [Cr(NH₃)₄BrCl]Cl

complex [Cr(NH₃)₄Cl₂]Br gives pale yellow precipitate soluble in concentrated ammonia . Hence, isomer B is [Cr(NH₃)₄Cl₂]Br

(ii) hybridisation of chromium in isomer A and B is d²SP³

(iii) number of unpaired electrons in chromium ion(Cr³⁺) = 3, [Ar]3d³

We know, magnetic moment is $\sqrt{n(n+2)}$ BM

Where n is number of unpaired electrons

So, magnetic moment of isomer A = $\sqrt{3(3+2)}=\sqrt{15}$ =3.87BM

Answer 2

A metal complex having composition Cr(NH₃)₄Cl₂Br has been isolated in two forms A and B.According to question,complex [Cr(NH₃)₄BrCl]Cl reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia.Hence chloride must reside in outer co-ordination sphere.

Isomer A is [Cr(NH₃)₄BrCl]Cl

Complex Cr(NH₃)₄Cl₂Br gives pale yellow precipitate soluble in concentrated ammonia. Hence, isomer B is [Cr(NH₃)₄Hence bromide must reside in outer co-ordination sphere.Cl₂]Br

(ii) hybridisation of chromium in isomer A and B is SP3d2

(iii) number of unpaired electrons in complex is 1

We know, magnetic moment is expressed in  BM

Where n is number of unpaired electrons

So, magnetic moment of isomer A = √1*(1+2)=1.732 B.M