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 ppt readily soluble in dil aq nh3 whereas b gives a pale yellow soluble in conc nh3. (a) write formula of a and b (b) state hybridization of cr in each of them (c) calculate mag moment (spin only value) of isomer a

Answer 1

Hey dear,

● Answers with explaination-
(a) Structural formulas-
I. Compound A reacts with AgNO3 to give a white ppt readily soluble in dil. aq.NH3
[Cr(NH3)4ClBr]Cl + AgNO3 ---> AgCl(white ppt) + [Cr(NH3)4ClBr] + NO3-
AgCl + 2NH4OH ---> [Ag(NH3)2Cl] + 2H20

Therefore, compound A must be [Cr(NH3)4ClBr]Cl.

II. Compound B reacts with AgNO3 to give a pale yellow soluble precipitate in conc.NH3
[Cr(NH3)4Cl2]Br + AgNO3 ---> AgBr(pale yello ppt) + [Cr(NH3)4Cl2] + NO3-
AgBr + 2NH4OH ---> [Ag(NH3)2Br] + 2H20

Therefore, compound B must be [Cr(NH3)4Cl2]Br.

(b) State of hybridization-
In both compounds chromium is in Cr3+ state.

(c) Spin magnetic moment-
Unpaired electronz = 3
Spin magnetic moment = √(3×5)
Spin magnetic moment = √15
Spin magnetic moment = 3.873 BM


Hope this is helpful...