# A metal complex having composition Cr(NH3)4 CI2Br 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. Write the formula of (A) and (B) and state the hybridization of chromium in each. Calculate their magnetic moments (spin only value).

Jitender Pal
9 years ago
Hello Student,
Compound (A) on treatment with AgNO3 gives white precipitate of AgCI, which is readily soluble in dil. Aq. NH3. Therefore it has at least one CI- ion in the ionization sphere furthermore chromium has coordination number equal to 6. So its formula is [Cr(NH3)4BrCI]CI.
Compound (B) on treatment with AgNO3 gives pale yellow precipitate of AgBr soluble in conc. NH3. Therefore it has Br- in the ionization sphere. So its formula is [Cr(NH3)4CI2]Br.
State of hybridization of chromium in both (A) and (B) is d2sp3.
Spin magnetic moment of (A) or (B),
Μspin = √n(n + 2) = √3(3 + 2) = √15
= 3.87 BM

Thanks
jitender