Jitender Pal
Last Activity: 10 Years ago
Hello Student,
Please find the answer to your question
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
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