A metal complex having composition `Cr(NH_(3))_(4)CI_(2)` Br has been isolated in two forms `A` and `B` The `A` reacts with `AgNO_(3)` to give a white precipitate readily soluble in dilute queous ammonia, whereas `B` gives a pale-yellow precipitate soluble in concentrated ammonia Write the formula of `A` and `B` and state the hybridisation of chromium in each. Calculate their magnetic moment (spin only value) .

Explanation

`Cr(NH_(3)_(4),Cl_(2)Br` have two forms A and B. `[Cr(NH_(3))_(4)BrCl]Cl, (d^(2)sp^(3)), [Cr(NH_(3))_(4)CI_(2)]`Br.`d^(2)sp^(3))`, 3.87 BM, 3.87 BMForm .A. gives white ppt. wth `AgNO_(3)` hence it must have chloride ion in form of noncomplex ion i.e. outside the complex sphere as<
`Cr(NH_(3))_(4)CIBr]Cl`
`underset(From A)[Cr(NH_(3))_(4)CIBr]Cl+AğNO_(3)tounderset(White ppt)(AgCI)downarrow(+)[Cr(NH_(3))_(4)CIBr]+NO_(3)`
These precipitates of AgCl are soluble in `NH_(4) OH` due to formation of complex salt
`underset(whiteppt)(AgCI)+2NH_(4)OH tounderset(complex salt)([Ag(NH_(4))_(2))CI]+2H_(2)O`
Similarly, form B gives pale yellow precipitate of AgBr which are sparingly soluble in `NH_(4)OH` Form .B. is `[Cr(NH_(3))_(4)CI_(2)]Br`
`AgNO_(3)+[Cr(NH_(3))_(4)C1_(2)]Br tounderset(pale yellow ppt) (AgBr)downarrow(+)+[Cr(NH_(3))_(4)Cl_(2)]^(+)+ NO_(3)^(-)`
`underset(pale yellow ppt)(AgBr)+2NH_(4)OHto [Ag(NH_(3))_(2)Br]+2H_(2)O`
In both complexes, chromium is present as central ion and its oxidation number is +3 So in these
`_(24) Cr = 1s^(2), 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5), 4s^(1)`
`Cr6(3+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
Number of ligands are six and `Cr^(3+)` shows `d^(2)sp^(3)` hybridization in both complexes A and B
Hence in it, number of unpaired electrons are 3. So magnetic moment `(mu) = sqrt(n(n + 2)).= sqrt(3(3+2)) = sqrt(15)` = 3.872 BM`