A metal complex having composition `Cr(NH_(3))_(4)CI_(2)Br` has been isolated in two forms A and B. The form A reacts with `AgNO_(3)` 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) of the isomer A.

Explanation

A metal complex having composition `Cr(NH_(3))_(4)CI_(2)` Br has been isolated in two forms `A` and `B` Form A reacts with `AgNO_(3)` to give a white precipitate readily soluble in dilute aqueous ammonia This is a test for chloride ion `B` gives a pale-yellow precipitate soluble in concentrated ammonia This is a test for bromide ion A is `[Cr(NH_(3))_(4)CIBr]CI`
`Cr(NH_(3))_(4)CIBr]CI +AgNo_(3)`
`rarrAgCI darr + [Cr(NH_(3))_(4)CIBr]^(o+)NO_(3)^(Θ)`
`AgCI + 2NH_(4)OHrarr [Ag(NH_(3))_(2)CI]+2H_(2)O`
`B is [Cr(NH_(3))_(4)CI_(2)]Br`
`Cr(NH_(3))_(4)CI_(2)]Br+AgNO_(3) rarrAgBrdarr+[Cr(NH_(3))_(4)CI_(2)NO_(3)^(Θ)`
`AgBr+ 2NH_(4)OH rarr[Ag(NH_(3))_(2)Br]+2H_(2)O`
In both `A` and `B`
The oxidation number of Cr is +3
Atomic number `=24`
`Cr=[Ar]3d^(5) 4s^(1)`
`Cr =[Ag] 3d^(3) 4s^(0)`
`d^(2)sp^(3)` hybridisation (octahedral) There are three unpaired electrons so hte complex is paramagnetic Spin magnetic moment
`mu =sqrt(n(n +2)) = sqrt(3(3+2)) = sqrt15 BM`
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