Q. 245.0( 3 Votes )
A metal complex h
(i) The isomer A = [Cr(NH3)4BrCl]Cl
The isomer B = [Cr(NH3)4Cl2]Br
(ii) Hybridization of Cr in both is d2sp3
(iii) 3. 87 BM
(i) Since, [Cr(NH3)4BrCl]Cl + AgNO3AgCl
WE know that AgCl is a white precipitate which is readily soluble in dilute aqueous ammonia we can tell that it is isomer A
Since, [Cr(NH3)4Cl2]Br + AgNO3AgBr
We know that AgBr gives a pale yellow precipitate which is soluble in concentrated ammonia we can tell that isomer B is the right answer.
(ii) Cr3+ is in 3d3 system; we can tell that its hybridization is d2sp3.
[NOTE:read about the different systems to know what falls in which system and its hybridization to learn this concept easily]
(iii) To find the magnetic moment:
Where, n = the number of unpaired electrons
In Cr3 + we have d3 outermost shell configuration =
1s2 2s2 2[6 3s2 3p6 3d5 4s1 [Ar]3d5 4s1 [Ar]3d3.
Moreover, therefore it has three unpaired electrons
Substitute 3 in the formula
= 3. 87 BM
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