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0.001 mol of Cr(NH(3))(5)(NO(3))(SO(4)) ...

0.001 mol of `Cr(NH_(3))_(5)(NO_(3))(SO_(4))` was passed through a cation exchanger the acid coming out of it reguired `20mL` of 0.1M NaoH for netralisation Hence the complex is
`[Cr(NH_(3))_(5)SO_(4)]NO_(3)`
(b) `[Cr(NH_(3))_(5)NO_(3)]SO_(4)`
(c ) `[Cr(NH_(3))_(5)](SO_(4))(NO_(3)`
(d) `[Co(NH_(3))_(5)Br]SO_(4)` .

Text Solution

Verified by Experts

The correct Answer is:
b
`M_(eq)` of acid `=M_(eq)` of Base `=20 xx 0.1 = 2`
Compounds is `Cr(NH_(3))_(5)(SO_(4))(0.001 mole)`
For A `[Cr(NH)_(3))_(5)(SO_(4))]NO_(3)rarrit` violate Werner' s theory Also `m_(eq)` of acid obtained `=1 xx 1 ne 2`
`implies(A)` is wrong
For `B` Acid obtained is `H_(2)SO_(4):M_(eqacid) = 1 xx 2 =2`
Also it does not violate Werner' s theory and hence Answer is `B` .
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