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(1) [Cd(H(2)O)(4)]^(2+)+4NH(3)iff [Cd(...

(1) `[Cd(H_(2)O)_(4)]^(2+)+4NH_(3)iff` `[Cd(NH_(3))_(4)]^(2+) +4H_(2)O,"log"beta_(1)`
(2) `[Cd(H_(2)O)_(4)]^(2+)+2en iff` `[Cd(en)_(2)]^(2+)+4H_(2)O"log"beta_(2)`
`"log"beta_(2) gt "log"beta_(1),` is due to

A

Enthalpy change

B

Entropy change

C

Ring strain in the chelate

D

Large change in `Delta_(0)`

Text Solution

Verified by Experts

The correct Answer is:
b

Both the reactions involve Cd-N bonds and so no much change in enthalpy . In (1) four water molecules are replaced by four `NH_(3)` molecules . But in (2) four water molecules are replaced by two en molecules . This increase the entropy and is called as Entropy effect .
`DeltaG^(0) = DeltaH^(0) - T DeltaS^(0)`
`DeltaG^(0)= -2.303 RT "log" beta`
In (2) , `DeltaG^(0)` varies due to `DeltaS^(0)` and so the log `beta_(2)` has a higher value . For (1) , `DeltaS^(0) = -"ve"` and (2) `DeltaS^(0) = + "ve"`
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