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The enthalpy of hydration of Cr^(+2) is ...

The enthalpy of hydration of `Cr^(+2)` is `-460` kcal `mo1^(-1)` In the absence of `CFSE` the value for `DeltaH=-424` kcal `mo1^(-1)` What is the value of `Delta_(0)` for `[Cr(H_(2)O)_(6))]^(2+)` .

A

`60kcal "mole"^(-1)`

B

`-60kcal "mole"^(-1)`

C

`25.7 kcal "mole"^(-1)`

D

`-25.7kcal "mole"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
a

`Cr = 3d^(5)4s^(1) , Cr^(+2) = 3d^(4)` Structure is `(t_(2g)3e_(g)1)`
`CFSE (-0.4 xx 3 + 0.6)Delta_(0) = - 0.6 Delta_(0))`
`- 0.6 Delta_(0) = (- 460) - (-424) = - 36k cal mo1^(-1)`
`Delta_(0) = (-36)/(-0.6) = 60 kcal "mole"^(-1)`
` 60 xx 350cm^(-1) = 21000cm^(1-)`
`(1 kcal mo1^(-1) = 350cm^(-1))` .
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