Experimental enthalpy of hydration of Cr 2 + is − 460 k cal mol − 1 . In the absence of crystal field splitting the value of enthalpy of hydration of Cr 2 + is − 436 k cal mol − 1 . (Use 1k cal mol -1

Question

Experimental enthalpy of hydration of Cr 2 + is − 460 k cal mol − 1 . In the absence of crystal field splitting the value of enthalpy of hydration of Cr 2 + is − 436 k cal mol − 1 . (Use 1k cal mol -1 = 350 cm -1 ) Find the value of ( 10 Δ 0 ​ ​ ) for [Cr(H 2 O) 6 ] 2+ in cm -1 :–

TestHub · Accepted Answer

The crystal field stabilization energy (CFSE) is calculated as:

$CFSE = - 460 - (- 436) = - 24 kcal mol^{- 1}$

This value can also be expressed in wavenumbers:

$- 24 kcal mol^{- 1} = - 24 \times 350 cm^{- 1}$

For $Cr^{2 +}$ , which has a $3 d^{4}$ electron configuration, the electron distribution in an octahedral field is $t_{2 g}^{1, 1, 1} e_{g}^{1, 0}$ .

The relationship between CFSE and the crystal field splitting energy ( $Δ_{0}$ ) is:

$- 0.6 Δ_{0} = - 24 \times 350$

Solving for $Δ_{0}$ :

$Δ_{0} = \frac{- 24 \times 350}{- 0.6} = 14000 cm^{- 1}$

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