If [Cu(H_{2}O)_{4}]^{2+} absorbs light of wav | vedclass

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(A) The energy of the absorbed photon for the $d-d$ transition in the tetrahedral complex is given by $\Delta_{t} = \frac{hc}{\lambda}$.Substituting t

Vedclass · Accepted Answer

$766$

Vedclass · Answer

(A) The energy of the absorbed photon for the $d-d$ transition in the tetrahedral complex is given by $\Delta_{t} = \frac{hc}{\lambda}$.Substituting the given values: $\Delta_{t} = \frac{6.63 	imes 10^{-34} 	imes 3.08 	imes 10^{8}}{600 	imes 10^{-9}} \ J$.$\Delta_{t} = \frac{20.4204 	imes 10^{-26}}{600 	imes 10^{-9}} = 0.034034 	imes 10^{-17} \ J = 340.34 	imes 10^{-21} \ J$.The relationship between octahedral splitting energy $(\Delta_{o})$ and tetrahedral splitting energy $(\Delta_{t})$ is $\Delta_{o} = \frac{9}{4} \Delta_{t}$.$\Delta_{o} = \frac{9}{4} 	imes 340.34 	imes 10^{-21} \ J = 765.765 	imes 10^{-21} \ J$.Rounding to the nearest integer,we get $766 	imes 10^{-21} \ J$.

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