left[ Ti left( H_{2} O right)_{6} right]^{3+} | vedclass

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(A) The energy of the absorbed photon corresponds to the octahedral splitting energy $\Delta_{0}$.Using the formula $E = \frac{hc}{\lambda}$:Given $\l

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$4$

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(A) The energy of the absorbed photon corresponds to the octahedral splitting energy $\Delta_{0}$.Using the formula $E = \frac{hc}{\lambda}$:Given $\lambda = 498 \, nm = 498 	imes 10^{-9} \, m$,$h = 6.626 	imes 10^{-34} \, Js$,and $c = 3 	imes 10^{8} \, ms^{-1}$.$\Delta_{0} = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^{8}}{498 	imes 10^{-9}} \, J$$\Delta_{0} = \frac{19.878 	imes 10^{-26}}{498 	imes 10^{-9}} \, J$$\Delta_{0} \approx 0.039915 	imes 10^{-17} \, J = 3.9915 	imes 10^{-19} \, J$.Rounding off to the nearest integer,we get $4 	imes 10^{-19} \, J$.

$I$. $H_2O$	$II$. $CO$
$III$. $NH_3$	$IV$. $I^{-}$
$V$. $F^{-}$

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