The correct order of the complexes [Co(NH_3)_ | vedclass

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(A) The energy of absorbed light is inversely proportional to the wavelength: $E = \frac{hc}{\lambda}$,which means $E \propto \frac{1}{\lambda}$.In al

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$D > A > B > C$

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(A) The energy of absorbed light is inversely proportional to the wavelength: $E = \frac{hc}{\lambda}$,which means $E \propto \frac{1}{\lambda}$.In all these complexes,the central metal ion is $Co^{3+}$.As the ligand field strength increases,the Crystal Field Splitting Energy ($CFSE$ or $\Delta_o$) increases.The spectrochemical series for the given ligands is: $CN^- > NH_3 > H_2O > Cl^-$.Therefore,the order of $CFSE$ is: $[Co(CN)_6]^{3-} > [Co(NH_3)_6]^{3+} > [Co(NH_3)_5(H_2O)]^{3+} > [CoCl(NH_3)_5]^{2+}$,which corresponds to $C > B > A > D$.Since $E \propto \frac{1}{\lambda}$,the order of wavelength $(\lambda)$ absorbed is the inverse of the $CFSE$ order: $D > A > B > C$.

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

List-$I$ (Coordination entity)	List-$II$ (Wavelength of light absorbed in $nm$)
$A$. $[CoCl(NH_3)_5]^{2+}$	$I$. $310$
$B$. $[Co(NH_3)_6]^{3+}$	$II$. $475$
$C$. $[Co(CN)_6]^{3-}$	$III$. $535$
$D$. $[Cu(H_2O)_4]^{2+}$	$IV$. $600$

The correct order of the complexes $[Co(NH_3)_5(H_2O)]^{3+}$ $(A)$,$[Co(NH_3)_6]^{3+}$ $(B)$,$[Co(CN)_6]^{3-}$ $(C)$,and $[CoCl(NH_3)_5]^{2+}$ $(D)$ in terms of the wavelength of light absorbed is:

Similar Questions

Write the correct order of wavelength of absorption in the visible region for the following complexes:

Number of complexes with an even number of electrons in $t_{2g}$ orbitals is -
$[Fe(H_2O)_6]^{2+}, [Co(H_2O)_6]^{2+}, [Co(H_2O)_6]^{3+}, [Cu(H_2O)_6]^{2+}, [Cr(H_2O)_6]^{2+}$

Arrange the following ligands in the order of increasing field strength.
$I$. $H_2O$ $II$. $CO$
$III$. $NH_3$ $IV$. $I^{-}$
$V$. $F^{-}$

In which case is a $d^7$ complex species more stable?

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