The order of absorption wavelength in the vis | vedclass

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Question

(A) The energy of absorption $(E)$ is inversely proportional to the wavelength of absorption $(\lambda)$,i.e.,$E = \frac{hc}{\lambda}$.According to th

Vedclass · Accepted Answer

$[Ni(NO_2)_6]^{4-} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+}$

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(A) The energy of absorption $(E)$ is inversely proportional to the wavelength of absorption $(\lambda)$,i.e.,$E = \frac{hc}{\lambda}$.According to the spectrochemical series,the field strength of ligands follows the order: $H_2O Stronger ligands cause a larger crystal field splitting energy $(\Delta_o)$,which means they absorb light of higher energy (shorter wavelength).Conversely,weaker ligands cause smaller splitting,absorbing light of lower energy (longer wavelength).Therefore,the order of absorption wavelength $(\lambda)$ is the reverse of the order of ligand field strength: $[Ni(NO_2)_6]^{4-} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+}$.

The order of absorption wavelength in the visible region for the following complexes is:

Similar Questions

What is the electronic arrangement of a metal atom/ion in an octahedral complex with $d^4$ configuration,if $\Delta_{0} < P$ (pairing energy)?

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

The crystal field stabilization energy of $[Co(NH_3)_6]Cl_3$ is $......... \Delta_0$

In the octahedral crystal field,the correct order of splitting for $I^{-}, H_2O, NH_3$ and $CN^{-}$ ligands is:

The values of the crystal field stabilization energies for a high spin $d^{6}$ metal ion in octahedral and tetrahedral fields,respectively,are :

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