Given the maximum absorption for \[d - d\]transition in ${[Ti{({H_2}O)_6}]^{3 + }}$occurs at $20300c{m^{ - 1}}$, predict where the peak $(d - dtransition)$will occur for ${[Ti{(CN)_6}]^{3 - }}$and ${[Ti{(Cl)_6}]^{3 - }}$respectively:
$
A)23000c{m^{ - 1}},32000c{m^{ - 1}} \\
B)18000c{m^{ - 1}},21300c{m^{ - 1}} \\
C)23000c{m^{ - 1}},17300c{m^{ - 1}} \\
D)18000C{M^{ - 1}},17300c{m^{ - 1}} \\
$
Answer
277.8k+ views
Hint: In a $d - d$ transition, an electron in $d$ orbital on the metal is excited by a photon to another $d$ orbital of higher energy. In complexes of the transition metals, the $d$ orbitals do not all have the same energy. In centrosymmetric complexes, $d - d$ transitions are forbidden by the Laporte rule.
Complete answer:
A spectrochemical series is a list of ligands ordered on ligand strength and a list of metal ions based on oxidation number, group and its identity. In crystal field theory, ligands modify the difference in energy between the $d$ orbitals called the ligand-field splitting parameter for ligands or the crystal-field splitting parameter, which is mainly reflected in differences in color of similar metal-ligand complexes.
According to spectrochemical series:
$Cl < {H_2}O < C{N^ - }$.
So, as the ligand field increases wave number also increases.
So, the correct answer is $A)23000c{m^{ - 1}},32000c{m^{ - 1}}$
Additional information:
The order of the spectrochemical series can be derived from the understanding that ligands are frequently classified by their donor or acceptor abilities. Some, like $N{H_3}$, are $\sigma $bond donors only, with no orbitals of appropriate symmetry for$\pi $bonding interactions. Bonding by these ligands to metals is relatively simple, using only the$\sigma $ bonds to create relatively weak interactions. Another example of a $\sigma $bonding ligand would be ethylenediamine; however, ethylenediamine has a stronger effect than ammonia, generating a larger ligand field split.
Note:
The metal ions can also be arranged in order of increasing $\Delta $, and this order is largely independent of the identity of the ligand. In general, it is not possible to say whether a given ligand will exert a strong field or a weak field on a given metal ion. However, when we consider the metal ion, $\Delta $ increases with increasing oxidation number, and $\Delta $ increases down a group.
Complete answer:
A spectrochemical series is a list of ligands ordered on ligand strength and a list of metal ions based on oxidation number, group and its identity. In crystal field theory, ligands modify the difference in energy between the $d$ orbitals called the ligand-field splitting parameter for ligands or the crystal-field splitting parameter, which is mainly reflected in differences in color of similar metal-ligand complexes.
According to spectrochemical series:
$Cl < {H_2}O < C{N^ - }$.
So, as the ligand field increases wave number also increases.
So, the correct answer is $A)23000c{m^{ - 1}},32000c{m^{ - 1}}$
Additional information:
The order of the spectrochemical series can be derived from the understanding that ligands are frequently classified by their donor or acceptor abilities. Some, like $N{H_3}$, are $\sigma $bond donors only, with no orbitals of appropriate symmetry for$\pi $bonding interactions. Bonding by these ligands to metals is relatively simple, using only the$\sigma $ bonds to create relatively weak interactions. Another example of a $\sigma $bonding ligand would be ethylenediamine; however, ethylenediamine has a stronger effect than ammonia, generating a larger ligand field split.
Note:
The metal ions can also be arranged in order of increasing $\Delta $, and this order is largely independent of the identity of the ligand. In general, it is not possible to say whether a given ligand will exert a strong field or a weak field on a given metal ion. However, when we consider the metal ion, $\Delta $ increases with increasing oxidation number, and $\Delta $ increases down a group.
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