
Explain the violet colour of \[{\left[ {Ti{{\left( {{H_2}O} \right)}_6}} \right]^{\,3 + }}\] complex on basis of the crystal field theory$?$
Answer
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Hint: The complex given is a coordination complex which has a $3\,d$ transition metal $Ti$ as the central metal atom. The central $Ti$ has an oxidation state of $\left( { + 3} \right)$ and is surrounded by six ${H_2}O$ molecules which are arranged in an octahedral geometry. Calculate the number of electrons present in the complex and interpret the colour.
Complete step by step solution:
The complex given is \[{\left[ {Ti{{\left( {{H_2}O} \right)}_6}} \right]^{\,3 + }}\], which is a coordination complex.
A coordination complex is a complex which consists of a central atom or ion, which is generally metallic and is called the coordination centre, and a surrounding array of bound molecules or ions, that are known as ligands or complexing agents.
In the given complex, $Ti$ is the central atom or coordination centre and six ${H_2}O$ molecules surround the central metal atom in an octahedral geometry.
Now, ${H_2}O$ molecules do not carry any charge and hence the oxidation state of $Ti$ is $\left( { + 3} \right)$.
The electronic configuration of $Ti$ is $\left[ {Ar} \right]\,3{d^2}\,4{s^2}$, after losing three electrons the configuration becomes $\left[ {Ar} \right]\,3{d^1}$.
Due to the presence of one unpaired electron $d - d$ transition occurs. The unpaired electron is excited from the ${t_{2g}}$ to ${e_g}$ level by absorbing yellow light. Hence the complex appears violet, the complementary colour of yellow.
Additional Information: A coordination complex appears coloured when the transition corresponding to the colour is Laporte allowed and Spin allowed. For being Laporte allowed $\Delta \,l\,\, = \, \pm 1$, and for being spin allowed $\Delta S\, = \,0$.
Note:
You must have a basic knowledge about the transition metals and their atomic numbers for doing this kind of question. Also you must know whether the given ligand is neutral or negatively charged or positively charged for calculating the oxidation state of the metal correctly.
Complete step by step solution:
The complex given is \[{\left[ {Ti{{\left( {{H_2}O} \right)}_6}} \right]^{\,3 + }}\], which is a coordination complex.
A coordination complex is a complex which consists of a central atom or ion, which is generally metallic and is called the coordination centre, and a surrounding array of bound molecules or ions, that are known as ligands or complexing agents.
In the given complex, $Ti$ is the central atom or coordination centre and six ${H_2}O$ molecules surround the central metal atom in an octahedral geometry.
Now, ${H_2}O$ molecules do not carry any charge and hence the oxidation state of $Ti$ is $\left( { + 3} \right)$.
The electronic configuration of $Ti$ is $\left[ {Ar} \right]\,3{d^2}\,4{s^2}$, after losing three electrons the configuration becomes $\left[ {Ar} \right]\,3{d^1}$.
Due to the presence of one unpaired electron $d - d$ transition occurs. The unpaired electron is excited from the ${t_{2g}}$ to ${e_g}$ level by absorbing yellow light. Hence the complex appears violet, the complementary colour of yellow.
Additional Information: A coordination complex appears coloured when the transition corresponding to the colour is Laporte allowed and Spin allowed. For being Laporte allowed $\Delta \,l\,\, = \, \pm 1$, and for being spin allowed $\Delta S\, = \,0$.
Note:
You must have a basic knowledge about the transition metals and their atomic numbers for doing this kind of question. Also you must know whether the given ligand is neutral or negatively charged or positively charged for calculating the oxidation state of the metal correctly.
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