
The color of \[F{e^{2 + }}\] the ion is:
A.blue
B.light green
C.red
D.none of the above
Answer
563.4k+ views
Hint:Transition metals form a complex with a different ligand using its d-orbital electrons. They can combine with a different number of ligands and form different kinds of geometry. The color of the ions of transition metals is found because of \[d - d\] transitions and charge transfer.
Complete step by step answer:
When a metal forms complex, due to different extent of interaction with ligands the d orbitals are divided into several energy states. The \[d - d\] transition of the transition metals is the excitation of electrons by absorbing energy from lower energy levels to higher energy levels. After these transitions, electrons immediately come back with the emission of some amount of radiation, for which the complex shows color. This color is the complementary color of the light absorbed in excitation.
Now if there is no mention of ligand, then assume the ligand is water. That is that a particular ion of transition metal is hydrated.
Now in the case of \[F{e^{2 + }}\] ion, the number of the electrons in d orbital is 6. With the configuration \[t_{2g}^4e_g^2\] . Now according to the selection rule, the \[d - d\] transitions are only allowed when spin multiplicity remains constant. In this case, the spin multiplicity before and after the transition is the same. Therefore, it shows color due to the \[d - d\] transition.
Now, during the \[d - d\] - transition the metal absorbs a particular wavelength of light and emits the complementary color of that light. In case of \[F{e^{2 + }}\] during the \[d - d\] - the transition has absorbed the light of color red, as a result, it emits its complementary color light green. That is why the color of \[F{e^{2 + }}\] is light green in color.
So, the correct answer is B.
Note:Charge transfer transition is another reason behind the color of the transition metal complex. There are two kinds of charge transfer.
1.LMCT (ligand to metal charge transfer): when metals are in high oxidation states charge from metal to ligands transfers. As a result, it shows color. For example, \[{\left[ {Mn{0_4}} \right]^{ - 2}}\] shows an intense violet color. Where oxygen to Mn+7 charge transfer takes place.
2.MLCT(metal to ligand charge transfer): when metals are in low oxidation state, charge transfer from metal to ligand takes place. For example, \[\left[ {W{{(CO)}_4}(Phen)} \right]\]
3.Metal to metal charge transfer: Metal to metal charge transfer happens when metal is present in a complex with two different oxidation states. The transition of charge takes place from lower oxidation to a higher oxidation state. In the case of Prussian blue, charge transfer takes place from Fe(II) to Fe(III).
Complete step by step answer:
When a metal forms complex, due to different extent of interaction with ligands the d orbitals are divided into several energy states. The \[d - d\] transition of the transition metals is the excitation of electrons by absorbing energy from lower energy levels to higher energy levels. After these transitions, electrons immediately come back with the emission of some amount of radiation, for which the complex shows color. This color is the complementary color of the light absorbed in excitation.
Now if there is no mention of ligand, then assume the ligand is water. That is that a particular ion of transition metal is hydrated.
Now in the case of \[F{e^{2 + }}\] ion, the number of the electrons in d orbital is 6. With the configuration \[t_{2g}^4e_g^2\] . Now according to the selection rule, the \[d - d\] transitions are only allowed when spin multiplicity remains constant. In this case, the spin multiplicity before and after the transition is the same. Therefore, it shows color due to the \[d - d\] transition.
Now, during the \[d - d\] - transition the metal absorbs a particular wavelength of light and emits the complementary color of that light. In case of \[F{e^{2 + }}\] during the \[d - d\] - the transition has absorbed the light of color red, as a result, it emits its complementary color light green. That is why the color of \[F{e^{2 + }}\] is light green in color.
So, the correct answer is B.
Note:Charge transfer transition is another reason behind the color of the transition metal complex. There are two kinds of charge transfer.
1.LMCT (ligand to metal charge transfer): when metals are in high oxidation states charge from metal to ligands transfers. As a result, it shows color. For example, \[{\left[ {Mn{0_4}} \right]^{ - 2}}\] shows an intense violet color. Where oxygen to Mn+7 charge transfer takes place.
2.MLCT(metal to ligand charge transfer): when metals are in low oxidation state, charge transfer from metal to ligand takes place. For example, \[\left[ {W{{(CO)}_4}(Phen)} \right]\]
3.Metal to metal charge transfer: Metal to metal charge transfer happens when metal is present in a complex with two different oxidation states. The transition of charge takes place from lower oxidation to a higher oxidation state. In the case of Prussian blue, charge transfer takes place from Fe(II) to Fe(III).
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