IIT JAM Exam > IIT JAM Questions > Yellow colour of Ce4+ and blood red colour of...
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Yellow colour of Ce4+ and blood red colour of Sm2+ due to respectively:
- a)Charge transfer and f-f transition
- b)Charge transfer in both
- c)f-f transition in both
- d)f-f transition and change transfer
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Yellow colour of Ce4+ and blood red colour of Sm2+ due to respectively...
Explanation:
The yellow colour of Ce4+ and blood-red colour of Sm2+ is due to the absorption of visible light in the respective ions. The reason for the absorption of visible light in these ions is different.
Charge Transfer:
When an electron is transferred from one atom to another, it results in the formation of an ionic bond. The transfer of an electron from one atom to another results in the formation of two ions, one positively charged and the other negatively charged. This transfer of an electron is called a charge transfer. In Ce4+ ion, the yellow colour is due to the charge transfer from oxygen to cerium ion.
f-f Transition:
In rare earth ions, the f-orbitals are partially filled, and these electrons are responsible for the absorption of visible light. When an electron in a f-orbital absorbs a photon of visible light, it jumps to a higher energy level. This jump is called f-f transition. In Sm2+ ion, the blood-red colour is due to the f-f transition of electrons from the partially filled f-orbitals.
Both Charge Transfer and f-f Transition:
In the case of Ce4+ and Sm2+ ions, the absorption of visible light is due to both charge transfer and f-f transition. Ce4+ ion shows a yellow colour due to the charge transfer from oxygen to cerium ion and also due to the f-f transition of electrons in the partially filled f-orbitals. Sm2+ ion shows a blood-red colour due to the f-f transition of electrons in the partially filled f-orbitals and also due to the charge transfer from ligands to Sm2+ ion.
Conclusion:
Hence, the correct option is B, i.e., the yellow colour of Ce4+ and the blood-red colour of Sm2+ is due to charge transfer in both ions.
The yellow colour of Ce4+ and blood-red colour of Sm2+ is due to the absorption of visible light in the respective ions. The reason for the absorption of visible light in these ions is different.
Charge Transfer:
When an electron is transferred from one atom to another, it results in the formation of an ionic bond. The transfer of an electron from one atom to another results in the formation of two ions, one positively charged and the other negatively charged. This transfer of an electron is called a charge transfer. In Ce4+ ion, the yellow colour is due to the charge transfer from oxygen to cerium ion.
f-f Transition:
In rare earth ions, the f-orbitals are partially filled, and these electrons are responsible for the absorption of visible light. When an electron in a f-orbital absorbs a photon of visible light, it jumps to a higher energy level. This jump is called f-f transition. In Sm2+ ion, the blood-red colour is due to the f-f transition of electrons from the partially filled f-orbitals.
Both Charge Transfer and f-f Transition:
In the case of Ce4+ and Sm2+ ions, the absorption of visible light is due to both charge transfer and f-f transition. Ce4+ ion shows a yellow colour due to the charge transfer from oxygen to cerium ion and also due to the f-f transition of electrons in the partially filled f-orbitals. Sm2+ ion shows a blood-red colour due to the f-f transition of electrons in the partially filled f-orbitals and also due to the charge transfer from ligands to Sm2+ ion.
Conclusion:
Hence, the correct option is B, i.e., the yellow colour of Ce4+ and the blood-red colour of Sm2+ is due to charge transfer in both ions.
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Community Answer
Yellow colour of Ce4+ and blood red colour of Sm2+ due to respectively...
Ce4+ easily reduces to Ce3+ and for orbital gets 1 electron
So charge transfer is possible and it is yellow coloured.
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Yellow colour of Ce4+ and blood red colour of Sm2+ due to respectively:a)Charge transfer and f-f transitionb)Charge transfer in bothc)f-f transition in bothd)f-f transition and change transferCorrect answer is option 'B'. Can you explain this answer?
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