In which of the following oxidation state Cerium achieves the noble gas configuration?a)+5b)+2c)+7d)+4Correct answer is option 'D'. Can you explain this answer?

Question

Answer

The reason for this behaviour is a result of the stability of half filled, empty or fulfilled F orbitals that these elements achieve in these Oxidation State.Thats it.

Answer

Any atom exhibit oxidation state in which it is found to be stable. It is second element of Lanthanide series having atomic number 58. General electronic configuration is
[Xe] 4f¹ 5d¹ 6s².
Lanthanide are very less electronegative. Their electronegative value is nearly equal to s-block elements. So they can lose electrons to form cation. When Cerium loses 4 e-, it will acquire fully field electronic configuration of Xenon (2,8,18,18,8). As fully filed or half filed electronic configuration possesses extra stability.
Therefore, Cerium exhibits +4 oxidation state.

Answer

To determine the oxidation state in which Cerium achieves the noble gas configuration, we need to understand the electronic configuration of Cerium and its tendency to gain or lose electrons.

Electronic Configuration of Cerium:
The atomic number of Cerium is 58, which means it has 58 electrons. The electron configuration of Cerium can be written as [Xe]4f^15d^16s^2, where [Xe] represents the noble gas configuration of Xenon.

Noble Gas Configuration:
The noble gas configuration refers to the electron configuration of the nearest noble gas element. In the case of Cerium, the nearest noble gas is Xenon, which has an electron configuration of [Xe]6s^24f^145d^10.

Determining the Oxidation State:
To achieve noble gas configuration, Cerium will either lose or gain electrons to match the electron configuration of Xenon.

Cerium can lose electrons from its outermost shell, which is the 6s orbital, or gain electrons to fill its 5d and 4f orbitals.

Given the options, we can analyze each oxidation state:

a) Oxidation state of 5:
If Cerium has an oxidation state of +5, it would lose all the electrons from its 6s and 4f orbitals. The resulting electron configuration would be [Xe]4f^0, which does not match the noble gas configuration of Xenon. Therefore, option (a) is incorrect.

b) Oxidation state of 2:
If Cerium has an oxidation state of +2, it would lose only the two electrons from its 6s orbital. The resulting electron configuration would be [Xe]4f^15d^0, which does not match the noble gas configuration of Xenon. Therefore, option (b) is incorrect.

c) Oxidation state of 7:
If Cerium has an oxidation state of +7, it would gain electrons to fill its 5d and 4f orbitals. The resulting electron configuration would be [Xe]4f^26s^25d^1, which does not match the noble gas configuration of Xenon. Therefore, option (c) is incorrect.

d) Oxidation state of 4:
If Cerium has an oxidation state of +4, it would lose all the electrons from its 6s orbital and one electron from its 4f orbital. The resulting electron configuration would be [Xe]4f^05d^0, which matches the noble gas configuration of Xenon. Therefore, option (d) is correct.

Therefore, Cerium achieves the noble gas configuration in its +4 oxidation state.

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Cerium can show the oxidation state of +4 because

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In which of the following oxidation states La achieve the noble gas configuration?

Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statement about cerium is incorrect -[AIEEE-04]

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