NEET · Chemistry · STD 12 - 4. d and f- block elements
Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because :
- A After losing one more electron, it acquires \(4 f^{14}\) electronic configuration.
- B Its nearest inert gas is Radon.
- C Its atomic number is 61
- D After losing one more electron, it acquires \(4 f^0\) electronic configuration.
Answer & Solution
Correct Answer
(D) After losing one more electron, it acquires \(4 f^0\) electronic configuration.
Step-by-step Solution
Detailed explanation
(D) After losing one more electron, it acquires \(4 f^0\) electronic configuration.
The atomic number of Cerium (Ce) is 58.
The electronic configuration of Ce is \([X e] 4 f^1 5 d^1 6 s^2\).
In +3 oxidation state, the electronic configuration of \(Ce ^{3+}\) is \([X e] 4 f^1\).
By losing one more electron, it forms \(Ce ^{4+}\) ion.
The electronic configuration of \(C e^{4+}\) is \([X e] 4 f^0\), which is a highly stable noble gas configuration.
Thus, Cerium shows +4 oxidation state because after losing one more electron from +3 state, it acquires \(4 f^0\) electronic configuration.
The atomic number of Cerium (Ce) is 58.
The electronic configuration of Ce is \([X e] 4 f^1 5 d^1 6 s^2\).
In +3 oxidation state, the electronic configuration of \(Ce ^{3+}\) is \([X e] 4 f^1\).
By losing one more electron, it forms \(Ce ^{4+}\) ion.
The electronic configuration of \(C e^{4+}\) is \([X e] 4 f^0\), which is a highly stable noble gas configuration.
Thus, Cerium shows +4 oxidation state because after losing one more electron from +3 state, it acquires \(4 f^0\) electronic configuration.
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