Cerium IV has a noble gas configuration. Which of the following is the correct statement about it?

Cerium(IV), i.e., $$Ce^{4+}$$, has the electronic configuration $$[Xe]$$, which is a noble gas configuration.

Electronic configuration of Cerium: Cerium (Ce) has atomic number 58. Its ground state configuration is $$[Xe] 4f^1 5d^1 6s^2$$.

When Ce loses 4 electrons to form $$Ce^{4+}$$, it achieves the configuration $$[Xe]$$ — a noble gas configuration.

Predicting the chemical behavior of $$Ce^{4+}$$: Although $$Ce^{4+}$$ has achieved a stable noble gas configuration, the +4 oxidation state represents a very high charge density on the cerium ion. This makes $$Ce^{4+}$$ a strong oxidizing agent because:

$$\bullet$$ It has a strong tendency to gain an electron (get reduced) to form $$Ce^{3+}$$.

$$\bullet$$ The standard reduction potential $$E^\circ(Ce^{4+}/Ce^{3+}) = +1.74$$ V is very high, confirming its strong oxidizing power.

$$\bullet$$ In aqueous solution, $$Ce^{4+}$$ readily accepts electrons from reducing agents.

Evaluate the options: Option A: It will prefer to gain electron and act as an oxidizing agent — Correct. $$Ce^{4+}$$ is a well-known strong oxidizing agent in analytical chemistry (ceric ammonium nitrate is widely used as an oxidant).

Option B: It will prefer to give away an electron and behave as reducing agent — Incorrect. $$Ce^{4+}$$ does not lose electrons easily.

Option C: It will not prefer to undergo redox reactions — Incorrect. $$Ce^{4+}$$ readily participates in redox reactions as an oxidant.

Option D: It acts as both, oxidizing and reducing agent — Incorrect. It predominantly acts as an oxidizing agent.

The correct answer is Option A: It will prefer to gain electron and act as an oxidizing agent.