1. Electronic configuration of cerium

Cerium has atomic number $58$.

Its approximate configuration is: $\mathrm{Ce}: [\mathrm{Xe}]\,4f^1 5d^1 6s^2$

For $\mathrm{Ce}^{4+}$, four electrons are removed, giving: $\mathrm{Ce}^{4+} = [\mathrm{Xe}]$

So $\mathrm{Ce}^{4+}$ has a noble gas configuration.

2. Implication of noble gas configuration

A species with noble gas configuration is especially stable. Therefore, $\mathrm{Ce}^{4+}$ will tend to remain as it is and resist losing electrons further.

If it reacts, it will more readily gain an electron to become $\mathrm{Ce}^{3+}$ rather than lose another electron.

That means: $\mathrm{Ce}^{4+} + e^- \rightarrow \mathrm{Ce}^{3+}$

So $\mathrm{Ce}^{4+}$ gets reduced, and hence it causes oxidation of another species.

Therefore, it behaves as an oxidizing agent.

3. Check each option

• A: It will not prefer to undergo redox reactions.
  Incorrect. $\mathrm{Ce}^{4+}$ is actually a well-known oxidizing ion.

• B: It will prefer to gain electron and act as an oxidizing agent.
  Correct. Because of its stable noble gas configuration tendency, it accepts an electron: $\mathrm{Ce}^{4+} + e^- \rightarrow \mathrm{Ce}^{3+}$

• C: It will prefer to give away an electron and behave as reducing agent.
  Incorrect. A highly charged ion like $\mathrm{Ce}^{4+}$ does not prefer to lose another electron.

• D: It acts as both, oxidizing and reducing agent.
  Incorrect in this context. The dominant behavior of $\mathrm{Ce}^{4+}$ is as an oxidizing agent.

4. Final answer

The correct option is: $\boxed{\text{B}}$