The correct answer is (A): Ce(At. No. 58)
Explanation:
Ce → [Xe]4f1 5d1 6s2
Ce+4 → [xe] 4f° 5d° 6s°
Cerium in +4 oxidation state acquires inert gas configuration.
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: