The electronic configuration 4f2 corresponds to the configuration of the Pr3+ ion, which is the ionized form of the element praseodymium (Pr). When an atom loses three electrons to become Pr3+, its electronic configuration changes. The loss of three electrons from the neutral praseodymium atom's electronic configuration of 4f35d06s2 results in the 4f2 configuration for the Pr3+ ion.
The correct answer is option (B) : Pr3+
Let α,β be the roots of the equation, ax2+bx+c=0.a,b,c are real and sn=αn+βn and \(\begin{vmatrix}3 &1+s_1 &1+s_2\\1+s_1&1+s_2 &1+s_3\\1+s_2&1+s_3 &1+s_4\end{vmatrix}=\frac{k(a+b+c)^2}{a^4}\) then k=
Lanthanoids are at the top of these two-row, while actinoids are at the bottom row.
Lanthanoids are inclusive of 14 elements, with atomic numbers 58-71:
These elements are also called rare earth elements. They are found naturally on the earth, and they're all radioactively stable except promethium, which is radioactive. A trend is one of the interesting properties of the lanthanoid elements, called lanthanide contraction.