Step 1: Write the electronic configurations.
\(_{63}\)Eu\(^{2+}\) - [Xe] 4f\(^7\) 6s\(^0\)
\(_{64}\)Gd\(^{3+}\) - [Xe] 4f\(^7\) 5d\(^0\) 6s\(^0\)
\(_{63}\)Eu\(^{3+}\) - [Xe] 4f\(^6\) 6s\(^0\)
\(_{65}\)Tb\(^{3+}\) - [Xe] 4f\(^8\) 6s\(^0\)
\(_{62}\)Sm\(^{2+}\) - [Xe] 4f\(^6\) 6s\(^0\)
Step 2: Identify the ions with 4f\(^7\) configuration.
From the electronic configurations, we can see that Eu\(^{2+}\) and Gd\(^{3+}\) have 4f\(^7\) configurations.
Step 3: Select the correct option.
Therefore, the correct answer is (A) and (B) only.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: