
To solve this question, we need to match the complex ions from List I with their correct electronic configurations in List II.
We'll examine each complex ion and determine its electronic configuration based on the oxidation state of the central metal and the coordination chemistry.
Based on the analysis, the correct matching is:
A-II, B-III, C-IV, D-I
Thus, the correct answer is:
A-II, B-III, C-IV, D-I
$[\text{Cr(H}_2\text{O)}_6]^{3+} \text{ contains } \text{Cr}^{3+}: [\text{Ar}]3d^3 \cdot t_{2g}^3 e_g^0$
$[\text{Fe(H}_2\text{O)}_6]^{3+} \text{ contains } \text{Fe}^{3+}: [\text{Ar}]3d^5 \cdot t_{2g}^3 e_g^2$
$[\text{Ni(H}_2\text{O)}_6]^{2+} \text{ contains } \text{Ni}^{2+}: [\text{Ar}]3d^8 \cdot t_{2g}^6 e_g^2$
$[\text{V(H}_2\text{O)}_6]^{3+} \text{ contains } \text{V}^{3+}: [\text{Ar}]3d^2 \cdot t_{2g}^2 e_g^0$
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Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 
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