
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
$[\text{Cr(H}_2\text{O)}_6]^{3+} \text{ contains } \text{Cr}^{3+}: [\text{Ar}]3d^3 \cdot t_{2g}^3 e_g^0$
$[\text{Fe(H}_2\text{O)}_6]^{3+} \text{ contains } \text{Fe}^{3+}: [\text{Ar}]3d^5 \cdot t_{2g}^3 e_g^2$
$[\text{Ni(H}_2\text{O)}_6]^{2+} \text{ contains } \text{Ni}^{2+}: [\text{Ar}]3d^8 \cdot t_{2g}^6 e_g^2$
$[\text{V(H}_2\text{O)}_6]^{3+} \text{ contains } \text{V}^{3+}: [\text{Ar}]3d^2 \cdot t_{2g}^2 e_g^0$
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| List-I Tetrahedral Complex | List-II Electronic configuration |
|---|---|
| (A) TiCl4 | (I) e2, t20 |
| (B) [FeO4]2- | (II) e4, t23 |
| (C) [FeCl4]- | (III) e0, t22 |
| (D) [CoCl4]2- | (IV) e2, t23 |
When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are ____.
Outermost electronic configurations of four elements A,B,C,D are given below:
(A) 3s2 (B) s23p1 (C) 3s23p3 (D) 3s23p4
The correct order of fist ionization enthalpy for them is:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)