Chemistry Q63 – JEE-MAIN 2025

Home Chemistry Coordination CompoundsQuestion #352

Match List - I with List - II.

\begin{array}{lcl} List - I & List - II \\ Complex & Primary valency and Secondary valency \\ (A) [C o (e n)_{2} C l_{2}] C l & (I) 3 6 \\ (B) [P t (N H_{3})_{2} C l (N O_{2})] & (II) 3 4 \\ (C) H g [C o (S C N)_{4}] & (III) 2 6 \\ (D) [M g (E D T A)]^{2 -} & (IV) 2 4 \end{array}

Choose the correct answer from the options given below :

Clue 1 of 3: Werner's Theory of Coordination Compounds

Recall that primary valency corresponds to the oxidation state of the central metal ion, and secondary valency corresponds to its coordination number.

🥷

Ninja Strategy•Sequential Elimination

Calculate the primary and secondary valencies for complex (A) first. This immediately eliminates options that do not match (A) with (I). Then, calculate for (B) to confirm the correct option among the remaining choices.

🧠 Full Solution Path

Step 1: Calculate Primary and Secondary Valency for (A) and (B)✦ Active

For complex (A) $[C o (e n)_{2} C l_{2}] C l$ : The central metal is $C o$ . Ligands are $e n$ (neutral, bidentate) and $C l^{-}$ (monodentate, $- 1$ charge). The counter ion is $C l^{-}$ . Let the oxidation state of $C o$ be $x$ . $x + 2 (0) + 2 (- 1) + (- 1) = 0 ⟹ x = + 3$ . Primary valency is 3. Coordination number is $2 \times 2 (f o r e n) + 2 \times 1 (f o r C l) = 6$ . Secondary valency is 6. So, (A) matches (I).

For complex (B) $[P t (N H_{3})_{2} C l (N O_{2})]$ : The central metal is $P t$ . Ligands are $N H_{3}$ (neutral, monodentate), $C l^{-}$ (monodentate, $- 1$ charge), and $N O_{2}^{-}$ (monodentate, $- 1$ charge). Let the oxidation state of $P t$ be $x$ . $x + 2 (0) + 1 (- 1) + 1 (- 1) = 0 ⟹ x = + 2$ . Primary valency is 2. Coordination number is $2 \times 1 (f o r N H_{3}) + 1 \times 1 (f o r C l) + 1 \times 1 (f o r N O_{2}) = 4$ . Secondary valency is 4. So, (B) matches (IV).

Step 2: Calculate Primary and Secondary Valency for (C) and (D)○ Expand

For complex (C) $H g [C o (S C N)_{4}]$ : This is a complex salt. The complex ion is $[C o (S C N)_{4}]^{2 -}$ . The central metal is $C o$ . Ligand is $S C N^{-}$ (monodentate, $- 1$ charge). Let the oxidation state of $C o$ be $x$ . $x + 4 (- 1) = - 2 ⟹ x = + 2$ . Primary valency is 2. Coordination number is $4 \times 1 (f o r S C N) = 4$ . Secondary valency is 4. So, (C) matches (II).

For complex (D) $[M g (E D T A)]^{2 -}$ : The central metal is $M g$ . Ligand is $E D T A^{4 -}$ (hexadentate, $- 4$ charge). Let the oxidation state of $M g$ be $x$ . $x + 1 (- 4) = - 2 ⟹ x = + 2$ . Primary valency is 2. Coordination number is $1 \times 6 (f o r E D T A) = 6$ . Secondary valency is 6. So, (D) matches (III).

Step 3: Match the Complexes with their Valencies○ Expand

Based on the calculations: (A) $[C o (e n)_{2} C l_{2}] C l$ has primary valency 3 and secondary valency 6. This matches (I). (B) $[P t (N H_{3})_{2} C l (N O_{2})]$ has primary valency 2 and secondary valency 4. This matches (IV). (C) $H g [C o (S C N)_{4}]$ has primary valency 2 and secondary valency 4. This matches (II). (D) $[M g (E D T A)]^{2 -}$ has primary valency 2 and secondary valency 6. This matches (III). Therefore, the correct matching is (A)-(I), (B)-(IV), (C)-(II), (D)-(III), which corresponds to option 3.

💡 Teacher's Secret Hint

Double-check the denticity of each ligand, especially for polydentate ligands like 'en' and 'EDTA'.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 63 – JEE-MAIN 2025

Clue 1 of 3: Werner's Theory of Coordination Compounds

🧠 Full Solution Path