In the compound \( \ce{PdCl2.4NH3} \), palladium shows a secondary valence of 4. An aqueous solution of this compound is treated with excess silver nitrate solution in the cold. Which of the following equations correctly represents the reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \ce{[Pd(NH3)4]Cl2 + 2AgNO3} \xrightarrow{\text{cold}} \ce{[Pd(NH3)4](NO3)2 + 2AgCl v} \)
By Werner’s theory, the secondary valence is the number of groups bound directly to the metal; these groups are non-ionisable. The primary valences are ionisable and are satisfied by negative ions, which stay outside the bracket and can be precipitated.
Secondary valence 4: the four NH3 molecules are bound to Pd, giving the entity \( \ce{[Pd(NH3)4]^2+} \).
Both Cl− ions satisfy the primary valence (2) of Pd(II), so they lie outside the bracket: \( \ce{[Pd(NH3)4]Cl2} \).
With excess AgNO3, both ionisable chloride ions precipitate, so 1 mol of the compound gives 2 mol of AgCl, while the entity \( \ce{[Pd(NH3)4]^2+} \) stays in solution.
\( \displaystyle \ce{[Pd(NH3)4]Cl2 + 2AgNO3} \xrightarrow{\text{cold}} \ce{[Pd(NH3)4](NO3)2 + 2AgCl v} \)
Why the other options are incorrect
Option (2) \( \ce{[PdCl2(NH3)4] + 2AgNO3 -> [Pd(NH3)4](NO3)2 + 2AgCl} \) — Incorrect. This puts six groups on Pd (secondary valence 6, not 4). Chloride ions inside the bracket are non-ionisable, so they could not be precipitated as AgCl.
Option (3) \( \ce{[Pd(NH3)4]Cl2 + AgNO3 -> [Pd(NH3)4](NO3)Cl + AgCl} \) — Incorrect. Both chloride ions are outside the bracket and ionisable. With excess AgNO3 both are precipitated, not just one.
Option (4) \( \ce{PdCl2.4NH3 + 2AgNO3 -> Pd(NO3)2 + 4NH3 + 2AgCl} \) — Incorrect.
This treats the compound like a double salt that breaks up completely into simple ions.
In a complex the groups inside the bracket form a single entity that does not dissociate, so NH3 is not set free.
NCERT (2026-27 edition):
5.1 Werner’s Theory of Coordination Compounds – “2. The primary valences are normally ionisable and are satisfied by negative ions.”
“3. The secondary valences are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal.”
“…the atoms within the square brackets form a single entity which does not dissociate under the reaction conditions. Werner proposed the term secondary valence for the number of groups bound directly to the metal ion…”
Example 5.1 – “On the basis of the following observations made with aqueous solutions, assign secondary valences to metals in the following compounds:” … “(i) PdCl2.4NH3 … 2” [moles of AgCl precipitated per mole of the compound with excess AgNO3] … Solution: “(i) Secondary 4”