In which of the following does the metal show a secondary valency of…

The correct answer is Option (1) → (B), (C), (D) only

In Werner’s theory the primary valence (ionisable) is satisfied by counter ions, which are precipitated as AgCl by \( \ce{AgNO3} \); the secondary valence (non-ionisable) equals the coordination number. The chlorides that are not precipitated are inside the coordination sphere:

• (A) PdCl₂.4NH₃, 2 AgCl — Incorrect. Both \( \ce{Cl^{-}} \) are ionisable: \( \ce{[Pd(NH3)4]Cl2} \); secondary valence 4.

• (B) NiCl₂.6H₂O, 2 AgCl — Correct. \( \ce{[Ni(H2O)6]Cl2} \); six water ligands, secondary valence 6.

• (C) PtCl₄.2HCl, no AgCl — Correct. All six \( \ce{Cl^{-}} \) are ligands: \( \ce{H2[PtCl6]} \); secondary valence 6.

• (D) CoCl₃.4NH₃, 1 AgCl — Correct. \( \ce{[Co(NH3)4Cl2]Cl} \); 4 \( \ce{NH3} \) + 2 \( \ce{Cl^{-}} \) = secondary valence 6.

• (E) PtCl₂.2NH₃, no AgCl — Incorrect. \( \ce{[Pt(NH3)2Cl2]} \); secondary valence 4.

NCERT (2026-27 edition):
Werner’s Theory of Coordination Compounds – “… On the basis of the following observations made with aqueous solutions, assign secondary valences to metals in the following compounds:
Formula — Moles of AgCl precipitated per mole of the compounds with excess AgNO₃
(i) PdCl₂.4NH₃ — 2
(ii) NiCl₂.6H₂O — 2
(iii) PtCl₄.2HCl — 0
(iv) CoCl₃.4NH₃ — 1
(v) PtCl₂.2NH₃ — 0
Solution: (i) Secondary 4 (ii) Secondary 6 (iii) Secondary 6 (iv) Secondary 6 (v) Secondary 4”