Arrange the following complexes in increasing order of number of moles of silver chloride precipitated when one mole of each is reacted with excess of silver nitrate.
(A) \( [Co(NH_3)_4Cl_2]Cl \)
(B) \( [Co(NH_3)_5Cl]Cl_2 \)
(C) \( [Co(NH_3)_3Cl_3] \)
(D) \( [Co(NH_3)_6]Cl_3 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C), (A), (B), (D)
In a complex, the chloride ions inside the coordination sphere are bound to the metal and do not ionise. Only the chloride ions outside the sphere (the counter ions) are free in solution and are precipitated by silver nitrate, one mole of AgCl per mole of \( \ce{Cl^{-}} \):
\( \displaystyle \ce{Ag^+ + Cl^- -> AgCl v} \)
(A) \( \ce{[Co(NH3)4Cl2]Cl -> [Co(NH3)4Cl2]^+ + Cl^-} \): 1 free \( \ce{Cl^{-}} \), so 1 mol AgCl.
(B) \( \ce{[Co(NH3)5Cl]Cl2 -> [Co(NH3)5Cl]^2+ + 2Cl^-} \): 2 free \( \ce{Cl^{-}} \), so 2 mol AgCl.
(C) \( \ce{[Co(NH3)3Cl3]} \) is neutral; all three Cl are in the coordination sphere, so 0 mol AgCl.
(D) \( \ce{[Co(NH3)6]Cl3 -> [Co(NH3)6]^3+ + 3Cl^-} \): 3 free \( \ce{Cl^{-}} \), so 3 mol AgCl.
Shortcut: The neutral complex (C) gives no precipitate and \( \ce{[Co(NH3)6]Cl3} \) (D) gives the most; only Option (1) starts with (C) and ends with (D).
NCERT (2026-27 edition):
Werner’s Theory of Coordination Compounds – “… In a series of compounds of cobalt(III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess silver nitrate solution in cold but some remained in solution.”
| 1 mol CoCl3.6NH3 (Yellow) | gave | 3 mol AgCl |
| 1 mol CoCl3.5NH3 (Purple) | gave | 2 mol AgCl |
| 1 mol CoCl3.4NH3 (Green) | gave | 1 mol AgCl |
| 1 mol CoCl3.4NH3 (Violet) | gave | 1 mol AgCl |
“… the compounds are formulated as shown in Table 5.1, where the atoms within the square brackets form a single entity which does not dissociate under the reaction conditions.”