Read the passage carefully and answer the Questions.
Werner in 1898, propounded his theory of coordination compounds. In coordination compounds metals show two types of linkages (valences)-primary and secondary. The primary valences are normally ionisable and are satisfied by negative ions. 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 ions or group bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to the different coordination numbers. Both double salts as well as complexes are formed by the combination of two or more stable compounds in stoichiometric ratio. However, they differ in the fact that double salts such as carnallite, Mohr's salt, potash alum etc. dissociate into simple ions completely when dissolved in water. But complex ions do not dissociate completely into simple ions.
When 0.1 mol of \( CoCl_3(NH_3)_5 \) is treated with excess of \( AgNO_3 \), 0.2 mol of AgCl are obtained. The conductivity of the solution will correspond to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1:2 electrolyte
The chloride ions outside the coordination sphere are free in solution and are precipitated by silver nitrate; those bound to the metal are not. So the moles of AgCl per mole of compound give the number of free chloride ions. The same count decides the conductivity, which depends on the number of ions each formula unit gives: a 1:1 electrolyte gives 2 ions, a 1:2 electrolyte 3 ions and a 1:3 electrolyte 4 ions (one cation with one, two or three anions).
Here 0.1 mol of \( \ce{CoCl3(NH3)5} \) gives 0.2 mol of AgCl:
\( \displaystyle \frac{0.2\ \text{mol AgCl}}{0.1\ \text{mol complex}} = 2\ \text{free } \ce{Cl^-} \text{ per formula unit} \)
Cobalt is in the +3 state (x + 0 + 3(−1) = 0) with coordination number 6. Five \( \ce{NH3} \) molecules and the one chloride that is not precipitated fill the six positions, so the formula is \( \ce{[CoCl(NH3)5]Cl2} \):
\( \displaystyle \ce{[CoCl(NH3)5]Cl2 -> [CoCl(NH3)5]^2+ + 2Cl^-} \)
One complex cation and two chloride anions, 3 ions in all: the solution conducts like a 1:2 electrolyte.