An octahedral complex having molecular composition Co.5NH3.ClSO4 has two isomers A and B. The solution of A gives a…

🧠 Two Different Precipitates ⇒ Ionisation Isomerism

Same formula, different test reactions: this is the signature of ionisation isomerism. The same anion can be either inside the coordination sphere (silent) or outside as a free counter-ion (test-active). One isomer keeps $\mathrm{Cl^-}$ free; the other keeps $\mathrm{SO_4^{2-}}$ free.

🗺️ Standard Approach

The empirical formula $\mathrm{Co \cdot 5NH_3 \cdot Cl \cdot SO_4}$ has two anions ($\mathrm{Cl^-}$ and $\mathrm{SO_4^{2-}}$). Co is octahedral with 5 NH₃ + one anion inside (sixth slot). One anion sits inside, the other outside.

Isomer A + $\mathrm{AgNO_3}$ → white ppt (AgCl). For $\mathrm{Ag^+}$ to find $\mathrm{Cl^-}$, the chloride must be free (outside): $A = [\mathrm{Co(NH_3)_5 (SO_4)}]\mathrm{Cl}$

Isomer B + $\mathrm{BaCl_2}$ → white ppt ($\mathrm{BaSO_4}$). For $\mathrm{Ba^{2+}}$ to find $\mathrm{SO_4^{2-}}$, the sulphate must be free: $B = [\mathrm{Co(NH_3)_5 Cl}]\mathrm{SO_4}$

Same formula, different ions inside/outside ⇒ ionisation isomers.

⚡ Pattern: Two Test Ions, Different Precipitates

This template repeats almost yearly in JEE Main:

Two isomers, same formula. $\mathrm{AgNO_3}$ test on one, $\mathrm{BaCl_2}$ (or another anion test) on the other. Different precipitates ⇒ ionisation isomerism.

If both gave the same precipitate, it'd be coordination isomerism. If different colours rather than different ions, it'd be linkage.

⚠️ Don't Confuse With Other Types

• Coordination isomerism — needs both cation AND anion to be complex units, e.g. $[\mathrm{Co(NH_3)_6}][\mathrm{Cr(CN)_6}]$ vs $[\mathrm{Cr(NH_3)_6}][\mathrm{Co(CN)_6}]$. Not the case here.
• Linkage isomerism — requires an ambidentate ligand ($\mathrm{NO_2^-}/\mathrm{ONO^-}$, $\mathrm{SCN^-}/\mathrm{NCS^-}$). Not present.
• Geometric isomerism — $[\mathrm{Ma_5 b}]$ has only one arrangement (no cis/trans).

$\boxed{\text{Answer: (c) Ionisation isomerism}}$