Q.(i) How would you differentiate between the two complex compounds [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br by a suitable chemical test?

The NO$_2^-$ ligand can bind through N (nitrito-N, $-NO_2$) or through O (nitrito-O, $-ONO$); complexes differing only in this attachment are linkage isomers.

$[Co(NH_3)_5(NO_2)]^{2+}$ has the nitrite ligand bonded to cobalt through its nitrogen atom (nitro form), while $[Co(NH_3)_5(ONO)]^{2+}$ has the same ligand bonded through one of its oxygen atoms (nitrito form).

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When two ligands (here Cl- and SO4^2-) exchange places between 'inside the coordination sphere' and 'outside as counter ion', the compounds are ionisation isomers.

In [Co(NH3)5Cl]SO4, Cl- is a ligand bound to Co, and SO4^2- is the free counter ion — this compound gives a precipitate with BaCl2 (free SO4^2-) but not readily with AgNO3.

In [Co(NH3)5SO4]Cl, SO4^2- is bound inside the coordination sphere and Cl- is the free counter ion — this compound gives an immediate precipitate with AgNO3 (free Cl-) but not readily with BaCl2.

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Swapping which ligands sit on the cation vs. the anion between two complex-salt formulas is coordination isomerism.

In $[Cu(NH_3)_4][PtCl_4]$ and $[Pt(NH_3)_4][CuCl_4]$, both compounds have the same overall composition (2 metal atoms, 4 $NH_3$, 4 $Cl^-$), but the distribution of ligands between the two complex ions is exchanged: in the first, $NH_3$ ligands are bound to Cu (cation) and $Cl^-$ to Pt (anion); in the second, this is reversed. This …