Which of the following is an example of homoleptic complex?

First, let us recall the meaning of the term homoleptic complex. A coordination complex is said to be homoleptic when the central metal ion is surrounded by only one kind of ligand. In other words, every coordination position around the metal is occupied by identical donor groups. If even one ligand is different, the complex becomes heteroleptic.

We now examine each option one by one, carefully listing every ligand that is actually bonded to the metal ion inside the square brackets.

For Option A we have $$[Co(NH_3)_6]Cl_3.$$ Inside the coordination sphere the species present is $$[Co(NH_3)_6]^{3+}.$$ We see six ligands and each ligand is $$NH_3.$$ Since all six ligands are the same ($$NH_3$$ only), the complex inside the brackets is homoleptic.

For Option B we have $$[Pt(NH_3)_2Cl_2].$$ Inside the brackets the platinum(II) ion is directly bonded to two $$NH_3$$ ligands and two $$Cl^-$$ ligands. Here two different kinds of ligands ($$NH_3$$ and $$Cl^-$$) are present, so this complex is heteroleptic, not homoleptic.

For Option C we have $$[Co(NH_3)_4Cl_2].$$ The cobalt(III) ion inside the brackets is attached to four $$NH_3$$ ligands and two $$Cl^-$$ ligands. Once again two different kinds of ligands are present, so this is heteroleptic.

For Option D we have $$[Co(NH_3)_5Cl]Cl_2.$$ Focusing on what is inside the square brackets, we get $$[Co(NH_3)_5Cl]^{2+},$$ in which cobalt(III) is bonded to five $$NH_3$$ ligands and one $$Cl^-$$ ligand. Because both $$NH_3$$ and $$Cl^-$$ appear, this complex is also heteroleptic.

Comparing all four possibilities, only Option A satisfies the strict requirement that a homoleptic complex contain only one type of ligand inside its coordination sphere.

Hence, the correct answer is Option A.