Which one of the following metal complexes is most stable?

We start by recalling the statement of the chelate effect. It says, “When a metal ion forms a coordination compound with multidentate (polydentate) ligands, the resulting chelate complex is thermodynamically more stable than a similar complex that contains the same number of monodentate ligands.” The increase in stability arises from both an enthalpy advantage (several bonds to the metal instead of one) and an entropy advantage (fewer particles after complex formation).

Now we examine the ligands present in each option.

We know that $$en$$ represents ethylenediamine, $$H_2N\!-\!CH_2CH_2\!-\!NH_2$$. It supplies two lone-pair donor nitrogen atoms, so it is a bidentate ligand. Hence each $$en$$ molecule forms one five-membered chelate ring with the central metal ion.

In contrast, $$NH_3$$ is a monodentate ligand; it gives only one donor nitrogen atom and cannot form a ring.

Next we list the total number of chelate rings (and thereby the relative stability) for each complex.

• In option A we have $$[Co(en)(NH_3)_4]^{2+}$$. There is $$1$$ $$en$$ ligand, so the number of chelate rings is

$$1(en)\times 1 \;=\;1\text{ ring}.$$

• In option B we have $$[Co(en)_3]^{2+}$$. There are $$3$$ $$en$$ ligands, hence

$$3(en)\times 1 \;=\;3\text{ rings}.$$

• In option C we have $$[Co(en)_2(NH_3)_2]^{2+}$$. There are $$2$$ $$en$$ ligands, giving

$$2(en)\times 1 \;=\;2\text{ rings}.$$

• In option D we have $$[Co(NH_3)_6]^{2+}$$. All six ligands are monodentate, so

$$0\text{ chelate rings (no }en\text{ present)}.$$

It is clear that the complex in option B contains the highest number of chelate rings (three), while options A, C, and D contain fewer or none. According to the chelate effect, more chelate rings imply greater thermodynamic stability.

Hence, the complex $$[Co(en)_3]Cl_2$$ (option B) is the most stable among the given choices.

Hence, the correct answer is Option B.