A metal complex with a formula $$MCl_4 \cdot 3NH_3$$ is involved in $$sp^3d^2$$ hybridisation. It upon reaction with excess of $$AgNO_3$$ solution gives 'x' moles of AgCl. Consider 'x' is equal to the number of lone pairs of electron present in central atom of $$BrF_5$$. Then the number of geometrical isomers exhibited by the complex is ______.

The hybridisation $$sp^3d^2$$ corresponds to an octahedral geometry, so the metal ion must have coordination number $$6$$.

In the given compound $$MCl_4\cdot 3NH_3$$ the six possible ligands are provided by $$3$$ molecules of $$NH_3$$ and some of the chloride ions. Let the number of chloride ions inside the coordination sphere be $$y$$ and the number of chloride ions present as counter-ions be $$(4-y)$$.

When the salt is treated with excess $$AgNO_3$$, only the ionisable (counter) chloride ions precipitate as $$AgCl$$. The number of moles of $$AgCl$$ obtained per mole of the complex is given to be $$x$$.

The problem states that $$x$$ equals the number of lone pairs on the central atom in $$BrF_5$$. In $$BrF_5$$, bromine is surrounded by $$5$$ bond pairs and has $$1$$ lone pair (total steric number $$6$$), so

$$x = 1$$.

Hence, only one chloride ion is outside the coordination sphere:

$$(4-y)=1 \;\;\Longrightarrow\;\; y = 3.$$

Thus the correct structural formula is

$$[\,M(NH_3)_3Cl_3\,]Cl.$$

This is of the general type $$MA_3B_3$$ (where $$A = NH_3$$ and $$B = Cl^-$$) in an octahedral environment.

For an octahedral complex of the form $$MA_3B_3$$, two geometrical (cis-trans) arrangements are possible: Case 1: The three identical ligands occupy one face of the octahedron (fac-isomer).
Case 2: The three identical ligands lie in a meridian passing through the metal (mer-isomer).

Therefore, the number of geometrical isomers exhibited by the complex is $$2$$.

Answer : 2