Find $x$ from $\mathrm{BrF}_5$

In $\mathrm{BrF}_5$, bromine is the central atom.

• Valence electrons on Br $= 7$
• It forms $5$ Br–F bonds
• Hence one lone pair remains on Br

Therefore, the number of lone pairs on the central atom in $\mathrm{BrF}_5$ is $x=1$

Use AgNO$_3$ test to identify ionisable chloride

The complex has molecular formula $\mathrm{MCl}_4\cdot 3\mathrm{NH}_3$

On treatment with excess $\mathrm{AgNO}_3$, it gives $x=1$ mole of AgCl.

This means only one chloride ion is outside the coordination sphere and is precipitated as AgCl.

So the complex must be written as $[\mathrm{MCl}_3(\mathrm{NH}_3)_3]\mathrm{Cl}$

Determine coordination number and geometry

Inside the coordination sphere, the ligands are:

• $3$ chloride ligands
• $3$ ammonia ligands

Total coordination number $=6$.

Since it is stated to involve $\mathrm{sp}^3d^2$ hybridisation, the geometry is octahedral.

Find geometrical isomers of $[\mathrm{MCl}_3(\mathrm{NH}_3)_3]^+$

This is an octahedral complex of type $[\mathrm{MA}_3\mathrm{B}_3]$

Such complexes show two geometrical isomers:

1. fac isomer: all three identical ligands occupy one face
2. mer isomer: three identical ligands lie in a meridional plane

Therefore, the number of geometrical isomers is $2$

Comparison with stored answer

Derived answer $=2$.

Stored correct answer $=2$.

Hence, the derived answer agrees with the stored answer.