Question

A metal complex with a formula MC$\ell_4$3NH$_3$ is involved in sp$^3$d$^2$ 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 ________

Hint:

When dealing with octahedral complexes, the number of geometrical isomers is determined by how the ligands are arranged. For complexes with three identical ligands and three other different ligands, two possible isomers (fac and mer) exist.

Answer 1

Correct Answer: 2

The formula MC$\ell_4$3NH$_3$ suggests a coordination complex with metal M, where coordination arises mainly through ligands C$\ell^-$ and NH$_3$. The sp$^3$d$^2$ hybridization implies an octahedral configuration. In octahedral complexes with four monodentate ligands of one type and two of another, geometrical isomerism can occur.

First, we determine ‘x’, the moles of AgCl produced. The reaction with AgNO$_3$ indicates that all Cl ligands are available to form AgCl. With 4 Cl ligands, x=4.

The reference to BrF$_5$, a pentagonal bipyramidal molecule with sp$^3$d hybridization, specifies that it has one lone pair, mirroring the requirement for x=4 derived from MC$\ell_4$3NH$_3$ reaction.

We now assess the complex's geometric isomerism: In an octahedral complex MA$_4$B$_2$, two geometrical isomers arise due to the arrangement of B ligands (cis or trans around the metal center). Hence, the number of geometrical isomers is 2.

In conclusion, the number of geometrical isomers exhibited by MC$\ell_4$3NH$_3$ is 2, matching the given range of 2,2.

Answer 2

The complex MC$\ell_4$3NH$_3$ undergoes sp$^3$d$^2$ hybridisation. 
The central atom of BrF$_5$ has 1 lone pair, so x = 1. 
The complex MC$\ell_4$3NH$_3$ has octahedral geometry, leading to 2 possible geometrical isomers: fac and mer. 
Thus, the number of geometrical isomers is 2.