Question

An aqueous solution of titanium chloride, when subjected to magnetic measurement, measured zero magnetic moments.
Assuming the complex as octahedral in aqueous solution, the formula of the complex is :
A) $[Ti{({H_2}O)_6}]C{l_2}$
B) $[Ti{({H_2}O)_6}]C{l_4}$
C) ${[TiC{l_6}]^{3 - }}$
D) $[Ti{({H_2}O)_4}C{l_2}]$

Answer 1

Hint: The aqueous part will indicate that the complex has water molecules as ligands. The magnetic moment of a compound can be written as -
$µ =$ $\sqrt {n(n + 1)} $
where n stands for the number of unpaired electrons.
For a species to possess zero magnetic moments, the number of unpaired electrons should be zero.

Complete step by step solution:
We know that Titanium belongs to transition metals. It has atomic number $22$. So, the electronic configuration for Titanium will be as -
$Ti (22) =$ $[Ar]3{d^2}4{s^2}$
We are given that it is an aqueous solution of Titanium chloride. This means the complex will contain water as the ligand. Further, the complex is octahedral meaning there will be six ligands. So, the complex will be - $[Ti{({H_2}O)_6}]$
The magnetic moment of a compound is given by -
$µ =$ $\sqrt {n(n + 1)} $
where n stands for the number of unpaired electrons.
As there are two unpaired electrons in the 3d orbital in case of Titanium. So, the magnetic moment will have some value. But in question, we have been asked for the value of magnetic moment to be 0. So, for the magnetic moment to be zero, the number of unpaired electrons should be zero. This means we have to donate four electrons to achieve $Ti (+4)$. This will have zero electron 3d or we can say zero unpaired electrons. This is possible only when the $Cl$ atoms are outside the bracket i.e. $Cl$ atoms do not behave as primary ligands.
But there will be four $Cl$ atoms to balance the charge.
So, the complex would be $[Ti{({H_2}O)_6}]C{l_4}$.

Thus, the option D) is the correct answer.

Note: It must be noted we are donating four electrons while we need to remove only two electrons from 3d. This is because first, the removal of two electrons of 4s will take place and then 3d. So, four electrons need to be removed.
Further, the water is a neutral compound. Thus, it will not contribute to any charge. But $Cl$ has a -1 charge.