Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of ${ MCl }_{ 2 }.6{ H }_{ 2 }O$ (X) and ${ NH }_{ 4 }Cl$ gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1:3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.
Among the following options which statements are correct?
(This question has multiple correct options)
(A) Z is a tetrahedral complex
(B) The hybridisation of the central metal ion in Y is ${ d }^{ 2 }{ sp }^{ 3 }$
(C) When X and Z are in equilibrium at $0^{ o }{ C }$, the colour of the solution is pink.
(D) Addition of silver nitrate to Y gives only two equivalents of silver chloride.

Answer 1

Hint: In order to solve this question we should know the meaning of high spin complex and a low spin complex. In a high spin complex, the pairing energy of the electrons is more than the crystal field splitting energy while in a low spin complex the crystal field splitting energy is more than the pairing energy.

Complete step by step solution:
In this question, we are given that compound X is pink in colour. It reacts with the ${ NH }_{ 4 }Cl$ to give an octahedral complex Y in the presence of air The complex Y behaves as 1:3 electrolyte in aqueous solution. The compound X is ${ MCl }_{ 2 }.6{ H }_{ 2 }O$ which can also be written as $\left[ M{ (H }_{ 2 }O{ ) }_{ 6 } \right] { Cl }_{ 2 }$. The metal cation M is in +2 oxidation state. The reaction of this complex with ammonium chloride will give the complex Y such that it has to be 1:3 electrolyte in aqueous solution. Also, the calculated spin only magnetic moment of X is 3.74 B.M. while for Y it is 0. Hence there are approximately three unpaired electrons present in complex X while there are no unpaired electrons present in complex Y. This means that complex X must be a high spin complex while complex Y must be a low spin complex. This can happen if in complex Y strong field ligands are present and the metal ion is in higher oxidation state. Hence if we say that all the water molecules in complex X are replaced by a strong field ligand like ammonia and the oxidation state of the metal ion changes from +2 to +3 in going from complex X to complex Y; the reactions will be:
$\begin{matrix} \left[ M{ (H }_{ 2 }O{ ) }_{ 6 } \right] { Cl }_{ 2 } \\ X \end{matrix}\xrightarrow [ { NH }_{ 4 }Cl+air ]{ Excess\quad { NH }_{ 3 }(aq) } \begin{matrix} \left[ M({ NH }_{ 3 }{ ) }_{ 6 } \right] { Cl }_{ 3 } \\ Y(gives\quad 1:3\quad electrolyte\quad in\quad aq.\quad solution) \end{matrix}$
Since it is given that Y is octahedral therefore its hybridisation will be ${ d }^{ 2 }{ sp }^{ 3 }$ since it is a low spin complex with ammonia as a strong field ligand. When a solution of silver nitrate is added to Y then we will get 3 equivalents of silver chloride since the aqueous solution of Y produces three chloride ions in solution.
Since the aqueous solution of X is pink in colour therefore the metal cation in X is most likely ${ Mn }^{ 2+ }$.
The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z which has a spin only magnetic moment of 3.87 B.M. which means that it is also a high spin complex. The reaction is given below:
$ \begin{matrix} \left[ Mn{ (H }_{ 2 }O{ ) }_{ 6 } \right] { Cl }_{ 2 } \\ X \end{matrix}\xrightarrow [ RT ]{ HCl\quad excess\quad } \begin{matrix} \left[ Co{ Cl }_{ 2 } \right] ^{ 2- } \\ Z(blue\quad in\quad colour) \end{matrix}$
Now, chloride ions are weak field ligands and hence Z is a high spin complex. High spin complex with only four ligands can only be formed if the complex has a tetrahedral geometry. Hence complex Z has $ { sp }^{ 3 }$ hybridisation.
When X and Z are present in equilibrium at $0^{ o }{ C }$, then the amount of X will be more than Z since X is an octahedral complex and octahedral complexes are more stable than tetrahedral complexes. Hence the solution will turn pink.

Hence, the correct answers are (A) Z is a tetrahedral complex, (B) The hybridisation of the central metal ion in Y is ${ d }^{ 2 }{ sp }^{ 3 }$ and (C) When X and Z are in equilibrium at $0^{ o }{ C }$, the colour of the solution is pink.

Note: When two different species are present in equilibrium with each other, then the species which is more stable predominates. Also, we can have a low spin complex where the number of ligands is four only in the case of square planar complexes.