Question

On the treatment of 100 ml of 0.1 M solution of \[CoC{l_3}.6{H_2}O\] with excess \[AgN{O_3}\] ; \[1.2 \times {10^{22}}\] ions are precipitated. The complex is:
A) \[[Co{({H_2}O)_6}]C{l_3}\]
B) \[[Co{({H_2}O)_5}Cl]C{l_2}.{H_2}O\]
C) \[[Co{({H_2}O)_4}C{l_2}]Cl.2{H_2}O\]
D) \[[Co{({H_2}O)_3}C{l_3}].3{H_2}O\]

Answer 1

Hint: In the above equation, \[AgN{O_3}\]will react with one of the ions which has a negative charge on it. Since the only charged negative ion in this complex is \[Cl\], the precipitate that would be formed will be \[Ag\] with \[Cl\]. It can be stated with the help of a equation as :
\[AgN{O_3} + \{ Cl\,Complex\} \to AgCl + Other\,product\,formed\]

Complete Step by Step Solution:
Since the Avogadro’s number is \[6.023 \times {10^{23}}\]
Therefore to calculate the moles of \[1.2 \times {10^{22}}\] ions will correspond to :
\[\dfrac{{1.2 \times {{10}^{22}}}}{{6.023 \times {{10}^{23}}}}\]
\[ = 0.02\,moles\]
The moles of 100 ml of 0.1M solution of \[CoC{l_3}.6{H_2}O\] will be equal to :
\[\dfrac{{100ml}}{{1000ml/L}} \times 0.1M\]
\[ = 0.01\,moles\]
This means that the \[0.01\,moles\] of \[CoC{l_3}.6{H_2}O\] is reacting with the excess of \[AgN{O_3}\] to form the \[0.02\,moles\]of \[AgCl\] precipitate, which means that it contains \[0.1\,moles\] of \[A{g^ + }\] and \[0.01\,moles\] of \[C{l^ - }\] ions.
Therefore, for the formation of one molecule of \[CoC{l_3}.6{H_2}O\] , one atom of chlorine will be displaced and so the complex that is formed will be \[[Co{({H_2}O)_5}Cl]C{l_2}.{H_2}O\] because the number of atoms that are present outside of the coordination sphere should be equal to one.
Hence, option B is the correct answer

Note: Coordination compounds are defined as any class of substances in which a central metal atom is surrounded by various non-metal atoms or groups of atoms. These non-metal atoms that surround the metal are termed as ligands and are joined to it by various chemical bonds. The major application of coordination compounds is that they act as a catalyst in a reaction, i.e. they help in the acceleration of a reaction.