If excess $AgC{O_3}$ solution is added to $100$$ ml $of $0.24$$M$ solution of dichlorobis (ethylene diamine) cobalt $\left( {III} \right)$ chloride . How many moles of $AgCl$ be precipitated?
A.0.0012
B.0.0016
C.0.0024
D.0.0048
Answer
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Hint: Here we have to find the charge on dichloro bis (ethylenediamine) cobalt ($III$) chloride. It is a problem based on limiting reagents.
Complete step by step solution:
This complex overall has an oxidation of $ + 1$. Since dichlorobi (ethlenediamine) cobalt () chloride has an oxidation state $ - 1$ in the inner coordination sphere. Its chemical formula is $\left[ {CoC{l_2}{{\left( {en} \right)}_2}} \right]Cl$, it contains two chloro ligands $\left( {C{l^ - }} \right)$ . The inner coordination sphere of two ethylenediamine ligands has oxidation $0$. Its chemical formula is $\left( {{H_2}N - C{H_2} - C{H_2} - N{H_2}} \right)$. Coordinated on the outer coordination sphere and cobalt($III$) a chloride $\left( {C{l^ - }} \right)$ ligand has an oxidation state of $3$. Therefore the coordination sphere contain $C{o^{ + 3}} + 2C{l^ - } + 2\left( {en} \right)$ .
$2 \times \left( { - 1} \right) + 2 \times 0 + \left( { + 3} \right)$ $ = + 1$, this gives us the total oxidation charge on the coordination sphere.
There is a double replacement reaction in the between $\left[ {CoC{l_2}{{\left( {en} \right)}_2}} \right]Cl$ and $AgC{O_3}$in the outer sphere it is as followed: $AgC{O_{3(aq)}} + \left[ {CoC{l_2}{{(en)}_2}} \right]C{l_{(aq)}} \to AgC{l_{(5)}} + \left[ {CoC{l_2}{{(en)}_2}} \right]C{O_3}$
Here the number of moles precipitated will be the product of molarity into volume in liters, since the complex is limiting reagent as per the above equation. Therefore the total number
$0.024 \times 0.1 = 0.0024$
So total number of moles precipitated are 0.0024, as shown the complex is $1:1$
So, the correct answer is option (C) , $0.0024$.
Note :
Strong field ligand is sigma donors. Complexes with these ligands will act as low spin complexes where the inner coordination sphere will be inert. Ethylenediamine ligands are bidentate.
Complete step by step solution:
This complex overall has an oxidation of $ + 1$. Since dichlorobi (ethlenediamine) cobalt () chloride has an oxidation state $ - 1$ in the inner coordination sphere. Its chemical formula is $\left[ {CoC{l_2}{{\left( {en} \right)}_2}} \right]Cl$, it contains two chloro ligands $\left( {C{l^ - }} \right)$ . The inner coordination sphere of two ethylenediamine ligands has oxidation $0$. Its chemical formula is $\left( {{H_2}N - C{H_2} - C{H_2} - N{H_2}} \right)$. Coordinated on the outer coordination sphere and cobalt($III$) a chloride $\left( {C{l^ - }} \right)$ ligand has an oxidation state of $3$. Therefore the coordination sphere contain $C{o^{ + 3}} + 2C{l^ - } + 2\left( {en} \right)$ .
$2 \times \left( { - 1} \right) + 2 \times 0 + \left( { + 3} \right)$ $ = + 1$, this gives us the total oxidation charge on the coordination sphere.
There is a double replacement reaction in the between $\left[ {CoC{l_2}{{\left( {en} \right)}_2}} \right]Cl$ and $AgC{O_3}$in the outer sphere it is as followed: $AgC{O_{3(aq)}} + \left[ {CoC{l_2}{{(en)}_2}} \right]C{l_{(aq)}} \to AgC{l_{(5)}} + \left[ {CoC{l_2}{{(en)}_2}} \right]C{O_3}$
Here the number of moles precipitated will be the product of molarity into volume in liters, since the complex is limiting reagent as per the above equation. Therefore the total number
$0.024 \times 0.1 = 0.0024$
So total number of moles precipitated are 0.0024, as shown the complex is $1:1$
So, the correct answer is option (C) , $0.0024$.
Note :
Strong field ligand is sigma donors. Complexes with these ligands will act as low spin complexes where the inner coordination sphere will be inert. Ethylenediamine ligands are bidentate.
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