
When $A{{g}^{+}}$ reacts with excess of sodium-thiosulphate then the obtained species having charge and geometry respectively:
A. -3, linear
B. -2, tetrahedral
C. -1, square planar
D. -3, square planar
Answer
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Hint: In order to solve this question, the first thing to know is the reaction between $A{{g}^{+}}$ and Sodium thiosulphate. We’ll get the charge on the product by this way only and to find the geometry of the product. We’ll find the oxidation state first and then the hybridisation of the product.
Complete Step by Step Solution:
To proceed with this question, we must know the reaction between$A{{g}^{+}}$ and Sodium thiosulphate.
\[A{{g}^{+}}+N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]+A{{g}^{+}}\],as we can see there is a formation of a coordination compound in this reaction.
A coordination complex is made up of a core atom or ion, known as the coordination centre, which is typically metallic, and an array of attached molecules or ions that are known as ligands or complexing agents. Now, \[{{S}_{2}}{{O}_{3}}\]is a ligand and is a monodentate ligand. So, we now know that the coordination number of the metal is 2.
Now, if we try to find the oxidation number of Ag. Then, if the complex completely ionises then it will be given as:
\[N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]\to 3N{{a}^{+}}+{{[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]}^{3-}}\]
So, the net charge on the complex came out to be -3. Now, let’s assume the oxidation number of Ag to be x. Now since \[({{S}_{2}}{{O}_{3}})\]carry a -2 charge and there are 2 molecules of it so the total negative charge would be -4.
$x+2(-2)=-3$
$x=-3+4=+1$
There the oxidation state of Ag in the compound is +1. Now as we can see that in the compound \[N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]\] Ag is forming bond with two molecules of\[({{S}_{2}}{{O}_{3}})\]. Therefore the hybridization must be sp. And the geometry associated with sp hybridisation is linear. Therefore, the charge on the compound is -3 and geometry is linear.
Hence, the correct option is A. -3, linear.
Note: The most important thing to note here is the reaction of $A{{g}^{+}}$ with sodium thiosulphate. You need to know about the coordination compounds and other things like oxidation number and ionisation of complexes. Another thing to note here is the shape and geometry of different hybridizations.
Complete Step by Step Solution:
To proceed with this question, we must know the reaction between$A{{g}^{+}}$ and Sodium thiosulphate.
\[A{{g}^{+}}+N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]+A{{g}^{+}}\],as we can see there is a formation of a coordination compound in this reaction.
A coordination complex is made up of a core atom or ion, known as the coordination centre, which is typically metallic, and an array of attached molecules or ions that are known as ligands or complexing agents. Now, \[{{S}_{2}}{{O}_{3}}\]is a ligand and is a monodentate ligand. So, we now know that the coordination number of the metal is 2.
Now, if we try to find the oxidation number of Ag. Then, if the complex completely ionises then it will be given as:
\[N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]\to 3N{{a}^{+}}+{{[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]}^{3-}}\]
So, the net charge on the complex came out to be -3. Now, let’s assume the oxidation number of Ag to be x. Now since \[({{S}_{2}}{{O}_{3}})\]carry a -2 charge and there are 2 molecules of it so the total negative charge would be -4.
$x+2(-2)=-3$
$x=-3+4=+1$
There the oxidation state of Ag in the compound is +1. Now as we can see that in the compound \[N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]\] Ag is forming bond with two molecules of\[({{S}_{2}}{{O}_{3}})\]. Therefore the hybridization must be sp. And the geometry associated with sp hybridisation is linear. Therefore, the charge on the compound is -3 and geometry is linear.
Hence, the correct option is A. -3, linear.
Note: The most important thing to note here is the reaction of $A{{g}^{+}}$ with sodium thiosulphate. You need to know about the coordination compounds and other things like oxidation number and ionisation of complexes. Another thing to note here is the shape and geometry of different hybridizations.
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