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Hint: To solve this we must first draw the structures of all the four given compounds. From the structures we can determine the compound having ${\text{X}} - {\text{O}} - {\text{X}}$ linkage. First of all we should draw the exact structure of the compound to find out the central atom.
Complete Step by step answer: ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$ is known as peroxydisulfate ion. Draw the structure of ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$:
In the structure of ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$, we cannot see ${\text{P}} - {\text{O}} - {\text{P}}$ linkage. Thus, ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$ does not have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (A) is not correct.
${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$ is known as thiosulphate ion. Draw the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$:
In the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$, we cannot see ${\text{S}} - {\text{O}} - {\text{S}}$ linkage. Thus, ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$ does not have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (B) is not correct.
$\gamma - {\text{S}}{{\text{O}}_3}$ is known as sulphur trioxide. Draw the structure of $\gamma - {\text{S}}{{\text{O}}_3}$:
In the structure of $\gamma - {\text{S}}{{\text{O}}_3}$, we can see ${\text{S}} - {\text{O}} - {\text{S}}$ linkage. The ${\text{S}} - {\text{O}} - {\text{S}}$ is shown in red colour. Thus, $\gamma - {\text{S}}{{\text{O}}_3}$ has ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (C) is correct.
${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$ is known as metabisulphite. Draw the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$:
In the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$, we cannot see ${\text{S}} - {\text{O}} - {\text{S}}$ linkage. Thus, ${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$ does not have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (D) is not correct.
Thus, the compound have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage where ‘${\text{X}}$’ is the so called central atom like ${\text{P}}$, ${\text{S}}$ etc is $\gamma - {\text{S}}{{\text{O}}_3}$.
Thus, the correct option is (C) $\gamma - {\text{S}}{{\text{O}}_3}$.
Note: Sulphur trioxide $\left( {{\text{S}}{{\text{O}}_3}} \right)$ has many structural changes and thus, it has many different forms. These changes are caused by the traces of water. $\gamma - {\text{S}}{{\text{O}}_3}$ is very pure gaseous sulphur trioxide. It is a trimeric structure i.e. it has three molecules of sulphur trioxide. $\gamma - {\text{S}}{{\text{O}}_3}$ forms a cyclic structure. It is colourless solid in appearance. Sulphur trioxide is highly hygroscopic in nature i.e. it can easily attract water molecules from its surroundings by absorption or adsorption.
Complete Step by step answer: ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$ is known as peroxydisulfate ion. Draw the structure of ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$:
In the structure of ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$, we cannot see ${\text{P}} - {\text{O}} - {\text{P}}$ linkage. Thus, ${{\text{P}}_{\text{2}}}{\text{O}}_8^{4 - }$ does not have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (A) is not correct.
${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$ is known as thiosulphate ion. Draw the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$:
In the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$, we cannot see ${\text{S}} - {\text{O}} - {\text{S}}$ linkage. Thus, ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$ does not have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (B) is not correct.
$\gamma - {\text{S}}{{\text{O}}_3}$ is known as sulphur trioxide. Draw the structure of $\gamma - {\text{S}}{{\text{O}}_3}$:
In the structure of $\gamma - {\text{S}}{{\text{O}}_3}$, we can see ${\text{S}} - {\text{O}} - {\text{S}}$ linkage. The ${\text{S}} - {\text{O}} - {\text{S}}$ is shown in red colour. Thus, $\gamma - {\text{S}}{{\text{O}}_3}$ has ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (C) is correct.
${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$ is known as metabisulphite. Draw the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$:
In the structure of ${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$, we cannot see ${\text{S}} - {\text{O}} - {\text{S}}$ linkage. Thus, ${{\text{S}}_{\text{2}}}{\text{O}}_5^{2 - }$ does not have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage.
Thus, option (D) is not correct.
Thus, the compound have ${\text{X}} - {\text{O}} - {\text{X}}$ linkage where ‘${\text{X}}$’ is the so called central atom like ${\text{P}}$, ${\text{S}}$ etc is $\gamma - {\text{S}}{{\text{O}}_3}$.
Thus, the correct option is (C) $\gamma - {\text{S}}{{\text{O}}_3}$.
Note: Sulphur trioxide $\left( {{\text{S}}{{\text{O}}_3}} \right)$ has many structural changes and thus, it has many different forms. These changes are caused by the traces of water. $\gamma - {\text{S}}{{\text{O}}_3}$ is very pure gaseous sulphur trioxide. It is a trimeric structure i.e. it has three molecules of sulphur trioxide. $\gamma - {\text{S}}{{\text{O}}_3}$ forms a cyclic structure. It is colourless solid in appearance. Sulphur trioxide is highly hygroscopic in nature i.e. it can easily attract water molecules from its surroundings by absorption or adsorption.
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