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# What is the total number of valence electrons in the peroxydisulfate, ${{S}_{2}}O_{8}^{2-}$ ion?(a) 58(b) 60(c) 62(d) 64

Last updated date: 23rd Jun 2024
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Hint: The valence electrons refers to the electrons present in the outermost shell of the atom and since, both the Sulphur and oxygen belong to the same group, so has the same number of the valence electrons. We can find it by using the formula as;
\begin{align} & total\text{ }valence\text{ }electrons\text{ }in\text{ }{{S}_{2}}O_{8}^{2-}=no.\text{ }of valence\text{ }electrons\text{ }in\text{ }sulphur +no.\text{ }of\text{ }valence\text{ }electrons\text{ }in\text{ }oxygen+ \\ & \text{ }no.\text{ }of\text{ }electrons\text{ }gained \\ \end{align}

Complete Solution :
First of all, let’s discuss the valence electrons. By the valence electrons we mean the number of electrons present in the valence shell of the atom i.e. in the outermost shell of the atom.
- Now considering the statement:
- In the peroxydisulfate ion i.e.${{S}_{2}}O_{8}^{2-}$, there are two Sulphur atoms and eight oxygen atoms and two electrons it has gained in its valence shell.
- As we know that the Sulphur and oxygen belong to the same family and group, so the number of valence electrons in both will be the same i.e. 6.
- Then the total number of valence electrons in ${{S}_{2}}O_{8}^{2-}$ ion is as:
\begin{align} & total\text{ }valence\text{ }electrons\text{ }in\text{ }{{S}_{2}}O_{8}^{2-}=no.\text{ } of valence\text{ }electrons\text{ }in\text{ }sulphur +no.\text{ }of\text{ }valence\text{ }electrons\text{ }in\text{ }oxygen+ \\ & \text{ }no.\text{ }of\text{ }electrons\text{ }gained \\ \end{align}
Since there are two Sulphur atoms in ${{S}_{2}}O_{8}^{2-}$. Then ;
number of valence electrons in Sulphur = $2\times 6 = 12$
Similarly, there are eight oxygen atoms in ${{S}_{2}}O_{8}^{2-}$. Then ;
number of valence electrons in oxygen = $8\times 6 = 48$
Total number of electrons gained = 2
Then :
\begin{align} & total\text{ }valence\text{ }electrons\text{ }in\text{ }{{S}_{2}}O_{8}^{2-}=12 + 48 + 2 \\ & \text{ =62} \\ \end{align}
So, the correct answer is “Option C”.

Note: An atom can lose the electrons in the valence shell to acquire the stable electronic configuration or gain electrons in the valence shell to complete the octet and attain nearest noble gas configuration.
- In case of metals, they lose the electrons in the valence shell to acquire the stable electronic configuration of nearest noble gas. Example: Sodium metal loses one electron present in its valence shell to acquire nearest gas configuration i.e. of neon.
- On the other hand, in case of non-metals, they gain the electrons in the valence shell to complete their octet and acquire the stable electronic configuration of nearest noble gas. Example: Chlorine element gains one electron in its valence shell to acquire nearest gas configuration i.e. of argon.