How many S-O-S linkage(s) is/ are present in ${H_2}{S_3}{O_6}$?

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Hint: In these types of problems we must remember the number of outermost electrons present in the valence shell of a particular element.
In any element each shell can contain only a fixed number of electrons. The first shell can hold up to 2 electrons. The second shell can hold up to 8 (2 + 6) electrons. The third shell can hold up to 18 (2 + 6 + 10) and so on.
The general formula is that the nth shell can in principle hold up to $2({n^2})$ electrons.

Complete answer:
In this problem, sulphur oxoacids are chemical compounds that contain sulphur, oxygen, and hydrogen.
We will first make the structure of ${H_2}{S_3}{O_6}$
In the valence shell of 1 sulphur atom there are 6 electrons as it belongs to Oxygen family members.
So, when we make its structure

$HO - \mathop S\limits_{\mathop \parallel \limits_0 }^{\mathop \parallel \limits^O } - S - \mathop S\limits_{\mathop \parallel \limits_0 }^{\mathop \parallel \limits^O } - OH$
So, from this structure we can make out that there is no S-O-S linkage present in the ${H_2}{S_3}{O_6}$

Note: For solving such types of problems we must remember the number of valence electrons present in a particular atom of an element. Then only we will be able to make the molecular structure of a compound. The understanding of periodic tables will help us in these types of topics. Also, the general formula for finding out the number of electrons in nth shell.