Answer
438.3k+ views
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.
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.
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