An element has two different sulphates in which its weight percentage is 28 and 37. What is the ratio of oxidation number of the element in these sulphates?
A.\[1:2\]
B.\[1:3\]
C.\[2:1\]
D.\[3:2\]

Hint: We have to know that the sulphate is a chemical compound having the formula, \[SO_4^{2 - }\]. Here, four oxygen atoms are surrounded by one sulphur atom. And this sulphate is ester or it may be the salt of sulphuric acid. The formation of sulphate ions is by the replacement of one or both hydrogen atoms which is present in sulphuric acid with an organic group or by using a metal cation.

Complete answer:
Here, the element containing two different sulphates with weight percentage is $28$ and $37$ . And the ratio of oxidation number of the element in these sulphates is not equal to\[1:2\]. Hence, option (A) is incorrect.
The ratio of the oxidation number of the element which is present in the sulphate is not equal to\[1:3\]. Hence, the option (B) is incorrect.
Here, the weight percentage is equal to $28$ and $37$ but the ratio of oxidation number of the element is not equal to \[2:1\]. Because, the oxidation number is not equal to two and one. Hence, option (C) is incorrect.
According to the question, there are two elements which have two different sulphates.
The formula of sulphate is \[SO_4^{2 - }\]
Let's consider two elements and that is, \[MxS{O_4}\] and \[MyS{O_4}\].
The molecular weight of sulphate ion is equal to \[96g\]
Thus, molecular weight of \[MxS{O_4} = xM + 96\]
Molecular weight of \[MyS{O_4} = yM + 96\]
Given, the weight percentage of both elements is equal to 28 and 37 respectively.
Consider the case of element, x
\[\% = \dfrac{{xM}}{{xM + 96}} \times 100 = 28\]
\[72 \times M = 28 \times 96\,\,\,\,\, \ldots \left( 1 \right)\]
For the element, y
\[\% = \dfrac{{yM}}{{yM + 96}} \times 100 = 37\]
\[63 \times M = 37 \times 96\]
By rearranging the equation one and two, will get
\[\dfrac{{72x}}{{63y}} = \dfrac{{28}}{{37}}\]
Therefore, \[\dfrac{x}{y} = \dfrac{1}{2}\]
Which means, the ratio of oxidation number of the element in this sulphates is equal to \[1:2\]

Hence, the option (D) is correct.

Note:
We have to know that sulphate ion is a polyatomic ion which has its empirical formula, \[SO_4^{2 - }\] with molecular weight $96g$. And the oxidation number of sulphate in sulphate ion is equal to \[ + 6\] and the oxidation number of oxygen is equal to \[ - 2\]. Here, the ratio of oxidation number of the element in these sulphates is equal to \[1:2\]. That is, the oxidation number of the element present in both elements is one and two respectively.