Question

If a compound has a formula of \[{A_2}{B_3}\] then
A. the formula of oxide of \[A\] is\[{A_2}{O_3}\]
B. nitrate of \[A\] is \[AN\]
C. bicarbonate of \[A\] is \[{\mathbf{A}}{\left( {{\mathbf{HC}}{{\mathbf{O}}_{\mathbf{3}}}} \right)_{\mathbf{3}}}\]
D. chlorate of\[A\] is \[{\mathbf{A}}{\left( {{\mathbf{Cl}}{{\mathbf{O}}_{\mathbf{2}}}} \right)_{\mathbf{2}}}\]

Answer 1

Hint: In the given question a compound is given and we have to give the oxide of the formula of compound along with its nitrate and bicarbonate also the carbonate. We can solve this given question by using a method known as criss cross method.

Complete step by step solution:

The crisscross method makes it in an easier way to find out the subscripts for each element in an ionic compound. By using this method, the number related to the charge of the first ion is assigned as the subscript of the second ion.
The ionic formula of a compound must be electrically neutral, that means it has no charge. When writing the formula for an ionic compound, the cation comes first which is followed by the anion and along with numeric subscripts to designate the number of atoms of each.

The number of the charges on the second ion is assigned as the subscript of the first ion.
According to the crisscross method valency of cation is \[ + 3\]and valency of anion is \[ - 2\].Thus a compound of formula of \[{A_2}{B_3}\] has the formula of oxide of \[A\]will be \[{A_2}{O_3}\] and the formula for its nitride is \[AN\] and bicarbonate is \[A\left( {HC{O_3}} \right)\].

Hence the correct option is A.

Note: To apply criss cross method. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second then Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. Thus the charge on Cation becomes the subscript of anion and the charge on anion becomes the subscript of Cation. And now by reducing it to the lowest ratio, finally write the final formula. Leave out all subscripts that are 1. If there is only 1 of the polyatomic ions, leave off parentheses.