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Hint :Sodium sulphate is a white solid powder highly soluble in water. It is usually employed in manufacturing of detergents as well as in the kraft process of paper pulping. This compound is generally synthesized by the reaction of sodium chloride and sulphuric acid (process known as Mannheim process).
Complete Step By Step Answer:
As the name suggests, the given compound sodium sulfate comprises a metal and a group of nonmetals. We know that any compound containing a metal and a non-metal is considered to be an ionic compound. In ionic compounds, we have to take into account the charges of each of the elements present in the compound. Now let us identify the formula for the given compound i.e. sodium sulfate. Sodium sulfate possesses a polyatomic ion (i.e. group of nonmetals (sulfate in this case) after the metal (Sodium) so we'll use a table of names for the common polyatomic ions, along with the Periodic Table. Rules to write formula of a ternary ionic compound are listed below:
1. Using the periodic table, write down the symbol for the metal along with its charge.
In the present case, the symbol of metal Sodium is $ Na $ having a charge +1.
2. Using the Common ion table, identify the symbol as well as charge of the polyatomic ion.
In the present case, polyatomic ion sulfate is written as $ S{O_4} $ having a charge -2.
3. Check if the charges are balanced
In the present case it becomes $ N{a^{1 + }}SO_4^{2 - } $ . The charges are not balanced as summation of charges is non-zero i.e. $ ( + 1) + ( - 2) = - 1 \ne 0 $
4. If unbalanced charges are there, add subscripts following the criss-cross method such that net charge for the compound is zero.
In the present case, we will add subscripts to both metal and non-metal such that the net charge becomes zero. So it becomes $ {(N{a^{1 + }})_2}{(SO_4^{2 - })_1} $ and now the net charge is balanced i.e. $ (2 \times ( + 1)) + (1 \times ( - 2)) = 0 $
Hence, the ionic compound formula for Sodium sulfate is $ N{a_2}S{O_4} $ .
Note :
While writing the chemical formula of an ionic compound, never write the subscript '1'. And it should be noted that we can also have two polyatomic ions like $ N{H_4}N{O_3} $ in the same compound. In this case, we have to find and write both of the names as identified from the Common Ion Table.
Complete Step By Step Answer:
As the name suggests, the given compound sodium sulfate comprises a metal and a group of nonmetals. We know that any compound containing a metal and a non-metal is considered to be an ionic compound. In ionic compounds, we have to take into account the charges of each of the elements present in the compound. Now let us identify the formula for the given compound i.e. sodium sulfate. Sodium sulfate possesses a polyatomic ion (i.e. group of nonmetals (sulfate in this case) after the metal (Sodium) so we'll use a table of names for the common polyatomic ions, along with the Periodic Table. Rules to write formula of a ternary ionic compound are listed below:
1. Using the periodic table, write down the symbol for the metal along with its charge.
In the present case, the symbol of metal Sodium is $ Na $ having a charge +1.
2. Using the Common ion table, identify the symbol as well as charge of the polyatomic ion.
In the present case, polyatomic ion sulfate is written as $ S{O_4} $ having a charge -2.
3. Check if the charges are balanced
In the present case it becomes $ N{a^{1 + }}SO_4^{2 - } $ . The charges are not balanced as summation of charges is non-zero i.e. $ ( + 1) + ( - 2) = - 1 \ne 0 $
4. If unbalanced charges are there, add subscripts following the criss-cross method such that net charge for the compound is zero.
In the present case, we will add subscripts to both metal and non-metal such that the net charge becomes zero. So it becomes $ {(N{a^{1 + }})_2}{(SO_4^{2 - })_1} $ and now the net charge is balanced i.e. $ (2 \times ( + 1)) + (1 \times ( - 2)) = 0 $
Hence, the ionic compound formula for Sodium sulfate is $ N{a_2}S{O_4} $ .
Note :
While writing the chemical formula of an ionic compound, never write the subscript '1'. And it should be noted that we can also have two polyatomic ions like $ N{H_4}N{O_3} $ in the same compound. In this case, we have to find and write both of the names as identified from the Common Ion Table.
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