Sodium when burnt in excess of oxygen gives sodium oxide. If true enter 1, else enter 0.
Answer
668.1k+ views
Hint: Sodium is metal, which belongs to the group I-A in the periodic table. It is based on the delocalization of charge over the sodium atom, and compares the metals of group I-A. Then you can know, the product formed by burning the sodium.
Complete step-by-step answer:
First, we know that sodium is larger in size as compared to the lithium.
Thus, Lithium forms monoxide because there is delocalization of positive over the small size, it will not attract the another oxide ion, only once it will be attracted, whereas in case of sodium the charge is delocalized over the large size of atom.
So, sodium will permit the oxide ion to form a bond with the other oxide ion. Thus, a peroxide bond is formed in case of sodium.
Now, if there is burning of sodium in excess of oxygen, i.e. at high temperature firstly it will form sodium oxide, but instantly it absorbs the oxygen leading to the formation of a peroxide ion.
We can write the chemical reaction as:
2Na + O$_2$ (excess) $\rightarrow$ Na$_2$O$_2$
All other elements of group I-A leads to the formation of superoxide, due to the increase in size.
So, we can conclude that the sodium on burning gives sodium peroxide. Thus, the given statement is not true.
Hence, you have to enter 0.
Note: There can be a confusion that if it forms sodium oxide in the first stage then why the statement is false. As sodium is burnt in excess oxygen, and this reaction happens in milliseconds, and as mentioned it readily absorbs the oxygen. Therefore, it is considered to be false.
Complete step-by-step answer:
First, we know that sodium is larger in size as compared to the lithium.
Thus, Lithium forms monoxide because there is delocalization of positive over the small size, it will not attract the another oxide ion, only once it will be attracted, whereas in case of sodium the charge is delocalized over the large size of atom.
So, sodium will permit the oxide ion to form a bond with the other oxide ion. Thus, a peroxide bond is formed in case of sodium.
Now, if there is burning of sodium in excess of oxygen, i.e. at high temperature firstly it will form sodium oxide, but instantly it absorbs the oxygen leading to the formation of a peroxide ion.
We can write the chemical reaction as:
2Na + O$_2$ (excess) $\rightarrow$ Na$_2$O$_2$
All other elements of group I-A leads to the formation of superoxide, due to the increase in size.
So, we can conclude that the sodium on burning gives sodium peroxide. Thus, the given statement is not true.
Hence, you have to enter 0.
Note: There can be a confusion that if it forms sodium oxide in the first stage then why the statement is false. As sodium is burnt in excess oxygen, and this reaction happens in milliseconds, and as mentioned it readily absorbs the oxygen. Therefore, it is considered to be false.
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