Question

When sodium reacts with excess of oxygen, the oxidation number of oxygen changes from:
\[A.\, \,0\,to\, - 1\]
\[B.\, \,0\,to\, - 2\]
\[C.\, \, - 1\,to\, - 2\]
\[D.\, \,no\,change\]

Answer 1

Hint:When the sodium reacts with the excess amount of the oxygen It leads to the formations of metal oxide that is the sodium peroxide. When the sodium is heated and put into a jar containing the oxygen it will lead to generating a lot of light and heat energy, and it forms white powder after the reaction i.e., sodium oxide.


Complete step by step solution:
The sodium metal is more reactive than the lithium (\[Li\]), and it readily reacts with water (\[{H_2}O\] ) to form a sodium hydroxide (\[NaOH\]) which is one the strong base and it is used in labs, industries, etc. This reaction is highly exothermic that liberates heat.
The equation is given below;
\[2Na\, + \,{H_2}O\, \to \,NaOH\, + \,\frac{1}{2}\,{H_2}\]
When the sodium (\[Na\]) metal is reacted with an excess amount of air or oxygen(\[{O_2}\] ) , it produces sodium peroxide.
The equation is given below;
\[2Na\, + \,{O_2}\, \to \,N{a_2}O_2^{1 - }\]
Let’s write the charge for each element,
\[2N{a^{1 + }}\, + \,O_2^0\, \to \,N{a_2}O_2^{1 - }\]
So, here the oxidation number of oxygen is getting changed from \[\] .

Therefore, the correct answer is option\[A.\,\,0\,to\, - 1\] .


Additional information:
When sodium metal reacts with oxygen\[{O_2}\] at the temperature of \[200 - {\text{ }}350{\text{ }}^\circ C\] , it produces sodium oxide \[N{a_2}O\]
\[4Na\, + \,{O_2}\, \to \,2N{a_2}O\]
When sodium metal reacts with excess oxygen \[{O_2}\] at the temperature of \[350 - {\text{ }}450{\text{ }}^\circ C\] , it produces sodium peroxide \[N{a_2}{O_2}\]
\[2N{a_2}O\, + \,{O_2}\, \to \,2N{a_2}{O_2}\]

Note:The sodium peroxide \[N{a_2}{O_2}\] is an inorganic compound which is yellowish in colour. In sodium oxide\[N{a_2}O\] , the oxygen is having \[ - 2\] , which is one of the very strong bases and has a high tendency to combine with the ions hydrogen.
When sodium peroxide \[N{a_2}{O_2}\] reacts with excess oxygen \[{O_2}\] at the high pressure, it produces sodium superoxide\[Na{O_2}\]
\[2N{a_2}{O_2}\, + \,{O_2}\, \to \,2Na{O_2}\]