What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3
The oxidation number of Tin (Sn) is +2. This is because it is equal to the charge on the ion. In this case Tin has a +2 charge.
Answer:
The oxidation number of tin (Sn) in the compound Na₂SnO₂ is +2 (option C)
Explanation:
First of all you should know that the oxidation state of a neutral element or molecule is 0. Then the sum of the oxidation states of the atoms that form a neutral molecule is zero.
On the other hand, the metals of group 1 (alkaline), as in the case of sodium Na, have an oxidation number of +1. And the oxidation state of oxygen in the compounds is -2.
Finally, you must take into account the amount of each element present in the compound, for which you observe the subindice.
Given this, it is possible to express:
2oxidation number of sodium+1oxidation number of tin+2oxidation number of oxigen=0
2(+1)+1oxidation number of tin+2(-2)=0
2+oxidation number of tin-4=0
-2+oxidation number of tin=0
oxidation number of tin=2
The oxidation number of tin (Sn) in the compound Na₂SnO₂ is +2 (option C)