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What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3

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Answer:

The oxidation number of tin (Sn) in the compound Na₂SnO₂ is +2 (option C)

Step-by-step 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:

2*oxidation number of sodium+1*oxidation number of tin+2*oxidation number of oxigen=0

2*(+1)+1*oxidation 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)

User Jsegal
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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.
User Joe Erinjeri
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