The oxidation variety of Tin (Sn) is +2. This is since it is equal to the fee on the ion. In this situation Tin has a +2 charge.

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

Explanation:

First of all you should know that the oxidation state that a neutral element or molecule is 0. Climate the sum of the oxidation claims of the atom that kind a neutral molecule is zero.

On the various other hand, the metals of group 1 (alkaline), together in the case of sodium Na, have an oxidation variety of +1. And the oxidation state the oxygen in the compound is -2.

Finally, you must take right into account the amount of each facet present in the compound, because that which you watch the subindice.

Given this, the is feasible to express:

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

2*(+1)+1*oxidation number of tin+2*(-2)=0

2+oxidation number of tin-4=0

-2+oxidation variety of tin=0

oxidation variety of tin=2

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

alternative (C) is the correct answer.

Explanation:

It is recognized that oxidation variety of sodium is +1 and the oxidation number of oxygen is -2.

Let the oxidation variety of tin (Sn) is x.

Therefore, calculation the oxidation variety of tin in the provided formula together follows.

= (2 imes oxidation variety of Na) + Oxidation variety of Sn + (2 imes oxidation number of O)

=

So, 2 + x - 4 = 0

therefore, x = +2

Thus, we have the right to conclude the the oxidation variety of tin (Sn) in the link is +2.

O.N. Of Na = +1

O.N. The O = -2

Let, O.N. Of believe = x

1*2 + x + -2*2 = 0

2+x-4 = 0

x-2 = 0

x = 2

SO alternative C IS her ANSWER

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