How do you calculate the oxidation number of $N a_{2} S O_{4}$ $Na_2SO_4$ ?

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How do you calculate the oxidation number of $N a_{2} S O_{4}$ $Na_2SO_4$ ?

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Answer 1 · 2016-08-15T13:32:38

One Sodium (+1) and one sulfate (-2)

Explanation:

Na2(+1) = 21 = 2
SO4(-2) = 1-2 =-2.

It means stable.

Answer 2 · 2016-08-15T13:43:07

Add up all of oxidation numbers of atoms and ions in the compound the result ( zero) is the oxidation number of the compound.

Explanation:

Na number 11 has an oxidation of +1
Sodium wants to lose one electron creating the electron structure of Neon which is stable.

Oxygen number 8 has an oxidation number of -2 Oxygen has an electro negativity of 3.5 the second highest of all elements. So Oxygen always gains ( takes) two electrons creating the electron structure of Neon which is stable.

Sulfur number 16 has oxidation numbers of -2, +2 +4 and +6.
When combined in the $S O 4^{- 2}$ $SO4^-2$ sulfur has a lower electro negativity than oxygen so will lose electrons to oxygen causing sulfur to be positive.

$S + 4 \times (- 2)$ $S + 4 xx (-2)$ == S + -8 = -2

S = +6

The $S O 4^{- 2}$ $SO4^-2$ sulfate ion is saturated containing the maximum number of oxygen so has the highest oxidation number for sulfur.

So $2 \times (+ 1) + 4 \times (- 2) + (+ 6)$ $2 xx(+1) + 4 xx(-2) + (+6)$ = 0

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How do you calculate the oxidation number of $N a_{2} S O_{4}$ $Na_2SO_4$ ?

2 Answers

Explanation:

Explanation:

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How do you calculate the oxidation number of $N a_{2} S O_{4}$ $Na_2SO_4$ ?