Question

oxidation number of S in Na2SO4​

Answer 1

Explanation:

Inside” the sodium sulfate, each sodium has an oxidation state of +1, the sulfur a +6 and each oxygen a -2.In Na2S4O6, the oxidation number of end sulphur atoms is +5 each and the oxidation number of middle sulphur atoms is 0 each.

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Answer 2

$\underline\mathfrak{ANSWER:-}$

In Na2SO4……..Na and O molecule doesnt have variable oxidation state or no.

In different compound….but S has variable oxidation state…..so to find out the correct oxidation no. For each molecule…….

We know that.this compound has no charge….therefore…..it is neutral compound..

So..Na = +1 ,O2 = -2 let ….S be x…

Since we hv only 2 moleculeof Na.but four moleculeof O2….and compound is neural so charge will be = 0.

Eqn will be…….

✬━━━━━━━━━━━━━━✬

= (+1)× 2 + (-2) × 4 + x = 0

= +2 + (-8) + x = 0

= -6 + x = 0

X = +6.

✬━━━━━━━━━━━━━━✬

So oxidation state of Na2 is +1….. O4 is -8 and S is +6.