Calculate the oxidation number of sulfur in na2s2o5. +5

We know that the oxidation number of sodium Na is +1, and oxygen O is -2, therefore:

2 (+1) + 2 x + 5 (-2) = 0

2 + 2x – 10 = 0

2x = 8

x = 4

The oxidation number is +4

The oxidation number of sulfur in Na2S2O5 is +4

HOW TO CALCULATE OXIDATION NUMBER:

Naturally;

Na2S2O5

+1 (2) + x(2) - 2(5) = 0

2 + 2x - 10 = 0

2 + 2x = 10

2x = 10 - 2

2x = 8

x = 4.

Therefore, the oxidation number of sulfur in Na2S2O5 is +4

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explanation:

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alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.