What is the oxidation number of sodium (Na) in pure sodium, Na?

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What is the oxidation number of sodium (Na) in pure sodium, Na?

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Answer 1 · 2016-07-09T05:54:57

In sodium metal, the oxidation number is $0$ $0$ . Elements are $zerovalent$ $"zerovalent"$ .

Explanation:

Redox reaction are conceived to involve the transfer of electrons. If a species gains electrons it is said to have been $REDUCED$ $"REDUCED"$ . If a species loses electrons is is said to have been $OXIDIZED$ $"OXIDIZED"$ . The oxidation number of an element in its standard state is thus manifestly $ZERO$ $"ZERO"$ . What are the oxidation numbers of $O_{2}$ $O_2$ , and $F_{2}$ $F_2$ , and $C a$ $Ca$ ?

The notions of oxidation and reduction have been flying pretty fast at you. Remember that these are conceptual notions, that do not have real significance. Assignment of oxidation numbers allows us to balance complicated redox equations. So it might be an idea to suspend your disbelief when we say that $M n^{2 +}$ $Mn^(2+)$ loses 5 electrons when it is oxidized to permanaganate ion, $M n (V I I +)$ $Mn(VII+)$ , and focus on the stoichiometry.

$M n^{2 +} + 4 H_{2} O \to M n O_{4}^{-} + 8 H^{+} + 5 e^{-}$ $Mn^(2+)+4H_2OrarrMnO_4^(-)+8H^(+)+5e^(-)$

Are mass and charge conserved here? If not, they should be.

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What is the oxidation number of sodium (Na) in pure sodium, Na?

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