Was ist die Oxidationszahl von Na?

Antworten:

Elementares Natrium hat eine Oxidationszahl von $0$ $0$ ... wie bei JEDEM Element ....

Erläuterung:

Die Oxidationszahl ist die Ladung, die mit einem bestimmten Atom verbunden ist, wenn es Elektronen nach vorgegebenen Regeln abgibt oder annimmt.

$1. The oxidation number of a free element is always 0.$ $"1. The oxidation number of a free element is always 0."$

$2. The oxidation number of a mono-atomic ion is equal$ $"2. The oxidation number of a mono-atomic ion is equal"$ $to the charge of the ion.$ $"to the charge of the ion."$

$3. For a given bond, X-Y, the bond is split to give X^{+}$ $"3. For a given bond, X-Y, the bond is split to give "X^+$ $and$ $"and"$ $Y^{-}$ $Y^-$ , $where Y is more electronegative than X.$ $"where Y is more electronegative than X."$

$4. The oxidation number of H is +1, but it is -1 in when$ $"4. The oxidation number of H is +1, but it is -1 in when"$ $combined with less electronegative elements.$ $"combined with less electronegative elements."$

$5. The oxidation number of O in its$ $"5. The oxidation number of O in its"$ Verbindungen $is usually -2, but it is -1 in peroxides.$ $"is usually -2, but it is -1 in peroxides."$

$6. The oxidation number of a Group 1 element$ $"6. The oxidation number of a Group 1 element"$ $in a compound is +1.$ $"in a compound is +1."$

$7. The oxidation number of a Group 2 element in$ $"7. The oxidation number of a Group 2 element in"$ $a compound is +2.$ $"a compound is +2."$

$8. The oxidation number of a Group 17 element in a binary compound is -1.$ $"8. The oxidation number of a Group 17 element in a binary compound is -1."$

$9. The sum of the oxidation numbers of all of the atoms$ $"9. The sum of the oxidation numbers of all of the atoms"$ $in a neutral compound is 0.$ $"in a neutral compound is 0."$

$10. The sum of the oxidation numbers in a polyatomic ion$ $"10. The sum of the oxidation numbers in a polyatomic ion"$ $is equal to the charge of the ion.$ $"is equal to the charge of the ion."$

Und so bekommen wir für elementares Natrium $N a (0)$ $Na(0)$ Natrium ist ein gutes Reduktionsmittel und für den größten Teil seiner Reaktionschemie erhalten wir $N a^{+} \equiv N a (I +)$ $Na^+ -=Na(I+)$ ... In dem gegebenen Szenario hat das Metall sein einwertiges Elektron NIEMALS angenommen oder abgegeben ...