Q. Oxidation number of Na.

Q: What is the oxidation number of sodium in compounds and in \text{Na} metal?

In elemental sodium \text{Na}, the oxidation number is 0. In essentially all compounds, sodium is +1 (because it forms \text{Na}^+).

Q: Why is sodium’s oxidation number always +1 in ionic compounds?

Sodium typically loses one valence electron to form \text{Na}^+. That means it has a charge of +1, so its oxidation number is +1.

Q: Find the oxidation number of Na in \text{NaCl}.

Chlorine is -1 in metal chlorides. Since the compound is neutral, sodium must be +1. So \text{Na} is +1 in \text{NaCl}.

Q: Find the oxidation number of Na in \text{Na}_2\text{SO}_4.

Sulfate \text{SO}_4 has charge -2. Two sodium ions balance it: 2\times(+1)+(-2)=0. So Na is +1.

Q: What is the oxidation number of Na in \text{Na}_2\text{O}?

Oxygen is -2 in oxides. The compound is neutral, so two Na atoms together must be +2. Therefore each Na has oxidation number +1.

Q: Can sodium have oxidation number other than +1 besides free metal?

For most chemistry problems, no. Sodium is +1 in its compounds and 0 in elemental form. Unusual cases are not typical at high school level.

Answer

In the neutral compound Na, the oxidation number of sodium (Na) is always \(+1\).

Detailed Explanation

Goal: Find the oxidation number of the element sodium, Na, in general.

Step 1: Recall the key rule for Group 1 metals.

Sodium (Na) is a Group 1 element. The standard rule in redox/oxidation-number problems is:

\(\text{Group 1 metals (including Na) have oxidation number } +1\) in their compounds.

Step 2: Apply the rule to sodium.

Therefore, the oxidation number of sodium is:

\[\boxed{+1}\]

Step 3: Note common exceptions (for completeness).

Oxidation numbers can change only in special cases, such as elemental form. But the question asks for oxidation number of Na (the usual meaning is in compounds).

In compounds: \(\boxed{+1}\)
In elemental sodium metal \(\left(\text{Na(s)}\right)\): \(\boxed{0}\)

Final answer: \(\boxed{+1}\) (for Na in compounds).

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General Chemistry FAQs

What is the oxidation number of sodium in compounds and in \(\text{Na}\) metal?

In elemental sodium \(\text{Na}\), the oxidation number is \(0\). In essentially all compounds, sodium is \(+1\) (because it forms \(\text{Na}^+\)).

Why is sodium’s oxidation number always \(+1\) in ionic compounds?

Sodium typically loses one valence electron to form \(\text{Na}^+\). That means it has a charge of \(+1\), so its oxidation number is \(+1\).

Find the oxidation number of Na in \(\text{NaCl}\).

Chlorine is \(-1\) in metal chlorides. Since the compound is neutral, sodium must be \(+1\). So \(\text{Na}\) is \(+1\) in \(\text{NaCl}\).

Find the oxidation number of Na in \(\text{Na}_2\text{SO}_4\).

Sulfate \(\text{SO}_4\) has charge \(-2\). Two sodium ions balance it: \(2\times(+1)+(-2)=0\). So Na is \(+1\).

What is the oxidation number of Na in \(\text{Na}_2\text{O}\)?

Oxygen is \(-2\) in oxides. The compound is neutral, so two Na atoms together must be \(+2\). Therefore each Na has oxidation number \(+1\).

Can sodium have oxidation number other than \(+1\) besides free metal?

For most chemistry problems, no. Sodium is \(+1\) in its compounds and \(0\) in elemental form. Unusual cases are not typical at high school level.

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