Determine the oxidation number of chromium in Cr2O7²⁻, NaCrO4, and Cr(s)

To determine the oxidation number of chromium in the given compounds, we will follow the rules for assigning oxidation states.

1. Cr2O7²⁻ (Dichromate Ion)

In the dichromate ion (Cr₂O₇)^2-, let’s assign oxidation states. Oxygen typically has an oxidation state of -2. Since there are 7 oxygens, their total contribution to the charge is 7 × (-2) = -14.

Letting the oxidation state of chromium be x, we write the equation for the total charge:

2x + (-14) = -2

Solving for x gives:

2x = -2 + 14
2x = 12
x = 6

Thus, the oxidation number of chromium in Cr₂O₇^2- is +6.

2. NaCrO4 (Sodium Chromate)

For sodium chromate (NaCrO₄), we again consider the oxidation states. Sodium (Na) has an oxidation state of +1, and oxygen is -2. The four oxygens contribute a total of 4 × (-2) = -8.

Letting the oxidation state of chromium be y, we set up the equation for the overall charge:

y + (-8) + 1 = 0

Solving for y gives:

y - 7 = 0
y = +6

Therefore, the oxidation number of chromium in NaCrO₄ is also +6.

3. Cr(s) (Solid Chromium)

In the case of solid chromium (Cr(s)), the element is in its standard state. Elements in their elemental form have an oxidation state of 0.

Therefore, the oxidation number of chromium in Cr(s) is 0.

Summary

To summarize, the oxidation numbers of chromium in the mentioned compounds are as follows:

• Cr₂O₇^2-: +6
• NaCrO₄: +6
• Cr(s): 0