Question

What is the oxidation number of chromium in Cr2O7 4-?

Answer 1

Answer: The oxidation state of chromium in the given compound is +5.

Explanation:

Oxidation state is defined as the number which is given to an atom when it looses or gains electron. It is written as a superscript.

If the element gains electron, it will attain a negative oxidation state and if the element looses electrons, it will attain a positive oxidation state.

We are given a chemical compound having chemical formula of $Cr_2O_7^{4-}$

We take the oxidation state of chromium atom be 'x'.

Oxidation state of oxygen atom = -2

Overall charge on chemical compound = -4

Evaluating the oxidation state of chlorine atom:

$2(x)+7(-2)=-4\\\\x=+5$

Hence, the oxidation state of chromium in the given compound is +5.

Answer 2

Answer:

C. +5

Explanation:

Oxygen is always -2. And the sum of the oxidation numbers must be equal to the charge. So:

2x + 7(-2) = -4

2x - 14 = -4

2x = 10

x=5