Question

What is the oxidation state of chromium in the compound K2Cr2O7?

Answer 1

• +6

Explanation:

Given compound:

• 
  `\sf K_2Cr_2O_7`

We have to find the oxidation state of chromium (Cr). Let us assume oxidation state of chromium be x.

Then in
`\sf K_2Cr_2O_7`

Oxidation state of K (potassium) is +1 and for two molecules of K, oxidation state will be +2 and oxidation state of O (oxygen) is -2 and for 7 molecules of O, oxidation state will be -14

`\sf 1(2) + 2(x) + -2(7) = 0`

`\sf 2 + 2x + (-14) = 0`

`\sf 2 + 2x - 14 = 0`

`\sf -12 + 2x = 0`

Add 12 on both sides

`\sf -12 + 2x +12 = 0 + 12`

`\sf 2x = 12`

Divide both sides by 2

`\sf x = + 6`

Hence the oxidation state of chromium in the compound
`\sf K_2Cr_2O_7` is +6.

Answer 2

Oxidation state of chromium in
`\sf K_2Cr_2O_7` is +6.

Step-by-step explanation:

In
`\sf K_2Cr_2O_7`, there are two chromium atoms, two potassium atoms, and seven oxygen atoms.

The oxidation state of potassium is +1, the oxidation state of oxygen is -2, and the overall charge of the compound is 0.

In this case:

Given:

Oxidation state of potassium (K) = +1

Oxidation state of oxygen (O) = -2

Overall charge of the compound = 0 (since it's neutral)

We can use the following equation based on the oxidation states and the charge:

`\sf \textsf{(Oxidation state of K) }+ (2 * \textsf{Oxidation state of Cr)}+\textsf{ (Oxidation state of O) }= 0`

Plugging in the values for potassium and oxygen:

`\sf (2 * +1) + (2 * \textsf{Oxidation state of Cr) }+ (7 * -2) = 0`

Solving for the oxidation state of chromium:

`\sf 2 + 2\textsf{(Oxidation state of Cr) }- 14 = 0`

`\sf 2\textsf{(Oxidation state of Cr) }= 12`

`\sf \sf \textsf{(Oxidation state of Cr) }=( 12)/(2)`

`\sf \textsf{Oxidation state of Cr }= +6`

Therefore, in
`\sf K_2Cr_2O_7`, the oxidation state of chromium is +6.