Question

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Answer 1

The oxidation state of O in OF₂ is -2

The oxidaiton state of C in CO is +2

The oxidation state of N in K₃N is -3

Step-by-step explanation:

In the OF₂ molecule, fluorine acts with -1 in this compound as the molecule has 2 atoms of F, the oxygen must act with -2.

Oxygen usually has -2 as oxidation state, except for the peroxides.

In CO, the oxygen has -2 as oxidation state so C works with +2

C usually has +4 as oxidation state but it can act with +2 and -4

For K₃N: oxidation state of K is +1,but we have 3 atoms (+3). Therefore the N must work with -3 as oxidation state. Nitrogen has many oxidation states, as the halogens (except for the F). It can work with 1, 2, +-3, 4 and 5.

Remember that the global charge of the molecule is 0

Answer 2

The oxidation state of oxygen (O) in
`OF_2` is, (+2)

The oxidation state of carbon (C) in
`CO` is, (+2)

The oxidation state of nitrogen (N) in
`K_3N` is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

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

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

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

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

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

(a) The given compound is,
`OF_2`

Let the oxidation state of 'O' be, 'x'

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

The oxidation state of oxygen (O) in
`OF_2` is, (+2)

(b) The given compound is,
`CO`

Let the oxidation state of 'C' be, 'x'

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

The oxidation state of carbon (C) in
`CO` is, (+2)

(c) The given compound is,
`K_3N`

Let the oxidation state of 'N' be, 'x'

`3(+1)+x=0\\\\3+x=0\\\\x=-3`

The oxidation state of nitrogen (N) in
`K_3N` is, (-3)