Answered

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What is the oxidation state of nitrogen in the
compound NH4Br?
(1) –1 (3) –3
(2) +2 (4) +4

Sagot :

lulu1991
The answer is (3) -3. We can know that typically Br has the oxidation state of -1. And the H always has the oxidation state of +1. The total formula needs to have zero oxidation state. So the nitrogen is -3.

Answer : The oxidation number of nitrogen (N) 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.

The given compound is, [tex]NH_4Br[/tex]

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

[tex]x+4(+1)+(-1)=0\\\\x+4-1=0\\\\x+3=0\\\\x=-3[/tex]

Therefore, the oxidation number of nitrogen (N) is, (-3)

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