Chemistry, 05.03.2020 01:53 sandy6265

Use the rules (in order) to assign oxidation numbers to each of the elements in the compounds below.

1. hydrogen phosphate ion HPO32-
H P O
2. aluminum oxide Al2O3
Al O
3. periodic acid HIO4
H I O

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Answer from: railuke7721

Answer :

(1) The oxidation number of P is, (+3)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

(2) The oxidation number of Al is, (+3)

The oxidation number of O is, (-2)

(3) The oxidation number of I is, (+7)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

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.

(1) The given compound is, $HPO_3^{2-}$

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

$(+1)+x+3(-2)=-2\\\\x=+3$

The oxidation number of P is, (+3)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

(2) The given compound is, Al_2O_3

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

$2x+3(-2)=0\\\\x=+3$

The oxidation number of Al is, (+3)

The oxidation number of O is, (-2)

(3) The given compound is, HIO_4

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

$(+1)+x+4(-2)=0\\\\x=+7$

The oxidation number of I is, (+7)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

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Use the rules (in order) to assign oxidation numbers to each of the elements in the compounds below....

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