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)

Answer from: Quest

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

Answer from: Quest

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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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