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Determine the oxidation number of P in H3PO3, Mn in KMnO4, Mn in MnO2, N in...

Determine the oxidation number of P in H3PO3, Mn in KMnO4, Mn in MnO2, N in Ca(NO3)2, O in H2O2.

Solutions

Expert Solution

The general oxidation state of O ( except in peroxides,super oxides & fluorine oxides) is -2

The general oxidation state of H (except in metal hydrides) is +1

The general oxidation state of IA(Li,Na,K,Rb,Cs) elements is +1

The general oxidation state of IIA(Be,Mg,Ca,Sr,Ba) elements is +2

(1) Let a be the oxidation state of P in H3PO3

We know that the sum of the oxidation states of elements in a compound is equal to its charge.

Applying this we have    3(-1) + a + 3(-2) = 0              ( since it is a neutral molecule)

                                                            a = +3

Therefore the oxidation state of P in H3PO3 is +3

(2) Let a be the oxidation state of Mn in KMnO4

We know that the sum of the oxidation states of elements in a compound is equal to its charge.

Applying this we have    (+1) + a + 4(-2) = 0              ( since it is a neutral molecule)

                                                            a = +7

Therefore the oxidation state of Mn in KMnO4 is +7

(3)Let a be the oxidation state of Mn in MnO2

We know that the sum of the oxidation states of elements in a compound is equal to its charge.

Applying this we have     a + 2(-2) = 0              ( since it is a neutral molecule)

                                                    a = +4

Therefore the oxidation state of Mn in MnO2 is +4

(4) Let a be the oxidation state of N in Ca(NO3)2

We know that the sum of the oxidation states of elements in a compound is equal to its charge.

Applying this we have    (+2) +2[ a + 3(-2)] = 0              ( since it is a neutral molecule)

                                                              a = +5

Therefore the oxidation state of N in Ca(NO3)2 is +5

(5) Let a be the oxidation state of O in H2O2

We know that the sum of the oxidation states of elements in a compound is equal to its charge.

Applying this we have    2(+1) + 2(a) = 0              ( since it is a neutral molecule)

                                                     a = -1

Therefore the oxidation state of O in H2O2 is -1

queen_honey_blossom answered 5 months ago
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