Find the oxidation number of F in HF.
A.\[ + 1\]
B.\[ - 1\]
C.\[0\]
D.\[ + 2\]
E.\[ + 3\]
Answer
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Hint: Let us first understand the concept of oxidation state or oxidation number. The oxidation state is also known as oxidation number which means the degree of oxidation (loss of electrons) of an atom in a chemical compound.
Complete answer:
We have to discuss some general rules to be followed when calculating the oxidation number of any element. They are discussed as follows:
(i) In a neutral compound, all oxidation numbers must add up to zero. A neutral compound does not have a plus or minus charge.
(ii)In an ion, all the oxidation numbers must add up to the charge of the ion.
(iii) The oxidation number of free elements is zero (e.g.\[Na,\]\[Fe,\]\[{H_2},\]\[{O_2},\]\[{S_8}\]).
(iv) Another straight-forward rule is Fluorine always has its oxidation number as \[ - 1\] as it is the most electronegative element and has one electron less to complete its octet.
(v)Elements in the Group 1 of periodic table have an oxidation number of \[ + 1\] and those of group 2 elements have an oxidation number of \[ + 2\].
(vi) Hydrogen when bonded to nonmetals has an oxidation number of +1 and hydrogen when bonded to metals(or Boron) have an oxidation number of \[ - 1\].
(vii) Oxygen usually has an oxidation state of $ - 2$ with two big exceptions, i.e, when bonded to Fluorine and in peroxides.
(viii)When none of the above rules seem applicable, we can use the group rules. In binary compounds with metals,
-Group 17 elements have oxidation state $ - 1$
-Group 16 elements have oxidation state $ - 2$
-Group 15 elements have oxidation state $ - 3$
As we know that the oxidation state of hydrogen is always $ + 1$ and applying the rule that Fluorine always its oxidation number as $ - 1$ as it the most electronegative element and has one electron less to complete its octet, it can be concluded that the oxidation number of fluorine in HF molecule is $ - 1$.
Therefore, the correct option is option B.
Note: We must know that the Hydrogen fluoride is a colourless liquid or gas which is the principal source of fluorine. HF is used as a super acid in the petrochemical industry. HF is prepared by the reaction of sulfuric acid on pure grades of the miner fluorite as,
$Ca{F_2} + {H_2}S{O_4} \to 2HF + CaS{O_4}$.
Complete answer:
We have to discuss some general rules to be followed when calculating the oxidation number of any element. They are discussed as follows:
(i) In a neutral compound, all oxidation numbers must add up to zero. A neutral compound does not have a plus or minus charge.
(ii)In an ion, all the oxidation numbers must add up to the charge of the ion.
(iii) The oxidation number of free elements is zero (e.g.\[Na,\]\[Fe,\]\[{H_2},\]\[{O_2},\]\[{S_8}\]).
(iv) Another straight-forward rule is Fluorine always has its oxidation number as \[ - 1\] as it is the most electronegative element and has one electron less to complete its octet.
(v)Elements in the Group 1 of periodic table have an oxidation number of \[ + 1\] and those of group 2 elements have an oxidation number of \[ + 2\].
(vi) Hydrogen when bonded to nonmetals has an oxidation number of +1 and hydrogen when bonded to metals(or Boron) have an oxidation number of \[ - 1\].
(vii) Oxygen usually has an oxidation state of $ - 2$ with two big exceptions, i.e, when bonded to Fluorine and in peroxides.
(viii)When none of the above rules seem applicable, we can use the group rules. In binary compounds with metals,
-Group 17 elements have oxidation state $ - 1$
-Group 16 elements have oxidation state $ - 2$
-Group 15 elements have oxidation state $ - 3$
As we know that the oxidation state of hydrogen is always $ + 1$ and applying the rule that Fluorine always its oxidation number as $ - 1$ as it the most electronegative element and has one electron less to complete its octet, it can be concluded that the oxidation number of fluorine in HF molecule is $ - 1$.
Therefore, the correct option is option B.
Note: We must know that the Hydrogen fluoride is a colourless liquid or gas which is the principal source of fluorine. HF is used as a super acid in the petrochemical industry. HF is prepared by the reaction of sulfuric acid on pure grades of the miner fluorite as,
$Ca{F_2} + {H_2}S{O_4} \to 2HF + CaS{O_4}$.
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