
An element, which never as a positive oxidation state in any of its compounds is:
(A) boron
(B) oxygen
(C) chlorine
(D) fluorine
Answer
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Hint: Oxidation state, also known as oxidation number it decreases. The degree of oxidation (loss of electron) of an atom in a chemical compound.
Complete step by step answer:
The Rules for finding oxidation state are as follows:
(1) The oxidation number of free elements is always zero.
For example: the oxidation number of $Na,K,{O_2},{N_2},{O_3}$is zero.
(2) Oxidation number of monatomic ions equal to charge on ion.
Example: Oxidation number of $N{a^ + } = ( + 1)$
$C{l^ - } = ( - 1)$
$O{H^ - } = ( - 1)$
(3) Oxidation number of H is +1 but it is -1 when combined with less electronegative elements.
Example: In ${H_2}O \to $it is $ + 1$
In $LiH \to $it is $ - 1$
(4) Oxidation number of O in compounds is (-2) but it is (-1) in peroxide.
Oxidation numbers also provide information about the ability of an element to lose or gain electrons during chemical reaction.
Let us discuss the oxidation state of the following element.
Boron: Electronic configuration is 2,3 loses 3 electrons.
Therefore, Oxidation state is +3.
Oxygen molecule: Electronic configuration is 2,6.
Accepts two electrons to form bonds.
Therefore, Oxidation state is (-2). It forms oxides.
But if it reacts with fluorine, forms ${F_2}O$
Oxidation state is (+2)
Chlorine molecule: Electronic configuration 2,8,7
It accepts one electron to form bonds.
Therefore, Oxidation state is $ = ( - 1)$
But if reacts with fluorine to form $Cl{F_3},Cl{F_3}$in which oxidation state is (+3) and (+5) fluorine is more electronegative than chlorine atom.
Fluorine: Electronic configuration 2,7
It accepts one electron form bond
Therefore, oxidation state is $ = ( - 1)$
Fluorine has the smallest atomic size and greatest electronegativity in the periodic table. Therefore, it always shows a negative oxidation state.
Therefore, from the above explanation is the correct option is (D) Fluorine.
Note:
Element shows positive oxidation state only when it combines with a more electronegative atom. Fluorine has the highest electronegativity therefore it never shows positive oxidation state.
Complete step by step answer:
The Rules for finding oxidation state are as follows:
(1) The oxidation number of free elements is always zero.
For example: the oxidation number of $Na,K,{O_2},{N_2},{O_3}$is zero.
(2) Oxidation number of monatomic ions equal to charge on ion.
Example: Oxidation number of $N{a^ + } = ( + 1)$
$C{l^ - } = ( - 1)$
$O{H^ - } = ( - 1)$
(3) Oxidation number of H is +1 but it is -1 when combined with less electronegative elements.
Example: In ${H_2}O \to $it is $ + 1$
In $LiH \to $it is $ - 1$
(4) Oxidation number of O in compounds is (-2) but it is (-1) in peroxide.
Oxidation numbers also provide information about the ability of an element to lose or gain electrons during chemical reaction.
Let us discuss the oxidation state of the following element.
Boron: Electronic configuration is 2,3 loses 3 electrons.
Therefore, Oxidation state is +3.
Oxygen molecule: Electronic configuration is 2,6.
Accepts two electrons to form bonds.
Therefore, Oxidation state is (-2). It forms oxides.
But if it reacts with fluorine, forms ${F_2}O$
Oxidation state is (+2)
Chlorine molecule: Electronic configuration 2,8,7
It accepts one electron to form bonds.
Therefore, Oxidation state is $ = ( - 1)$
But if reacts with fluorine to form $Cl{F_3},Cl{F_3}$in which oxidation state is (+3) and (+5) fluorine is more electronegative than chlorine atom.
Fluorine: Electronic configuration 2,7
It accepts one electron form bond
Therefore, oxidation state is $ = ( - 1)$
Fluorine has the smallest atomic size and greatest electronegativity in the periodic table. Therefore, it always shows a negative oxidation state.
Therefore, from the above explanation is the correct option is (D) Fluorine.
Note:
Element shows positive oxidation state only when it combines with a more electronegative atom. Fluorine has the highest electronegativity therefore it never shows positive oxidation state.
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