
The non-metal that does not exhibit positive oxidation state is:
A: fluorine
B: oxygen
C: iodine
D: chlorine
Answer
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Hint: The oxidation state or sometimes known as oxidation number, defines the degree of oxidation that an atom possesses in a chemical compound. In other words, oxidation number refers to the charge left on the central atom especially when all the bonding pairs of electrons get broken, with the charge allocated to the most electronegative atom.
Complete answer
Let us check the oxidation state of each element as follows:
->Option A: Fluorine possesses the highest electronegativity and it is small in size. Moreover, it does not have a vacant d orbital so it cannot depict a positive oxidation state.
->Option B: Oxygen possesses an oxidation state of −1 in case of all peroxides (containing oxygen-oxygen linkage). But in case of Oxygen difluoride (compound is neutral), fluoride possesses an oxidation state of -1 as oxygen being less electronegative compared to fluorine and thus, oxygen possesses an oxidation state of +2.
->Option C: Iodine also possesses both negative and positive oxidation states. The polarization as well as electronegativity of iodine is mainly responsible for its negative oxidation state such as \[NaI,{\text{ }}AgI\]. On the other hand, iodine possesses a positive oxidation state (e.g \[I{F_5}\]) when it combines with higher electronegative atoms.
->Option D: Chlorine can possess an oxidation state of -1, 0, and +1, +3, +4, +5, +7, depending upon the substance comprising chlorine. Mostly, chlorine possess an oxidation state of -1 (such as in case of \[HCl{\text{ }} \,and\, {\text{ }}NaCl\]) as well as 0 (as in case of \[C{l_2}\]).
Hence, the correct answer is Option A.
Note:
The more electronegative element in a substance is always allotted a negative oxidation state whereas, the less electronegative element is allotted a positive oxidation state. Always remember the fact that electronegativity is greatest at the top-right in the periodic table which declines toward the bottom-left.
Complete answer
Let us check the oxidation state of each element as follows:
->Option A: Fluorine possesses the highest electronegativity and it is small in size. Moreover, it does not have a vacant d orbital so it cannot depict a positive oxidation state.
->Option B: Oxygen possesses an oxidation state of −1 in case of all peroxides (containing oxygen-oxygen linkage). But in case of Oxygen difluoride (compound is neutral), fluoride possesses an oxidation state of -1 as oxygen being less electronegative compared to fluorine and thus, oxygen possesses an oxidation state of +2.
->Option C: Iodine also possesses both negative and positive oxidation states. The polarization as well as electronegativity of iodine is mainly responsible for its negative oxidation state such as \[NaI,{\text{ }}AgI\]. On the other hand, iodine possesses a positive oxidation state (e.g \[I{F_5}\]) when it combines with higher electronegative atoms.
->Option D: Chlorine can possess an oxidation state of -1, 0, and +1, +3, +4, +5, +7, depending upon the substance comprising chlorine. Mostly, chlorine possess an oxidation state of -1 (such as in case of \[HCl{\text{ }} \,and\, {\text{ }}NaCl\]) as well as 0 (as in case of \[C{l_2}\]).
Hence, the correct answer is Option A.
Note:
The more electronegative element in a substance is always allotted a negative oxidation state whereas, the less electronegative element is allotted a positive oxidation state. Always remember the fact that electronegativity is greatest at the top-right in the periodic table which declines toward the bottom-left.
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