Answer
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Hint: The oxidation number of an element depends on the electronic configuration of the element. The number of electrons required to complete the octet or the number of electrons lost to acquire the nearest noble gas configuration is the oxidation state. Elements show variable oxidation states and these variable oxidation states depend on certain conditions.
Complete step by step answer:
Phosphorus belongs to group 15, which is the nitrogen family.
It is a very reactive element and it is not found free. It is always found in combined states with other elements.
It is the eleventh most abundant element in the earth’s crust.
Its atomic number is 15.
Hence, its electronic configuration will be\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}\]
So, with the outer electronic configuration, we can see that the last shell number is 3.
And the outer shell has 5 valence electrons, 3 valence electrons in p-orbital, and 2 valence electrons in s-orbital.
So, to acquire the nearest noble gas configuration it can either gain 3 electrons or donate 5 electrons.
Although the gain of three electrons from more metallic elements to form \[{{P}^{3-}}\] ions requires a large amount of energy. Hence, phosphorus only form -3 oxidation state in calcium phosphide (\[C{{a}_{3}}{{P}_{2}}\]) and phosphine (\[P{{H}_{3}}\]).
It also shows the -2 oxidation state in diphosphine (\[{{P}_{2}}{{H}_{4}}\]).
Phosphorus can show positive oxidation till 5, by losing all the 5 electrons of the outer shell.
It shows +1 oxidation state in phosphinic acid (\[{{H}_{3}}P{{O}_{2}}\]), +3 in Orthophosphorus acid (\[{{H}_{3}}P{{O}_{3}}\]) and pyrophosphorus acid (\[{{H}_{4}}{{P}_{2}}{{O}_{5}}\]), +4 in hypophosphoric acid (\[{{H}_{4}}{{P}_{2}}{{O}_{6}}\]) and +5 in many oxides of phosphorus.
Hence, we can conclude that the oxidation state of phosphorus varies from -3 to +5.
The correct option is (a)- -3 to +5
Note: Gaining of three electrons requires a very high amount of energy, that's why phosphorus doesn't show this oxidation state in many compounds but nitrogen of this group can easily form \[{{M}^{3-}}\]because of its high electronegativity and small size.
Complete step by step answer:
Phosphorus belongs to group 15, which is the nitrogen family.
It is a very reactive element and it is not found free. It is always found in combined states with other elements.
It is the eleventh most abundant element in the earth’s crust.
Its atomic number is 15.
Hence, its electronic configuration will be\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}\]
So, with the outer electronic configuration, we can see that the last shell number is 3.
And the outer shell has 5 valence electrons, 3 valence electrons in p-orbital, and 2 valence electrons in s-orbital.
So, to acquire the nearest noble gas configuration it can either gain 3 electrons or donate 5 electrons.
Although the gain of three electrons from more metallic elements to form \[{{P}^{3-}}\] ions requires a large amount of energy. Hence, phosphorus only form -3 oxidation state in calcium phosphide (\[C{{a}_{3}}{{P}_{2}}\]) and phosphine (\[P{{H}_{3}}\]).
It also shows the -2 oxidation state in diphosphine (\[{{P}_{2}}{{H}_{4}}\]).
Phosphorus can show positive oxidation till 5, by losing all the 5 electrons of the outer shell.
It shows +1 oxidation state in phosphinic acid (\[{{H}_{3}}P{{O}_{2}}\]), +3 in Orthophosphorus acid (\[{{H}_{3}}P{{O}_{3}}\]) and pyrophosphorus acid (\[{{H}_{4}}{{P}_{2}}{{O}_{5}}\]), +4 in hypophosphoric acid (\[{{H}_{4}}{{P}_{2}}{{O}_{6}}\]) and +5 in many oxides of phosphorus.
Hence, we can conclude that the oxidation state of phosphorus varies from -3 to +5.
The correct option is (a)- -3 to +5
Note: Gaining of three electrons requires a very high amount of energy, that's why phosphorus doesn't show this oxidation state in many compounds but nitrogen of this group can easily form \[{{M}^{3-}}\]because of its high electronegativity and small size.
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