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Nitrogen and phosphorus are both in Group 15 of the periodic table. Phosphorus forms a chloride with the formula $P{Cl}_{5}$ but nitrogen does not form $N{Cl}_{5}$. Which of the statements help to examine this?
(1) Nitrogen's outer shell cannot contain more than eight electrons.
(2) Nitrogen cannot have an oxidation state of +5.
(3) Nitrogen is less electronegative than phosphorus.
(A) 1, 2 and 3 are correct
(B) 1 and 2 only are correct
(C) 2 and 3 only are correct
(D) 1 only is correct

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Last updated date: 21st Jul 2024
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Answer
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Hint: The atomic number of nitrogen is 7 and the atomic number of phosphorus is 15. They both belong to group 15 of the periodic table but are of different periods. Nitrogen is of period 2 and phosphorus is of period 3.

Complete step by step answer:
>The electronic configuration of nitrogen is $1{s}^{2}2{s}^{2}2{p}^{3}$. This shows that nitrogen can lose 5 electrons from its outer shell to form compounds and can gain 3 electrons to its outer shell. Therefore, it shows that nitrogen can only have 8 electrons in its outer shell. It also shows that nitrogen can have an oxidation state in the range -3 to +5.
> Whereas, if we talk about phosphorus, its electronic configuration is $1{s}^{2}2{s}^{2}2{p}^{6}2{s}^{2}3{p}^{3}$. This shows that phosphorus can also lose 5 electrons from its outer shell to form compounds and can gain 3 electrons to its outer shell.
- But when phosphorus forms a compound, it has enough energy that one electron jumps from the p-orbital to the vacant d-orbital. This gives phosphorus two extra valencies. This is called expanded octet and thus can form $P{Cl}_{5}$.
- But nitrogen does not have an empty d-orbital and thus, cannot form $N{Cl}_{5}$.
> As we know, that electronegativity decreases down a group due to the decrease in the atomic size of the elements down the group. This means that nitrogen is more electronegative than phosphorus.
- Thus, considering the above points, we can say that statements 2 and 3 from that question are incorrect. Thus, only statement 1 is correct.
Hence the correct option is option (d).

Note: It is important to note that the expanded octet occurs only in the elements of the ${3}^{rd}$ period. Period 2 elements cannot have an expanded orbital due to the unavailability $3d$ subshell .