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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

Last updated date: 22nd Feb 2024
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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 .