Step 1: Analyzing the Assertion (A):
- Phosphorus (P) is in the third period. Its electronic configuration is \([Ne] 3s^2 3p^3\). It also has vacant 3d orbitals. To form five bonds (as in PCl₅), phosphorus can promote one of its 3s electrons to an empty 3d orbital, resulting in five unpaired electrons and allowing it to exhibit a +5 oxidation state. It can also form PCl₃ using its three unpaired p-electrons (+3 state). So, the statement that phosphorus forms PCl₃ and PCl₅ is correct.
- Nitrogen (N) is in the second period. Its electronic configuration is \([He] 2s^2 2p^3\). Nitrogen does not have any d-orbitals in its valence shell (n=2). Therefore, it cannot expand its octet by promoting an electron to a higher energy orbital. It can only form a maximum of four covalent bonds (e.g., in NH₄⁺ using its lone pair). It cannot form five covalent bonds, so NCl₅ does not exist.
- Therefore, Assertion (A) is a correct statement.
Step 2: Analyzing the Reason (R):
- The statement "The electronegativity of nitrogen is more than that of phosphorus" is a correct fact. Electronegativity generally decreases down a group in the periodic table. Nitrogen is above phosphorus in Group 15, so it is more electronegative.
- Therefore, Reason (R) is also a correct statement.
Step 3: Evaluating if (R) explains (A):
- The reason why nitrogen cannot form NCl₅ is the absence of vacant d-orbitals in its valence shell, which prevents it from expanding its octet to accommodate five bonds.
- The reason given, that nitrogen is more electronegative than phosphorus, is not the explanation for its inability to form a pentachloride. Electronegativity affects the polarity and nature of the bonds formed, but the primary constraint on the number of bonds is the availability of orbitals for bonding.
- Thus, while both (A) and (R) are individually correct statements, (R) is not the correct explanation for (A).
Step 4: Final Answer:
Both (A) and (R) are correct, but (R) is not the correct explanation of (A). This corresponds to option (C).