$PCl_5$ is well known, but $NCl_5$ is not. Because,

Question

Which law is illustrated by compounds H$_2$O and H$_2$O$_2$ formed from two different elements, H and O?

Answer 1

To understand why $NCl_5$ is not known while $PCl_5$ is well known, we need to consider the electronic configurations and the availability of orbitals in the valence shell of nitrogen and phosphorus.

Electronic Configuration:
- Nitrogen ($N$): The electronic configuration is $1s^2 2s^2 2p^3$. The valence shell comprises 2s and 2p orbitals.
- Phosphorus ($P$): The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^3$. In the valence shell, it has 3s, 3p, and access to 3d orbitals.
Valency and Orbital Availability:
- Nitrogen does not have d-orbitals in its valence shell. Therefore, it cannot expand its octet to form compounds like $NCl_5$. Typically, nitrogen can form three covalent bonds using its three unpaired p-orbitals (as in $NH_3$).
- Phosphorus, on the other hand, has access to vacant d-orbitals, which allows it to expand its valency beyond 3 to 5. This enables phosphorus to form $PCl_5$ by using the d-orbitals to accommodate five covalent bonds.
Conclusion: Based on the above explanation, the correct answer is: "nitrogen doesn’t have d-orbitals in its valence shell." This absence prevents nitrogen from forming $NCl_5$.

Thus, phosphorus can participate in a greater range of chemical bonding due to the availability of d-orbitals compared to nitrogen, which is limited by not having such orbitals.

Answer 2

To understand why $NCl_5$ is not known while $PCl_5$ is well known, we need to consider the electronic configurations and the availability of orbitals in the valence shell of nitrogen and phosphorus.

Electronic Configuration:
- Nitrogen ($N$): The electronic configuration is $1s^2 2s^2 2p^3$. The valence shell comprises 2s and 2p orbitals.
- Phosphorus ($P$): The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^3$. In the valence shell, it has 3s, 3p, and access to 3d orbitals.
Valency and Orbital Availability:
- Nitrogen does not have d-orbitals in its valence shell. Therefore, it cannot expand its octet to form compounds like $NCl_5$. Typically, nitrogen can form three covalent bonds using its three unpaired p-orbitals (as in $NH_3$).
- Phosphorus, on the other hand, has access to vacant d-orbitals, which allows it to expand its valency beyond 3 to 5. This enables phosphorus to form $PCl_5$ by using the d-orbitals to accommodate five covalent bonds.
Conclusion: Based on the above explanation, the correct answer is: "nitrogen doesn’t have d-orbitals in its valence shell." This absence prevents nitrogen from forming $NCl_5$.

Thus, phosphorus can participate in a greater range of chemical bonding due to the availability of d-orbitals compared to nitrogen, which is limited by not having such orbitals.

\(PCl_5\) is well known, but \(NCl_5\) is not. Because,

The Correct Option is B

Solution and Explanation

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