Question:

In solid state, PCl5 is a/an

Updated On: Apr 10, 2025
  • Octahedral structure
  • Ionic solid with [PCl6]+ and [PCl4]-
  • Ionic solid with [PCl4]+ and [PCl6]-
  • Covalent solid present in the form of P2Cl10
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The Correct Option is C

Approach Solution - 1

In solid state, PCl5 forms an ionic compound. The phosphorus atom (P) has an empty 3d orbital and can expand its octet to form a compound with more than four bonds. In PCl5, it forms five bonds with chlorine atoms, but in the solid state, it forms an ionic compound by losing one electron to form a cation [PCl4]+ and gaining an electron to form an anion [PCl6]-.

Why is option (C) correct?

  • PCl5 forms an ionic compound in the solid state.
  • The cation formed is [PCl4]+ and the anion formed is [PCl6]-.

Why are the other options incorrect?

  • Option (A): PCl5 does not have an octahedral structure in the solid state.
  • Option (B): The ions are reversed. The correct ions are [PCl4]+ and [PCl6]-.
  • Option (D): PCl5 does not form a covalent solid in the form of P2Cl10.

Conclusion: The correct answer is (C) Ionic solid with [PCl4]+ and [PCl6]- because PCl5 forms an ionic compound in the solid state with these specific ions.

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Approach Solution -2

In the solid state, PCl₅ exhibits ionic character and forms a crystal structure in which it dissociates into ions. The structure of PCl₅ in the solid state consists of both \( [PCl_4]^+ \) and \( [PCl_6]^- \) ions.

Option A (Octahedral structure)**: This is incorrect because PCl₅ in its solid state does not adopt an octahedral structure.
Option B (Ionic solid with \( [PCl_6]^+ \) and \( [PCl_4]^- \))**: This is incorrect because the charges on the ions are reversed.
Option C (Ionic solid with \( [PCl_4]^+ \) and \( [PCl_6]^- \))**: This is correct. In the solid-state structure of PCl₅, it exists as an ionic solid with these ions.
Option D (Covalent solid present in the form of \( P_2Cl_{10} \))**: This is incorrect because PCl₅ in the solid state exists as ionic solids, not covalent molecules like \( P_2Cl_{10} \).

Thus, the correct answer is: \({\text{(C) Ionic solid with } [PCl_4]^+ \text{ and } [PCl_6]^-}\)

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