How do the P-Ci single bond lengths in PCl5, PCl4+, and PCl6- generally compare

How do the P-Ci single bond lengths in PCl5, PCl4+, and PCl6- generally compare? 4. Identify the type of bonding the central atom undergoes in PCl5 and PCl6-.

The Correct Answer and Explanation is:

Answer:

  • P–Cl single bond lengths trend:
    PCl₄⁺ < PCl₅ < PCl₆⁻
  • Type of bonding on the central atom:
    • PCl₅: sp³d hybridization (trigonal bipyramidal)
    • PCl₆⁻: sp³d² hybridization (octahedral)

Explanation

The phosphorus-chlorine (P–Cl) bond lengths in phosphorus halides such as PCl₄⁺, PCl₅, and PCl₆⁻ vary systematically based on the molecule’s structure, electron count, and bonding environment. These differences stem from changes in electron repulsion, oxidation state, and hybridization of the central phosphorus atom.

In PCl₄⁺, phosphorus forms four bonds and carries a positive charge, resulting in increased effective nuclear charge on the phosphorus center. This leads to stronger attraction between P and Cl atoms and thus shorter P–Cl bond lengths. The molecule adopts a tetrahedral geometry, and phosphorus undergoes sp³ hybridization.

In PCl₅, phosphorus forms five P–Cl bonds in a trigonal bipyramidal geometry, requiring sp³d hybridization. The axial P–Cl bonds are typically longer than the equatorial ones due to increased electron repulsion in the axial positions. Overall, the average bond length in PCl₅ is longer than in PCl₄⁺, due to reduced positive charge density and increased repulsion from the extra bonded pair.

In PCl₆⁻, phosphorus is surrounded by six chlorine atoms and an extra electron (due to the negative charge), further increasing electron–electron repulsion. The molecule adopts an octahedral geometry and undergoes sp³d² hybridization. The additional electron density causes P–Cl bond lengths to be the longest among the three species, as repulsion pushes the ligands farther apart.

Hence, the general trend in P–Cl bond length is:
PCl₄⁺ < PCl₅ < PCl₆⁻, reflecting increasing number of bonds and electron repulsion effects.

The bonding in PCl₅ and PCl₆⁻ requires the use of d-orbitals to accommodate more than an octet of electrons around phosphorus, enabling it to form five and six bonds, respectively. This leads to the sp³d and sp³d² hybridizations, necessary for the trigonal bipyramidal and octahedral shapes.

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