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What is the hybridization state of the cation part of solid \[PC{{l}_{5}}\].
(A) \[s{{p}^{3}}{{d}^{2}}\]
(B) \[s{{p}^{2}}\]
(C) \[s{{p}^{3}}\]
(D) \[s{{p}^{3}}d\]

Last updated date: 26th Feb 2024
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IVSAT 2024
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Hint: For writing the hybridization of \[PC{{l}_{5}}\] we have to draw the structure of \[PC{{l}_{5}}\] and then we have to count number of lone pairs, shared number of electrons and number of single bonds.

Step by step solution:
Hybridisation is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc.) suitable for the pairing of electrons to form chemical bonds in valence bond theory. Hybrid orbitals are very useful in the explanation of molecular geometry and atomic bonding properties and are symmetrically disposed in space.
Lattice energy is usually the largest factor in determining the stability of an ionic solid. The extra energy gained by the lattice energy more than compensates for the energy needed to transfer a chloride ion from one \[PC{{l}_{5}}\] molecule to another. Thus, \[PC{{l}_{5}}\] exists as an ionic solid and ions are \[PC{{l}_{4}}^{+}\] and \[PC{{l}_{6}}^{-}\].
Here the cation is \[PC{{l}_{4}}^{+}\]:
So, here ‘P’ has 5 valence electrons. In the cation part it has +1 charge on ‘P’, so it has only four electrons in its valence shell it has 4 electrons and these four electrons get paired with four chlorine atoms. And it has zero lone pair. So, the hybridization of \[PC{{l}_{4}}^{+}\] is \[s{{p}^{3}}\].

So, the correct answer is option “C”.

Note: Here \[PC{{l}_{5}}\] has \[s{{p}^{3}}d\] and \[PC{{l}_{6}}^{-}\] has \[s{{p}^{3}}{{d}^{2}}\]. These ions in solid are to increase interionic bond attraction. The P - Cl bond is formed by the overlap of a phosphorus \[s{{p}^{3}}\] orbital with a chlorine p orbital.