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Question: In a change from\[PC{l_3} \to PC{l_5}\]. The hybrid state of \[p\] change from: A. \[s{p^2}\] to \...

In a change fromPCl3PCl5PC{l_3} \to PC{l_5}. The hybrid state of pp change from:
A. sp2s{p^2} to sp3s{p^3}
B. sp3s{p^3} to sp2s{p^2}
C. sp3s{p^3} to sp3ds{p^3}d
D. sp3s{p^3} to dsp2ds{p^2}

Explanation

Solution

PCl3PC{l_3} is phosphorus trichloride and PCl5PC{l_5} is phosphorus pentachloride. The addition of Cl2C{l_2} with PCl3PC{l_3} produces PCl5PC{l_5}. The presence of vacant d orbitals in phosphorus allows it to add the Cl2C{l_2} atoms to the inner dd orbitals.

Complete step by step answer: Hybridization is the process of combining two different orbitals of atoms generating new orbitals which are different from atomic orbitals. These new orbitals are referred to as hybrid orbitals.
Phosphorus is an element in the periodic table with atomic number 16. The electronic configuration of element phosphorus (PP) is:
P : 1s22s22p63s23p3P{\text{ }}:{\text{ }}1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}
Thus phosphorus has three unpaired electrons in the outer shellpp. Further in phosphorus there are vacant dd orbitals present i.e. 3d3d. Due to these dd orbitals the valency of the central atom can be extended and more atoms can be added to the phosphorus atom.
In PCl3PC{l_3}, phosphorus is the central atom which is attached to three chlorine atoms. The phosphorus has one lone pair available in the form of ss electrons. The three pp electrons are present in 3Px3{P_x}, 3Py3{P_y}, 3Pz3{P_z} orbitals. The ss and pp orbitals undergo hybridization to produce sp3s{p^3} hybrid orbitals of equal energy.
In PCl5PC{l_5}, phosphorus the central atom is attached to five chlorine atoms. Here the vacant 3d3d orbitals take part in hybridization to produce five hybrid orbitals. One of the two ss electrons moves to the dd orbital to produce five singly filled orbitals. Each of the orbitals accepts an electron from chlorine to form five sigma bonds. Thus the hybridization of phosphorus pentachloride is sp3ds{p^3}d.
Thus, option C is the correct answer, the hybrid state of PP changes from sp3s{p^3} to sp3ds{p^3}d.

Note: The number of hybrid orbitals will be equal to the number of atomic orbitals. Only orbitals of the same atom will undergo hybridization. The shape of the orbital can be predicted by knowing the hybridization of orbitals.