Question

Assertion ${\text{P}}{{\text{H}}_{\text{3}}}$ exists but ${\text{P}}{{\text{H}}_{\text{5}}}$ does not exist while both ${\text{PC}}{{\text{l}}_{\text{3}}}$ and ${\text{PC}}{{\text{l}}_{\text{5}}}$ exist.
Reason ${\text{P}}{{\text{H}}_{\text{3}}}$ has ${\text{s}}{{\text{p}}^3}$ -hybridization.
A. The Assertion is true. Reason is true and Reason is the correct explanation for Assertion
B. The Assertion is true. Reason is true and Reason is NOT the correct explanation for Assertion
C. The Assertion is true. Reason is false.
D. The Assertion is false. Reason is true.

Answer 1

Hydrogen is less electronegative than chlorine. Hence, hydrogen cannot unpair and excite an electron from 3s atomic orbital of phosphorus to 3p atomic orbital.

Complete answer:
The atomic number of phosphorus is 15. The valence shell electronic configuration of phosphorus is ${\text{3}}{{\text{s}}^2}{\text{3}}{{\text{p}}^3}$ . In trivalent compounds of phosphorus, such as ${\text{P}}{{\text{H}}_{\text{3}}}$ and ${\text{PC}}{{\text{l}}_{\text{3}}}$ , the three electrons in 3p atomic orbitals of phosphorus can form bond with three hydrogen atoms or three chlorine atoms. In ${\text{P}}{{\text{H}}_{\text{3}}}$ molecules, no hybridization occurs according to Drago’s rule. Hence, the reason is incorrect.
In ${\text{PC}}{{\text{l}}_{\text{5}}}$ molecule, due to presence of highly electronegative chlorine atom, one electron from 3s atomic orbital of phosphorus atom is unpaired and excited to 3d orbital. Then the phosphorus atom undergoes $s{p^3}d$ hybridization. Five unpaired electrons are present in five degenerate $s{p^3}d$ orbitals. These five unpaired electrons are shared with five electrons of five chlorine atoms to form five phosphorus-chlorine bonds in ${\text{PC}}{{\text{l}}_{\text{5}}}$ molecule.
However, in case of ${\text{P}}{{\text{H}}_{\text{5}}}$ molecule, hydrogen cannot unpair and excite an electron from 3s atomic orbital of phosphorus to 3p atomic orbital. This is because hydrogen has much lower electronegativity than chlorine. Hence, $s{p^3}d$ hybridization is not possible in ${\text{P}}{{\text{H}}_{\text{5}}}$ molecule. Hence, ${\text{P}}{{\text{H}}_{\text{5}}}$ molecule is unstable and does not exist.
Hence, the assertion is correct.

**Hence, the correct option is the option C.

Note:**
You can give another reason based on the size of the overlapping atomic orbitals. The overlap between large sized 3p atomic orbital of phosphorus and small sized 1s atomic orbital of hydrogen in ${\text{P}}{{\text{H}}_{\text{5}}}$ is much less efficient that the overlap between 3p atomic orbital of phosphorus and 3p atomic orbital of chlorine in ${\text{PC}}{{\text{l}}_{\text{5}}}$