First slide

Phosphorous pentachloride

Question

Identify the correct statements from the following.
A) P-Cl_axial bond is shorter than P-Cl_equatorial bond in PCl₅(g) molecule.
B) Hybridization of Phosphorous in PCl₅(s) is sp³d.
C) POCl₃ is the intermediate formed during the hydrolysis of PCl₅.
D) PCl₅ oxidizes Sn to SnCl₄.

Moderate

A

A, B, C and D
B

A, B and D only
C

A, C and D only
D

C and D only

Solution

A) P-Cl_axial bond is longer than P-Cl_equatorial bond in ${PCl}_{5} (g)$ molecule.
Axial bonds experience more repulsions than equatorial bonds.
B) Hybridization of Phosphorous in PCl₅(s) is sp³ and sp³d².
Solid PCl₅ exists as ${[{PCl}_{4}]}^{+} {[{PCl}_{6}]}^{-}$
In ${[{PCl}_{4}]}^{+}$ the hybridization of ‘P’ is sp³ and in ${[{PCl}_{6}]}^{-}$ it is sp³d².
C) POCl₃ is the intermediate formed in the hydrolysis of PCl₅.
${PCl}_{5} + H_{2} O \to {POCl}_{3} + 2 HCl; {POCl}_{3} + 3 H_{2} O \to H_{3} {PO}_{4} + 3 HCl;$
D) PCl₅ oxidizes Sn to SnCl₄ at itself gets reduced to PCl₃.

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