| Directions: Based on following paragraph. |
| In certain polar solvents, \[\text{PC}{{\text{l}}_{\text{s}}}\]undergoes an ionization reaction in which \[\text{C}{{\text{l}}^{-}}\]ion leaves one \[\text{PC}{{\text{l}}_{\text{5}}}\]molecule and attaches itself to another. \[\text{2PC}{{\text{l}}_{\text{5}}}\text{PCl}_{\text{4}}^{\text{+}}\text{+PCl}_{\text{6}}^{\text{-}}\] |
A) The dissociation is a redox reaction
B) Hybridization changes from \[s{{p}^{3}}d\] to \[s{{p}^{3}}{{d}^{2}}\]\[\text{(PCl}_{\text{6}}^{\text{-}}\text{)}\]and\[\text{s}{{\text{p}}^{\text{3}}}\text{(PCl}_{4}^{+}\text{)}\]
C) Structure changes from trigonal bipyramidal to tetrahedral \[\text{(PCl}_{4}^{+}\text{)}\]and octahedral \[\text{(PCl}_{6}^{-}\text{)}\]
D) None of the above is incorrect
Correct Answer: A
Solution :
The reactions, in which one reactant is oxidized while the other is reduced, are called, redox reactions. \[\overset{+5}{\mathop{2P}}\,C{{l}_{5}}\overset{+5}{\mathop{PC}}\,l_{4}^{+}+\overset{+5}{\mathop{PC}}\,l_{6}^{-}\] In the above reaction the oxidation state of P remains the same even after dissociation, thus , it is not a redox reaction, (every dissociation reaction is not a redox reaction) \[\underset{s{{p}^{3}}d}{\mathop{2PC{{l}_{5}}}}\,\underset{s{{p}^{3}}}{\mathop{PCl_{4}^{+}}}\,+\underset{s{{p}^{3}}{{d}^{2}}}{\mathop{PCl_{6}^{-}}}\,\] (trigonal tiipyramidal) (tetrahedral) (octahedral)
You need to login to perform this action.
You will be redirected in
3 sec
