Answer
374.3k+ views
Hint: Phosphorus has five electrons in the outermost shell and it can make five bonds with other molecules whereas iodine has 7 electrons and has a tendency to accept a pair of electrons so it forms mainly a single bond.
Complete answer:
-In $\text{PC}{{\text{l}}_{5}}$, the electronic configuration of phosphorus is $\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{3}}$ to form five bonds with chlorine electron form s-orbital is excited to the vacant d-orbital due to which phosphorus gains the ability to make 5 bonds.
-Now, it has a total of five unpaired electrons.
-So, the hybridisation of phosphorus is $\text{s}{{\text{p}}^{3}}\text{d}$ and the geometry will be trigonal bipyramidal.
-Because it has 5 bond pairs and 0 lone pairs.
-In $\text{I}{{\text{F}}_{5}}$, according to the electronic configuration of iodine two electrons from 5p-orbital are excited to move to the 5d-orbital.
-So, now iodine tends to make 5 bonds and along with it have one lone pair.
-The lone pair will cause the repulsion towards the bond pair and increases the bond angles that’s why it is oriented in such a way that it causes less repulsion.
-So, the hybridisation will be $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$and the geometry will be square pyramidal.
-The structure is:
So, the correct answer is “Option A”.
Note: The lone pair- lone pair repulsion causes maximum repulsion than lone pair-bond pair and lone pair-bond pair repulsion is more than the bond pair-bond pair repulsion.
Complete answer:
-In $\text{PC}{{\text{l}}_{5}}$, the electronic configuration of phosphorus is $\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{3}}$ to form five bonds with chlorine electron form s-orbital is excited to the vacant d-orbital due to which phosphorus gains the ability to make 5 bonds.
-Now, it has a total of five unpaired electrons.
-So, the hybridisation of phosphorus is $\text{s}{{\text{p}}^{3}}\text{d}$ and the geometry will be trigonal bipyramidal.
-Because it has 5 bond pairs and 0 lone pairs.
![seo images](https://www.vedantu.com/question-sets/e8a5cdc9-0d56-42bc-abe1-81e23267da6d8813142188090803659.png)
-In $\text{I}{{\text{F}}_{5}}$, according to the electronic configuration of iodine two electrons from 5p-orbital are excited to move to the 5d-orbital.
-So, now iodine tends to make 5 bonds and along with it have one lone pair.
-The lone pair will cause the repulsion towards the bond pair and increases the bond angles that’s why it is oriented in such a way that it causes less repulsion.
-So, the hybridisation will be $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$and the geometry will be square pyramidal.
-The structure is:
![seo images](https://www.vedantu.com/question-sets/ae0d2f2b-3cbd-457a-8bc2-00bc8ebe6c262223000622641449502.png)
So, the correct answer is “Option A”.
Note: The lone pair- lone pair repulsion causes maximum repulsion than lone pair-bond pair and lone pair-bond pair repulsion is more than the bond pair-bond pair repulsion.
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