Home
Class 12
CHEMISTRY
A : PF5 and IF5 have similar shapes . ...

A : `PF_5` and `IF_5` have similar shapes .
R : All the bond lengths are equal `PF_(5)`

A

If both Assertion & Reason are true and the reason is the correct explanation of the assertion , then mark (1)

B

If both Assertion & Reason are true but the reason is not the correct explanation of the assertion , then mark (2)

C

If Assertion is true statement but Reason is false , then mark (3)

D

If both Assertion and Reason are false statements , then mark (4)

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the statements provided in the question, we will evaluate the Assertion (A) and Reason (R) step by step. ### Step 1: Evaluate the Assertion (A) The assertion states that `PF5` and `IF5` have similar shapes. - **PF5 (Phosphorus Pentafluoride)**: The phosphorus atom is the central atom and is surrounded by five fluorine atoms. The hybridization of phosphorus in PF5 is `sp3d`, leading to a trigonal bipyramidal geometry. In this geometry, three fluorine atoms occupy the equatorial positions, and two fluorine atoms occupy the axial positions. - **IF5 (Iodine Pentafluoride)**: The iodine atom is the central atom and is also surrounded by five fluorine atoms. However, iodine has a lone pair of electrons, leading to a hybridization of `sp3d2`, resulting in an octahedral geometry. Due to the presence of one lone pair, the shape of IF5 becomes square pyramidal. **Conclusion for Assertion**: Since PF5 has a trigonal bipyramidal shape and IF5 has a square pyramidal shape, the assertion that they have similar shapes is **false**. ### Step 2: Evaluate the Reason (R) The reason states that all the bond lengths in `PF5` are equal. - In PF5, the three equatorial bonds (between phosphorus and the three equatorial fluorine atoms) are indeed equal in length due to symmetry. However, the two axial bonds (between phosphorus and the two axial fluorine atoms) are longer than the equatorial bonds due to the difference in bond angles and the presence of lone pair repulsion. **Conclusion for Reason**: Therefore, the statement that all bond lengths in PF5 are equal is also **false**. ### Final Conclusion Both the assertion and the reason are false. Therefore, the correct answer is option 4: **Both Assertion and Reason are false.** ---
Promotional Banner

Similar Questions

Explore conceptually related problems

All eukaryotic cells have similar shape

In the following questions, a statement of assertion (A) is followed by a statement of reason (R ) A : All P-Cl bond lengths are equal in PCl_(3) but different in PCl_(5) R : Hybrid state of central atom is different in both molecules.

Statement-1 : PH_(3) and PF_(3) are pyramidal in shape with one lone pair on P . But PF_(3) has greatest bond angle than PH_(3) . and Statement-2 : Back bonding is present in PF_(3) but absent in PH_(3) .

Explain the structure of PCl_(5) according to hybridization. Why all P – Cl bonds lengths are not equivalent in PCl_(5) ?

Explain the structure of PCl_(5) according to hybridization. Why all P – Cl bonds lengths are not equivalent in PCl_(5) ?

Find the total number of compounds having equal all bond lengths equal : PCl_(5),PF_(5),SF_(4),XeF_(2), XeF_(4),XeF_(6),SF_(6),ClF_(3), BrF_(5),Ni(CO)_(4),IF_(7)