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Evaluate Statements on Bond Angle Trend and Ionic Nature

Given below are two statements:

Statement I : F2O<H2O<Cl2O\text{F}_2\text{O} < \text{H}_2\text{O} < \text{Cl}_2\text{O} is the correct trend in terms of bond angle.

Statement II : SiF4\text{SiF}_4, SnF4\text{SnF}_4 and PbF4\text{PbF}_4 are ionic in nature.

In the light of the above statements, choose the correct answer from the options given below:

Options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

Correct
D

Statement I is false but Statement II is true

Step-by-Step Solution

To determine the correct option, let us evaluate both statements step-by-step:

Evaluation of Statement I:

  • F2O\text{F}_2\text{O}: Fluorine is more electronegative than oxygen (ENF>ENOEN_{\text{F}} > EN_{\text{O}}). The shared electron pairs in the O-F\text{O-F} bonds are pulled away from the central oxygen atom toward the fluorine atoms. This decreases the bond-pair electron density near the oxygen atom, resulting in weaker bond-pair--bond-pair repulsion compared to lone-pair--bond-pair repulsion. Thus, the bond angle decreases to approximately 103103^\circ.
  • H2O\text{H}_2\text{O}: Oxygen is more electronegative than hydrogen (ENO>ENHEN_{\text{O}} > EN_{\text{H}}). The bonding electron pairs are drawn closer to the central oxygen atom, increasing bond-pair--bond-pair repulsion near the central atom, which gives a bond angle of 104.5104.5^\circ.
  • Cl2O\text{Cl}_2\text{O}: Although chlorine is electronegative, chlorine atoms are much larger in size compared to hydrogen and fluorine. The strong steric repulsion (van der Waals strain) between the two bulky chlorine atoms significantly widens the bond angle to approximately 110.8110.8^\circ.

Thus, the bond angle order is: F2O  (103)<H2O  (104.5)<Cl2O  (110.8)\text{F}_2\text{O} \;(103^\circ) < \text{H}_2\text{O} \;(104.5^\circ) < \text{Cl}_2\text{O} \;(110.8^\circ)

Hence, Statement I is true.


Evaluation of Statement II:

  • According to group 14 periodic trends, SnF4\text{SnF}_4 and PbF4\text{PbF}_4 are ionic solids due to the larger metallic size and lower ionization energies of Sn\text{Sn} and Pb\text{Pb}.
  • However, SiF4\text{SiF}_4 is a gaseous compound with predominant covalent character because silicon has a small atomic size and high electronegativity relative to heavier group 14 metals, making the formation of a discrete Si4+\text{Si}^{4+} ion energetically unfavorable.

Since SiF4\text{SiF}_4 is covalent, the claim that SiF4\text{SiF}_4, SnF4\text{SnF}_4, and PbF4\text{PbF}_4 are all ionic in nature is incorrect.

Hence, Statement II is false.


Conclusion:

  • Statement I is true.
  • Statement II is false.

This matches Option C.

Evaluate Statements on Bond Angle Trend and Ionic Nature | Chemistry PYQ Solution - JEE Challenger