Question

Which of the following statement is false:

• A. In $PC{l}_3$, the axial bond length is longer than the equatorial bond length
• B. The dipole moment of $C{H}_{3}F$ is greater than $C{H}_{3}Cl$
• C. In $XeO{F}_4$, the central atom $Xe$ doesnot lie on the plane of the square
• D. In $P{F}_{2}C{l}_3$, both the $F$ atoms lie on the axial bond

Answer 1

The correct option is B The dipole moment of $C{H}_{3}F$ is greater than $C{H}_{3}Cl$

$XeO{F}_4$ is square pyramidal in shape.

Due to lone pair - bond pair repulsion the four $F$ atoms move away from lone pair resulting in the movement of $Xe$ towards lone pair.

The most electronegative substituent lies at the axial position to minimize bond pair - lone pair repulsion.

$PC{l}_5$ shows up $s{p}^{3}d$ hybridisation with the involvement of axial $d_z^2$ orbital. Since, the axial $Cl$ is at ${90}^o$ hence suffers relatively greater repulsion increasing the axial bond length.

In $C{H}_{3}F$ and $C{H}_{3}Cl$, the size of $F$ is smaller than $Cl$ leading smaller $C-F$ bond length decreasing the dipole moment.