Question

The group of molecules having identical shapes:
A. $B{{F}_{3}},PC{{l}_{3}},Xe{{O}_{3}}$
B. $S{{F}_{4}},Xe{{F}_{4}},CC{{I}_{4}}$
C. $PC{{l}_{5}},I{{F}_{5}},Xe{{O}_{2}}{{F}_{2}}$
D. $CI{{F}_{3}},XeO{{F}_{2}},XeF_{3}^{+}$

Answer 1

T-shape is a molecular geometry that outcomes when there are 3 bonds and 2 lone pairs across the central atom in the molecule. The atoms bonded to the central atom lie at the ends of a T with ${{90}^{\circ }}$ angles between them.

Step by step answer: The group of molecules having identical shape is $CI{{F}_{3}},XeO{{F}_{2}},XeF_{3}^{+}$ .
In these molecules, the central atom has 2 lone pairs of electrons and 3 bonding domains. The electron geometry is trigonal bipyramidal geometry and the molecules are T shaped.
$CI{{F}_{3}}$ Chlorine Trifluoride:
These are aligned in a trigonal bipyramidal shape along with a ${{175}^{\circ }}$ $F$ (axial) $-Cl-F$ (axial) bond angle. The two lone pairs take equatorial positions because they command more space than the bonds. The result is a T-shaped molecule.
$XeO{{F}_{2}}$
The hybridization of $Xe$ in $XeO{{F}_{2}}$ is $s{{p}^{3}}d$ and structure is T-shaped having the 2 lone pairs and oxygen atom in the equatorial position and the two F atoms at the axial position,but if you consider the lone pairs during the determination of the structure then it will be TBP(trigonal bipyramidal).
$XeF_{3}^{+}$Xenon tetrafluoride cation:
$Xe$ has 8 valence electrons plus 1 for each $Cl-F$single bond less one for the charge Total 10 electrons ,three bond pairs and two lone pairs. Based on trigonal bipyramidal lone pair equatorial , so T-shaped structure.

Hence, the correct option is D. $CI{{F}_{3}},XeO{{F}_{2}},XeF_{3}^{+}$.

Note: The term trigonal reveals to us that the general shape is three-sided, similar to a triangle. Assembling the two terms, we can see that a three-sided bipyramidal particle has a trigonal shape with each side being a bipyramid.