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

Hint: 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.