Question

Label each molecule with the name of its shape.

See attachment.

Answer 1

The correct answers are shown in image

Answer 2

The molecular shape of ${\text{Si}}{{\text{F}}_4}$ is $\boxed{{\text{tetrahedral}}}$.

The molecular shape of ${\text{I}}{{\text{F}}_{\text{5}}}$ is $\boxed{{\text{square pyramidal}}}$.

The molecular shape of ${\text{S}}{{\text{F}}_{\text{4}}}$ is $\boxed{{\text{see - saw}}}$.

The molecular shape of ${\text{PC}}{{\text{l}}_{\text{5}}}$ is $\boxed{{\text{trigonal bipyramidal}}}$.

Further explanation:

The shape adopted by a molecule is predicted on the basis of Valence Shell Electron Pair Repulsion theory. This theory governs the type of electronic arrangement around the central atom. The shape is mostly the one that results in the least electronic repulsions.

The table of the molecular shape that corresponds to the different number of electron pairs is attached in the image.

In ${\mathbf{Si}}{{\mathbf{F}}_{\mathbf{4}}}$, silicon has four valence electrons that are utilized in the formation of four Si-F bonds. Such four bonds are best stabilized in the tetrahedral geometry in accordance with the VSPER theory.

In ${\mathbf{I}}{{\mathbf{F}}_{\mathbf{5}}}$, iodine has seven valence electrons it utilizes five of these to form bonds with five fluorine atoms. The remaining two electrons are accommodated as the lone pair. In order to have minimum repulsion as per the VSPER theory this fourth pair must lie in the plane above the four I-F bonds, and this constitutes the square pyramidal shape.

In ${\mathbf{S}}{{\mathbf{F}}_{\mathbf{4}}}$, sulfur has six valence electrons out of which it utilizes four electrons to form four S-F bonds the remaining one equatorial position is occupied with lone pair. This results in distortion from the usual trigonal pyramidal to a ”see-saw geometry”. That is essentially derived from original pyramidal geometry with one equatorial position occupied by the lone pair.

In ${\mathbf{PC}}{{\mathbf{l}}_{\mathbf{5}}}$, the phosphorus has five valence electrons and it forms two axial bonds with the chlorine and three equatorial bonds so as to have minimum repulsion. Therefore most favorable shape for ${\text{PC}}{{\text{l}}_{\text{5}}}$ is the trigonal pyramidal.

The structure of ${\mathbf{Si}}{{\mathbf{F}}_{\mathbf{4}}}$,${\mathbf{I}}{{\mathbf{F}}_{\mathbf{5}}}$, ${\mathbf{S}}{{\mathbf{F}}_{\mathbf{4}}}$ and ${\mathbf{PC}}{{\mathbf{l}}_{\mathbf{5}}}$ is given in the attached image.

Learn more:

1. Identification of ionic bonding: https://brainly.com/question/1603987

2. How ${\text{M}}{{\text{n}}^{{\text{3 + }}}}$ acts as an acid: https://brainly.com/question/2396649

Answer details:

Grade: High school

Subject: Chemistry

Chapter: Covalent bonding and molecular structure.

Keywords: SiF4, IF5, SF4, PCl5 molecular shape, central atom, linear, tetrahedral, trigonal pyramidal. equatorial, axial, see-saw and lone pair.