Q10.63CP - Chemistry: Molecular Nature Of Matter And Change

Question

Give the molecular shape of each species in Problem 10.62.

$\begin{array}{l}\left(a\right){\mathbf{PF}}_5;\left(b\right){\mathbf{CCl}}_4;\left(c\right){\mathbf{H}}_3{\mathbf{O}}^{-};\left(d\right){\mathbf{ICl}}_3;\left(e\right){\mathbf{BeH}}_2;\left(f\right){\mathbf{PH}}_2^{-};\\ \left(g\right){\mathbf{GeBr}}_4;\left(h\right){\mathbf{CH}}_3^{-};\left(i\right){\mathbf{BCl}}_3;\left(j\right){\mathbf{BrF}}_4^{-};\left(k\right){\mathbf{XeO}}_3;\left(l\right){\mathbf{TeF}}_4.\end{array}$

Step-by-Step Solution

Answer

The Lewis structure for the species are:

Step 1: Definition of Lewis Structure

Lewis structure is the two-dimensional structure which contains electron-dot symbols of every atom present and the bonding pairs that hold them together. The steps involved while writing Lewis structures are as follows:  

1. Consider the molecular formula of the molecule and place the atom which has the lowest electronegativity as the centre.
2. Place other atoms relative to each other. 
3. Find the total number of valence electrons. 
4. Draw single bonds, and subtract 2 electrons for each bond. 
5. From the remaining valence electrons give each atom 8 electrons. By doing this we get the Lewis structure for the molecule. 

 We know that the molecular shapes of the molecules from the Lewis structure are determined by employing Valence-Shell Electron-Pair Repulsion (VSEPR) theory. The main principle of this theory is that in order to reduce the repulsion, each group of valence electrons around the central atom are located far away from each other.  The steps involved in determining the molecular shapes are as follows:  

1. Consider the molecular formula of the molecule and write its Lewis Structure.
2. Count all the electron groups and arrange them around the central atom. 
3. Check for any lone pairs or double bond positions for finding the bond angle. 
4. Count the bonding and non-bonding electron groups separately. 
5. From which the molecular shape with specific designations is determined.

  The molecular shapes are assigned specific ${\mathrm{AX}}_{\mathrm{m}}{\mathrm{E}}_{\mathrm{n}}$ designations. Where m and n are integers

A - central atom 

X - Surrounding atom 

E - Nonbonding valence-electron group (usually a lone pair).

Step 2: Molecular Shapes of Species

Let us now draw molecular shapes for all the species given one by one.

${\mathbf{PF}}_5$ , Phosphorus pentafluoride

The lewis structure of ${\mathbf{PF}}_5$ would be as below:

As per VSEPR theory, ${\mathbf{PF}}_5$ has ${\mathbf{AX}}_5$ a designation, with 5 bonding atoms and no lone pairs on P. Three F atoms occupy the equatorial positions and two F atoms occupy the axial positions, giving the molecule a trigonal bipyramidal shape. The bond angle for F atoms at equatorial position P-F-P is ${120}^o$(axial-central-axial). Hence, the molecular shape of ${\mathbf{PF}}_5$ is:

${\mathbf{CCl}}_4$, Carbon tetrachloride

The lewis structure of ${\mathbf{CCl}}_4$ would be as below:

As per VSEPR theory, ${\mathbf{CCl}}_4$ has ${\mathbf{AX}}_4$ designation, with 4 bonding atoms and no lone pairs on C. But the lone pairs of electrons on Cl, repulsive in nature are in the plane. Hence the bond angle is formed between two chlorine atoms which is ${109.5}^o$ giving the molecule a tetrahedral shape. Hence, the molecular shape of ${\mathbf{CCl}}_4$is given below:

${\mathbf{H}}_3{\mathbf{O}}^{+}$, Hydronium ion

The lewis structure of ${\mathbf{H}}_3{\mathbf{O}}^{+}$ would be as below:

As per VSEPR theory, ${\mathbf{H}}_3{\mathbf{O}}^{+}$ has ${\mathbf{AX}}_3\mathbf{E}$ designation, with 3 bonding atoms and one lone pair of electrons on O. The lone pair of electrons on O results in strong repulsion which causes the bond angle less than ${109.5}^o$ bond angle. giving the molecule a trigonal pyramidal shape. Hence, the molecular shape of ${\mathbf{CCl}}_4$ is given below:

${\mathbf{ICl}}_3$, Iodine trichloride

The lewis structure of ${\mathbf{ICl}}_3$ would be as below:

As per VSEPR theory, ${\mathbf{ICl}}_3$ has ${\mathbf{AX}}_{\mathbf{3}}{\mathbf{E}}_{\mathbf{2}}$ designation, with three bonding atoms and two lone pairs on iodine. The two lone pairs of electrons on I, occupy the equatorial position to reduce bond pair-lone pair repulsive. Hence the Cl-I-Cl bond angle will be ${90}^o$ giving the molecule a T-shape. Hence, the molecular shape of is given below:

${\mathbf{BeH}}_2$

The lewis structure of ${\mathbf{BeH}}_2$ would be as below:

As per VSEPR theory, ${\mathbf{BeH}}_2$ has ${\mathbf{AX}}_3$ designation, with two bonding atoms, with no lone pairs. The two bond pairs are arranged as far as possible in opposite direction. Hence the bond angle will be ${180}^o$ giving the molecule a linear shape. Hence, the molecular shape of ${\mathbf{BeH}}_2$ is given below:

${\mathbf{PH}}_2^{-}$, Phosphanide

The lewis structure of ${\mathbf{PH}}_2^{-}$ would be as below:

As per VSEPR theory, ${\mathbf{PH}}_2^{-}$ has ${\mathbf{AX}}_{\mathbf{2}}{\mathbf{E}}_{\mathbf{2}}$ designation, with two bonding atoms, with two lone pairs on P. The repulsion from the two lone pairs are stronger and push down closer to the two bonding pairs of hydrogens. Hence the bond angle will be around ${107.7}^o$ giving the molecule a bent shape. Hence, the molecular shape of ${\mathbf{PH}}_2^{-}$ is given below:

${\mathbf{GeBr}}_4$, Germanium tetrabromide

The lewis structure of ${\mathbf{GeBr}}_4$ would be as below:

As per VSEPR theory, ${\mathbf{GeBr}}_{\mathbf{4}}$ has ${\mathbf{AX}}_4$ designation, with four bonding atoms and no lone pairs of atoms. It takes the tetrahedral shape by placing four Br to the four corners of the tetrahedron. Hence the Br-Ge-Br bond angle will be ${109.5}^o$. The molecular shape of ${\mathbf{GeBr}}_{\mathbf{4}}$ is given below:

${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$, Methyl anion

The Lewis structure of ${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$ would be as below:

As per VSEPR theory, ${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$ has ${\mathbf{AX}}_{\mathbf{3}}\mathbf{E}$ designation, with three bonding atoms and one lone pair on carbon. The presence of lone pair of electrons on C, reduces the H-C-H bond angle which ia less than ${109.5}^o$, giving the molecule a trigonal pyramidal shape. The molecular shape of ${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$ is given below:

${\mathbf{\text{BCl}}}_{\mathbf{3}}$, Boron trichloride

The Lewis structure of ${\mathbf{\text{BCl}}}_{\mathbf{3}}$ would be as below:

As per VSEPR theory, ${\mathbf{BCl}}_{\mathbf{3}}$ has ${\mathbf{AX}}_3$ designation, with three bonding atoms. Hence, the Cl-B-Cl bond angle will be ${120}^o$ giving the molecule a trigonal planar shape. The molecular shape of ${\mathbf{BCl}}_{\mathbf{3}}$ is given below:

${\mathbf{BrF}}_{\mathbf{4}}^{\mathbf{-}}$, Bromine tetrafluoride anion

The lewis structure of ${\mathbf{BrF}}_{\mathbf{4}}^{\mathbf{-}}$ would be as below:

As per VSEPR theory, ${{\mathbf{BrF}}^{\mathbf{-}}}_4$ has ${\mathbf{AX}}_{\mathbf{4}}{\mathbf{E}}_{\mathbf{2}}$ designation, with four bonding atoms and two lone pairs on bromine. The two lone pairs of electrons on Br, occupy the opposite position to avoid strong ${90}^o$ lone pair-lone pair repulsion. Hence the molecule has a square planar shape. Hence, the molecular shape of ${{\mathbf{BrF}}^{\mathbf{-}}}_4$ is given below:

${\mathbf{\text{XeO}}}_{\mathbf{3}}$, Xenon trioxide

The lewis structure of ${\mathbf{\text{XeO}}}_{\mathbf{3}}$ would be as below:  

As per VSEPR theory, ${\mathbf{\text{XeO}}}_{\mathbf{3}}$ has ${\mathbf{AX}}_{\mathbf{3}}\mathbf{E}$ designation, with three bonding atoms and one lone pair on xenon. The lone pair of electrons on Xe, results in stronger repulsion with a bond angle slightly less than ${109.5}^o$ giving the molecule a trigonal pyramidal shape. The molecular shape of ${\mathbf{\text{XeO}}}_{\mathbf{3}}$ is given below:

${\mathbf{\text{TeF}}}_{\mathbf{4}}$, Tellurium tetrafluoride

The lewis structure of ${\mathbf{\text{TeF}}}_{\mathbf{4}}$ would be as below:

As per VSEPR theory, ${\mathbf{\text{TeF}}}_{\mathbf{4}}$ has ${\mathbf{AX}}_{\mathbf{4}}\mathbf{E}$ designation, with four bonding atoms and one lone pair on Te. The lone pairs of electrons on Te, occupy the equatorial position as they exert stronger repulsion than the bond pair and reduces the bond angles. Hence the molecule has a seesaw shape. The molecular shape of ${\mathbf{\text{TeF}}}_{\mathbf{4}}$is given below:

Thus, the molecular shapes for the given molecules are drawn.