Chapter 10, Problem 10.95P

(a)

Interpretation Introduction

Interpretation:

The distance between $\text{H}$ atoms in ${\text{C}}_2{\text{H}}_2$ is to be calculated.

Concept introduction:

The bond length or bond distance is the average distance between the nuclei of two bonded atoms in a molecule. When two similar atoms are bonded together, half of the bond length is known as the covalent radius. The bond length depends on the number of bonded of two atoms. The unit of bond length is picometer.

In single, double and triple bonds, the order of bond length is as follows:

  $\text{Triple bond }<\text{Double bond }<\text{Single bond}$

Answer to Problem 10.95P

$\text{339 pm}$ is the distance between the two hydrogen atoms in acetylene.

Explanation of Solution

The total number of electrons in ${\text{C}}_2{\text{H}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}\left(\text{V}\text{.E}\right)=2\left(\text{V}\text{.E of C}\right)+2\left(\text{V}\text{.E of H}\right)\\ =2\left(4\right)+2\left(1\right)\\ =10\end{array}$

So ${\text{C}}_2{\text{H}}_2$ has a total of 10 valence electrons. Out of these electrons, 6 electrons are utilized in one $\text{C}-\text{C}$ single bond and two $\text{C}-\text{H}$ single bonds and therefore 4 electrons are left behind. If these electrons are given to one carbon atom to complete its octet, the octet of another carbon atom cannot be fulfilled. So the 4 electrons act as bonding pairs for the triple bond that exists between the two carbon atoms. Therefore, the shape of ${\text{C}}_2{\text{H}}_2$ is linear.

The Lewis structure of ${\text{C}}_2{\text{H}}_2$ is as follows:

The two hydrogen atoms are separated by two $\text{C}-\text{H}$ single bonds and one $\text{C}-\text{C}$ triple bond. The distance between the carbon and the hydrogen atom in $\text{C}-\text{H}$ single bond is $\text{109}\text{pm}$ and that between the two carbon atoms in $\text{C}-\text{C}$ triple bond is $\text{121}\text{pm}$.

The total distance between $\text{H}$ atoms in ${\text{C}}_2{\text{H}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Total distance}=\text{2}\left(109\text{pm}\right)+\text{121}\text{pm}\\ =\text{218 pm}+121\text{pm}\\ =\text{339 pm}\text{.}\end{array}$

Conclusion

The bond length between the two atoms in a molecule depends not only on the atoms but also on other factors like orbital hybridization and the steric nature of the substituents.

(b)

Interpretation Introduction

Interpretation:

The distance between $\text{F}$ atoms in ${\text{SF}}_6$ is to be determined.

Concept introduction:

The bond length or bond distance is the average distance between the nuclei of two bonded atoms in a molecule. When two similar atoms are bonded together, half of the bond length is known as the covalent radius. The bond length depends on the number of bonded of two atoms. The unit of bond length is picometer.

In single, double and triple bonds, the order of bond length is as follows:

  $\text{Triple bond }<\text{Double bond }<\text{Single bond}$

Answer to Problem 10.95P

$\text{316 pm}$ is the distance between the opposite fluorine atoms and $\text{223 pm}$ is the distance between the adjacent fluorine atoms.

Explanation of Solution

The total number of electrons in ${\text{SF}}_6$ can be calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}\left(\text{V}\text{.E}\right)=\text{V}\text{.E of S}+6\left(\text{V}\text{.E of F}\right)\\ =6+6\left(7\right)\\ =48\end{array}$

So ${\text{SF}}_6$ has total of 48 valence electrons. Out of these electrons, 12 electrons are utilized in six $\text{S}-\text{F}$ single bonds and 36 electrons are left behind. These electrons are given to the six fluorine atoms to complete their octets. So, the shape of ${\text{SF}}_6$ is octahedral.

The Lewis structure of ${\text{SF}}_6$ is as follows:

The two fluorine atoms that are opposite to the sulfur atom are separated by two $\text{S}-\text{F}$ single bonds. The distance between the sulfur and the fluorine atom in $\text{S}-\text{F}$ single bond is $\text{158}\text{pm}$.

The total distance between $\text{F}$ atoms opposite to the $\text{S}$ atom in ${\text{SF}}_6$ is calculated as follows:

  $\begin{array}{c}\text{Total distance}=\text{2}\left(158\text{pm}\right)\\ =\text{316 pm}\end{array}$

The adjacent fluorine atoms are present at the two corners of a right-angled triangle. The two sides of the triangle are the bond lengths of $\text{S}-\text{F}$ single bonds and the distance between the two adjacent fluorine atoms is the hypotenuse of the right-angled triangle.

The distance between the adjacent $\text{F}$ atoms in ${\text{SF}}_6$ is calculated by the Pythagorean Theorem as follows:

  ${\left(\text{F}-\text{F}\text{bond distance}\right)}^2={\left(\text{S}-\text{F}\text{bond distance}\right)}^2+{\left(\text{S}-\text{F}\text{bond distance}\right)}^2$        (1)

Or,

  $\text{F}-\text{F}\text{bond distance}=\sqrt{{\left(\text{S}-\text{F}\text{bond distance}\right)}^2+{\left(\text{S}-\text{F}\text{bond distance}\right)}^2}$        (2)

Substitute $\text{158}\text{pm}$ for the $\text{S}-\text{F}\text{bond distance}$ in equation (2).

  $\begin{array}{c}\text{F}-\text{F}\text{bond distance}=\sqrt{{\left(158\text{pm}\right)}^2+{\left(158\text{pm}\right)}^2}\\ =\sqrt{49928{\text{pm}}^2}\\ =223.4457\text{pm}\\ \approx \text{223 pm}\text{.}\end{array}$

Conclusion

The bond length between the two atoms in a molecule depends not only on the atoms but also on other factors like orbital hybridization and the steric nature of the substituents.

(c)

Interpretation Introduction

Interpretation:

The distance between equatorial $\text{F}$ atoms in ${\text{PF}}_5$ is to be determined.

Concept introduction:

The bond length or bond distance is the average distance between the nuclei of two bonded atoms in a molecule. When two similar atoms are bonded together, half of the bond length is known as the covalent radius. The bond length depends on the number of bonded of two atoms. The unit of bond length is picometer.

In single, double and triple bonds, the order of bond length is as follows:

  $\text{Triple bond }<\text{Double bond }<\text{Single bond}$

Answer to Problem 10.95P

$270\text{pm}$ is the distance between equatorial $\text{F}$ atoms in ${\text{PF}}_5$.

Explanation of Solution

The total number of electrons in ${\text{PF}}_5$ can be calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}\left(\text{V}\text{.E}\right)=\text{V}\text{.E of P}+5\left(\text{V}\text{.E of F}\right)\\ =5+5\left(7\right)\\ =40\end{array}$

So ${\text{PF}}_5$ has total of 40 valence electrons. Out of these electrons, 10 electrons are utilized in five $\text{P}-\text{F}$ single bonds and 30 electrons are left behind. These electrons are given to the five fluorine atoms to complete their octets. So, the shape of ${\text{PF}}_5$ is trigonal bipyramidal.

The Lewis structure of ${\text{PF}}_5$ is as follows:

The adjacent fluorine atoms are present at the corners of a triangle with $\text{F}-\text{P}-\text{F}$ bond angle of $120°$. The two sides of the triangle are the bond lengths of $\text{P}-\text{F}$ single bonds $\left(156\text{pm}\right)$.

The distance between the adjacent $\text{F}$ atoms in ${\text{PF}}_5$ is calculated by the Law of cosines as follows:

  $\begin{array}{c}{\left(\text{F}-\text{F}\text{bond distance}\right)}^2={\left(\text{P}-\text{F}\text{bond distance}\right)}^2+\\ {\left(\text{P}-\text{F}\text{bond distance}\right)}^2-\\ 2\left(\text{P}-\text{F}\text{bond distance}\right)\\ \left(\text{P}-\text{F}\text{bond distance}\right)\cos \text{θ}\end{array}$        (1)

Or,

  $\text{F}-\text{F}\text{bond distance}=\sqrt{\begin{array}{c}{\left(\text{P}-\text{F}\text{bond distance}\right)}^2+\\ {\left(\text{P}-\text{F}\text{bond distance}\right)}^2-\\ 2\left(\text{P}-\text{F}\text{bond distance}\right)\\ \left(\text{P}-\text{F}\text{bond distance}\right)\cos \text{θ}\end{array}}$        (2)

Substitute $156\text{pm}$ for the $\text{P}-\text{F}\text{bond distance}$ and $\text{120}°$ for $\text{θ}$ in equation (2).

  $\begin{array}{c}\text{F}-\text{F}\text{bond distance}=\sqrt{{\left(156\text{pm}\right)}^2+{\left(156\text{pm}\right)}^2-2\left(156\text{pm}\right)\left(156\text{pm}\right)\cos \text{120}°}\\ =\sqrt{24336{\text{pm}}^2+24336{\text{pm}}^2-48672{\text{pm}}^2\left(-0.5\right)}\\ =270.1999\text{pm}\\ \approx \text{270 pm}\text{.}\end{array}$

Conclusion

The bond length between the two atoms in a molecule depends not only on the atoms but also on other factors like orbital hybridization and the steric nature of the substituents.