Q10.98 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Using bond lengths in Table 9.3 (p. 353) and assuming ideal geometry, calculate each of the following distances:

1. Between H atoms in  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}$
2. Between F atoms in  ${\mathbf{SF}}_{\mathbf{6}}$ (two answers)
3. Between equatorial F atoms in  ${\mathbf{PF}}_{\mathbf{5}}$

Step-by-Step Solution

Answer

1. The distance between H atoms in  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}$ is 339pm
2. The distance between F atoms in  ${\mathbf{SF}}_{\mathbf{6}}$ is 223pm
3. The distance between F atoms in ${\mathbf{PF}}_{\mathbf{3}}$  is 270.1pm 

Step 1: Definition

Bond length is the distance between the centers of two atoms that are covalently bound.

Step 2: Ethyne

In an ethyne molecule, the distance between two hydrogens is calculated by adding bond lengths of Carbon-carbon (triple bond) and Carbon hydrogen (single bond).

The bond length between carbon-carbon (triple bond) is 121pm and the bond length between Carbon-hydrogen (single bond) is 109pm.

So, the distance between the two hydrogens is calculated as  

$\mathbf{2}\mathbf{ }\left(\mathbf{109}\mathbf{pm}\right)\mathbf{+}\mathbf{121}\mathbf{pm}\mathbf{=}\mathbf{339}\mathbf{pm}$

Step 3: Sulphur hexafluoride

The geometry of  ${\mathbf{SF}}_{\mathbf{6}}$ is octahedral. All S-F bond lengths are equal in  ${\mathbf{SF}}_{\mathbf{6}}$ molecule.  The value of Sulfur-fluorine (single bond) is 158pm.

So, the distance between the two equatorial or two axial fluorine are calculated as 

$\mathbf{2}\mathbf{\times }\left(\mathbf{158}\mathbf{pm}\right)\mathbf{=}\mathbf{316}\mathbf{pm}$

and

The distance between the two fluorine (one at equatorial and the other at axial) is 

$\mathbf{=}\sqrt{{\mathbf{158}}^{\mathbf{2}}\mathbf{+}{\mathbf{158}}^{\mathbf{2}}}\mathbf{pm}\mathbf{=}\mathbf{223}\mathbf{pm}$

Step 4: phosphorous pentafluoride

The geometry of phosphorous pentafluoride is trigonal bipyramidal in which equatorial and axial bond length are different. The bond angle between axial and equatorial P-F bond is ${\mathbf{90}}^{\mathbf{o}}$ and the bond angle between two axial P-F bond is  ${\mathbf{120}}^{\mathbf{o}}$. The axial P-F bond length is 158pm and the equatorial P-F bond length is 156pm.

So, the distance between the two equatorial fluorine is $\mathbf{=}\left|\mathbf{156}\mathbf{\times }\mathbf{tan}\mathbf{120}\mathbf{°}\right|\mathbf{ }\mathbf{=}\mathbf{270}\mathbf{.}\mathbf{1}\mathbf{pm}$