Question

Question: For each of the following pairs, use electronegativity differences to predict which compound has the higher boiling point.

(a) $MgB{r}_2$or$PB{r}_3$

(b) $Os{O}_4$or $SrO$

(c) $C{l}_{2}O$or $A{l}_2{O}_3$

Short answer

(a)$PB{r}_3$is a slightly polar covalent compound, so it has lower boiling point than ionic $MgB{r}_2$.

(b)$Os{O}_4$ is a polar covalent molecule, so it has lower boiling point than highly ionic $SrO$.

(c) $C{l}_{2}O$is a slightly polar covalent molecule, so it has lower boiling point than ionic $A{l}_2{O}_3$.

Step-by-step solution

Effect of electronegativity on boiling point

Compounds having hydrogen atoms coupled to strongly electronegative atoms have abnormally high boiling temperatures.

Step 2: Compare boiling points from electronegativity differences

The point of ionic nature in a bond is set on by the distinction in electronegativity among the components. If $∆\chi \ge 1.6$ the compound is considered ionic, if role="math" localid="1663400029404" $∆\chi \le 0.4$ the bond is mostly covalent; molecules that are in between have polar covalent bonds. The values may vary from text to text since the difference is not so strict.

The higher the electronegativity difference more ionic is the molecule

Strong intermolecular forces - electrostatic forces - keep ions together firmly, resulting in lower vapour pressures in ionic compounds. Van der Waals forces exist between covalent molecules; however, they are weak in comparison to electrostatic interactions.

The weak attraction forces in covalent molecules have higher vapour pressure, and therefore lower boiling point and melting point

a) $MgB{r}_2$and$PB{r}_3$

The relevant electronegativities are:

$$\begin{gathered} \chi \left(Br\right)=2.B_{\chi }\left(P\right) \\ \chi \left(Br\right)=2.1_{\chi }\left(Mg\right) \\ \chi \left(Br\right)=1.2 \end{gathered}$$

Electronegativity difference in present bonds ($P-Br$and $Mg-Br$):

$$\begin{gathered} ∆\chi \left(Br-P\right)=\left|2.8-2.1\right| \\ ∆\chi \left(Br-P\right)=0.7∆ \\ \chi \left(Mg-Br\right)=\left|2.8-0.8\right| \\ \chi \left(Mg-Br\right)=2.0 \end{gathered}$$

The weak attraction forces in covalent molecules have higher vapour pressure, and therefore lower boiling point and melting point

b) $Os{O}_4$and $SrO$

The relevant electronegativities are:

$$\begin{gathered} \chi \left(O\right)=3.5 \\ \chi \left(Sr\right)=1.0 \\ \chi \left(Os\right)=2.2 \end{gathered}$$$$\begin{gathered} \chi \left(O\right)=3.5 \\ \chi \left(Sr\right)=1,0 \\ \chi \left(Os\right)=2.2 \end{gathered}$$

Electronegativity difference in present bonds ( $Os-O$and $Sr-O$):

$$\begin{gathered} ∆\chi \left(Os-O\right)=\left|2.2-3.5\right| \\ =1.3 \\ ∆\chi \left(Sr-O\right)=\left|1.0-3.5\right| \\ =2.5 \end{gathered}$$

The weak attraction forces in covalent molecules have higher vapour pressure, and therefore lower boiling point and melting point 

c) $C{l}_{2}O$and $A{l}_2{O}_3$

The relevant electronegativities are:

$$\begin{gathered} \chi \left(O\right)=3.5 \\ \chi \left(Cl\right)=3.0 \\ \chi \left(Al\right)=1.5 \end{gathered}$$

Electronegativity difference in present bonds $\left(Al-OandCl-O\right)$:

$$\begin{gathered} ∆\chi \left(Al-O\right)=\left|1.5-3.5\right| \\ =2.0 \\ ∆\chi \left(Cl-O\right)=\left|3.0-3.5\right| \\ =0.5 \end{gathered}$$