Q. 6.70 - An Introduction to General, Organic, and Biological Chemistry

Question

For each of the following bonds, indicate the positive end with ${\delta }^{+}$and the negative end with ${\delta }^{-}$. Draw an arrow to show the dipole for each.
a. $\mathrm{P}$ and $\mathrm{Cl}$
b. $\mathrm{Se}$ and $\mathrm{F}$
c. $\mathrm{Br}$ and $\mathrm{F}$
d. $\mathrm{N}$ and $\mathrm{H}$
e. $\mathrm{B}$ and $\mathrm{Cl}$

Step-by-Step Solution

Answer

We get dipole of for bonds as:

a. $\mathrm{P}$ and $\mathrm{Cl}$ : ${\mathrm{P}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{Cl}}^{{\mathrm{\delta }}^{-}}$
b. $\mathrm{Se}$ and $\mathrm{F}$ : ${\mathrm{Se}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{F}}^{{\mathrm{\delta }}^{-}}$
c. $\mathrm{Br}$ and $\mathrm{F}$ : ${\mathrm{Br}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{F}}^{{\mathrm{\delta }}^{-}}$
d. $\mathrm{N}$ and $\mathrm{H}$ : ${\mathrm{H}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{N}}^{{\mathrm{\delta }}^{-}}$
e. $\mathrm{B}$ and $\mathrm{Cl}$ : ${\mathrm{B}}^{\mathrm{\delta }+}\mapsto {\mathrm{Cl}}^{{\mathrm{\delta }}^{-}}$

Part(a) Step 1: Given Information

We need to indicate the positive end with ${\delta }^{+}$ and the negative end with ${\delta }^{-}$. Draw an arrow to show the dipole for $\mathrm{P}$ and $\mathrm{Cl}$.

Part(a) Step 2: Explanation

• Electronegativity of $\mathrm{P}= 2.1$
• Electronegativity of $\mathrm{Cl}= 3.0$

• Electronegativity  difference= $3.0-2.1=0.9$

So, it forms polar covalent bond we will have dipole.

$\mathrm{Cl}$ is more electronegative than $\mathrm{P}$. So we get dipole as

${\mathrm{P}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{Cl}}^{{\mathrm{\delta }}^{-}}$

Part(b) Step 1: Given Information

We need to indicate the positive end with ${\delta }^{+}$ and the negative end with ${\delta }^{-}$. Draw an arrow to show the dipole for $\mathrm{Se}$ and $\mathrm{F}$.

Part(b) Step 2: Explanation

• Electronegativity of $\mathrm{Se}= 2.4$
• Electronegativity of $\mathrm{F}= 4.0$

• Electronegativity  difference= $4.0-2.4=1.6$

So, it forms polar covalent bond we will have dipole.

$\mathrm{F}$ is more electronegative than $\mathrm{Se}$. So we get dipole as

 ${\mathrm{Se}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{F}}^{{\mathrm{\delta }}^{-}}$

Part(c) Step 1: Given Information

We need to indicate the positive end with ${\delta }^{+}$ and the negative end with ${\delta }^{-}$. Draw an arrow to show the dipole for $\mathrm{Br}$ and $\mathrm{F}$.

Part(c) Step 2: Explanation

• Electronegativity of $\mathrm{Br}= 2.8$ 
• Electronegativity of $\mathrm{F}= 4.0$

• Electronegativity  difference= $4.0-2.8= 1.2$

So, it forms polar covalent bond we will have dipole.

$\mathrm{F}$ is more electronegative than $\mathrm{Br}$. So we get dipole as

 ${\mathrm{Br}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{F}}^{{\mathrm{\delta }}^{-}}$

Part(d) Step 1: Given Information

We need to indicate the positive end with ${\delta }^{+}$ and the negative end with ${\delta }^{-}$. Draw an arrow to show the dipole for $\mathrm{N}$ and $\mathrm{H}$.

Part(d) Step 2: Explanation

• Electronegativity of $\mathrm{N}= 3.0$
• Electronegativity of $\mathrm{H}=2.1$

• Electronegativity  difference= $3.0-2.1=0.9$

So, it forms polar covalent bond we will have dipole.

 $\mathrm{N}$ is more electronegative than $\mathrm{H}$. So we get dipole as

${\mathrm{H}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{N}}^{{\mathrm{\delta }}^{-}}$

Part(e) Step 1: Given Information

We need to indicate the positive end with ${\delta }^{+}$ and the negative end with ${\delta }^{-}$. Draw an arrow to show the dipole for $\mathrm{B}$ and $\mathrm{Cl}$.

Part(e) Step 2: Explanation

• Electronegativity of $\mathrm{B}= 2.0$
• Electronegativity of $\mathrm{Cl}= 3.0$
• Electronegativity  difference= $3.0-2.0=1.0$
  

So, it forms polar covalent bond we will have dipole.

$\mathrm{Cl}$ is more electronegative than $\mathrm{B}$. So we get dipole as

 ${\mathrm{B}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{Cl}}^{{\mathrm{\delta }}^{-}}$