Q. 6.69 - 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. N and F
b. Si and Br
c. C and O
d. P and Br
e. N and P

Step-by-Step Solution

Answer

We get dipole of for bonds as:

a. $\mathrm{N}$ and $\mathrm{F}$ : ${\mathrm{N}}^{{\mathrm{\delta }}^{+}}\to {\mathrm{F}}^{{\mathrm{\delta }}^{-}}$
b. $\mathrm{Si}$ and $\mathrm{Br}$ : ${\mathrm{Si}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{Br}}^{{\mathrm{\delta }}^{-}}$
c. $\mathrm{C}$ and $\mathrm{O}$ : ${\mathrm{C}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{O}}^{{\mathrm{\delta }}^{-}}$
d. $\mathrm{P}$ and $\mathrm{Br}$ : ${\mathrm{P}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{Br}}^{{\mathrm{\delta }}^{-}}$
e. $\mathrm{N}$ and $\mathrm{P}$ : ${\mathrm{P}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{N}}^{{\mathrm{\delta }}^{-}}$

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 $N$ and $F$.

Part(a) Step 2: Explanation

Electronegativity of $\mathrm{N}=3.0$

Electronegativity of $\mathrm{F}=4.0$

Electronegativity  difference= $4.0-3.0=1.0$

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

$\mathrm{F}$ is more electronegative than $\mathrm{N}$. So we get dipole as ${\mathrm{N}}^{{\mathrm{\delta }}^{+}}\to {\mathrm{F}}^{{\mathrm{\delta }}^{-}}$

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 $Si$ and $Br$.

Part(b) Step 2: Explanation

Electronegativity of $\mathrm{Si} = 1.8$

Electronegativity of $\mathrm{Br}= 2.8$

Electronegativity  difference= $2.8-1.8=1.0$

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

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

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

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 $C$ and $O$.

Part(c) Step 2: Explanation

Electronegativity of $\mathrm{C} = 2.5$

Electronegativity of  $\mathrm{O}= 3.5$

Electronegativity  difference= $3.5-2.5=1.0$

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

$\mathrm{O}$ is more electronegative than $\mathrm{C}$. So we get dipole as

${\mathrm{C}}^{{\mathrm{\delta }}^{+}}\mapsto {\mathrm{O}}^{{\mathrm{\delta }}^{-}}$ 

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 P and $Br$.

Part(d) Step 2: Explanation

Electronegativity of $\mathrm{P}= 2.1$

Electronegativity of $\mathrm{Br}=2.8$

Electronegativity  difference= $2.8-2.1=0.7$

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

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

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

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 $N$ and $P$.

Part(e) Step 2: Explanation

Electronegativity of $\mathrm{N}=3.0$

Electronegativity of $\mathrm{P}= 2.1$

Electronegativity  difference= $3.0-2.1=0.9$

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

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

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