91Ó°ÊÓ

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

• 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

${\mathrm{P}}^{{\mathrm{δ}}^{+}}↦{\mathrm{Cl}}^{{\mathrm{δ}}^{-}}$

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

• 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

${\mathrm{Se}}^{{\mathrm{δ}}^{+}}↦{\mathrm{F}}^{{\mathrm{δ}}^{-}}$

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

• 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

${\mathrm{Br}}^{{\mathrm{δ}}^{+}}↦{\mathrm{F}}^{{\mathrm{δ}}^{-}}$

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

• 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

${\mathrm{H}}^{{\mathrm{δ}}^{+}}↦{\mathrm{N}}^{{\mathrm{δ}}^{-}}$

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

• 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

${\mathrm{B}}^{{\mathrm{δ}}^{+}}↦{\mathrm{Cl}}^{{\mathrm{δ}}^{-}}$