Chapter 12, Problem 15A

(a)

Interpretation Introduction

Interpretation:

The charges attain by the given polar molecule HF are to be written

Concept introduction:

Dipole moment arrows are drawn from the atom with less electronegativity to atom with more electronegativity.

Less electronegative gets partial positive charge and more electronegative atom gets partial negative charge.

Answer to Problem 15A

Hydrogen is positive relative to fluorine in $\mathrm{H}-\mathrm{F}$

Explanation of Solution

Fluorine is more electronegative than hydrogen. So, arrow will be from hydrogen to fluorine. Dipole can be represented as:

  $\stackrel{\partial +}{\mathrm{H}}-\stackrel{\partial -}{\mathrm{F}}$

(b)

Interpretation Introduction

Interpretation:

The charges attain by the given polar molecule Cl-Fare to be written

Concept introduction:

Dipole moment arrows are drawn from the atom with less electronegativity to atom with more electronegativity.

Less electronegative gets partial positive charge and more electronegative atom gets partial negative charge.

Answer to Problem 15A

Chlorine is positive relative to fluorine in $\mathrm{C}\mathrm{l}-\mathrm{F}$

Explanation of Solution

Fluorine is more electronegative than chlorine. So, arrow will be from chlorine to Fluorine. Dipole can be represented as:

  $\stackrel{\partial +}{\mathrm{C}\mathrm{l}}-\stackrel{\partial -}{\mathrm{F}}$

(c)

Interpretation Introduction

Interpretation:

The charges attain by the given polar molecule I-Cl are to be written

Concept introduction:

Dipole moment arrows are drawn from the atom with less electronegativity to atom with more electronegativity.

Less electronegative gets partial positive charge and more electronegative atom gets partial negative charge.

Answer to Problem 15A

Iodine is positive relative to chlorine in $\mathrm{I}-\mathrm{C}\mathrm{l}$

Explanation of Solution

Chlorine is more electronegative than iodine. So, arrow will be from iodine to chlorine. Dipole can be represented as:

  $\stackrel{\partial +}{\mathrm{I}}-\stackrel{\partial -}{\mathrm{C}\mathrm{l}}$