Chapter 12, Problem 57A

(a)

Interpretation Introduction

Interpretation:

The charges attain by the given polar molecule 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 57A

Oxygen is negative relative to carbon in $\mathrm{C}-\mathrm{O}$

Explanation of Solution

Oxygen is more electronegative than carbon. So, arrow will be from carbon to oxygen. The partial charge is represented as follows:

  $\stackrel{\partial +}{\mathrm{C}}-\stackrel{\partial -}{\mathrm{O}}$

Thus, oxygen is negative relative to C.

(b)

Interpretation Introduction

Interpretation:

The charges attain by the given polar molecule 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 57A

Bromine is negative relative to Iodine in $\mathrm{I}-\mathrm{B}\mathrm{r}$

Explanation of Solution

Bromine is more electronegative than iodine. So, arrow will be from iodine to bromine. The partial charges will be represented as follows:

  $\stackrel{\partial +}{\mathrm{I}}-\stackrel{\partial -}{\mathrm{B}\mathrm{r}}$

Thus, bromine will be negative relative to I.

(c)

Interpretation Introduction

Interpretation:

The charges attain by the given polar molecule 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 57A

Iodine is negative relative to Hydrogen in $\mathrm{H}-\mathrm{I}$

Explanation of Solution

Iodine is more electronegative than hydrogen. So, arrow will be from hydrogen to iodine. The partial positive and negative charges will be represented as follows:

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

Here, I gets negative charge.