Chapter 8.5, Problem 74SSC

Interpretation Introduction

(a)

Interpretation:

The type of bond form between H and S to be predicted.

Concept introduction:

Power of an atom to attract electrons in a bond is electro negativity. Electro negativity is directly proportional to ionic character of the bond.

Answer to Problem 74SSC

Bond is non polar covalent.

Explanation of Solution

Electro negativity of H atom is 2 and electro negativity of S atom is 2.58. Thus electro negativity difference is as follows:

$\begin{array}{c}\Delta \text{EN}=2.58-2.2\\ =0.38\end{array}$

$\text{ΔEN}$< 0.5, thus the bond is non polar covalent.

Interpretation Introduction

(b)

Interpretation:

The type of bond form between C and H to be predicted.

Concept introduction:

Power of an atom to attract electrons in a bond is electro negativity. Electro negativity is directly proportional to ionic character of the bond.

Answer to Problem 74SSC

Bond is non polar covalent.

Explanation of Solution

Electro negativity of H atom is 2 and electro negativity of C atom is 2.55. Thus electro negativity difference is as follows:

$\begin{array}{c}\Delta \text{EN}=2.55-2.2\\ =0.35\end{array}$

$\text{ΔEN}$< 0.5, thus the bond is non polar covalent

Interpretation Introduction

(c)

Interpretation:

The type of bond form between Na and S to be predicted.

Concept introduction:

Power of an atom to attract electrons in a bond is electro negativity. Electro negativity is directly proportional to ionic character of the bond.

Answer to Problem 74SSC

Bond is polar covalent.

Explanation of Solution

Electro negativity of Na atom is 0.93 and electro negativity of S atom is 2.58. Thus electro negativity difference is as follows:

$\begin{array}{c}\Delta \text{EN}=2.58-0.93\\ =1.65\end{array}$

$\text{ΔEN}$ = 0.5 to 1.6, thus the bond is polar covalent