Chapter 8, Problem 120A

Rank the bonds according to increasing polarity.

1. $\text{C-H}$

$\text{N-H}$

$\text{Si-H}$

$\text{O-H}$

$\text{Cl-H}$

Interpretation Introduction

Interpretation:

The given bonds need to be ranked according to increasing polarity.

C-H b. N-H c. Si-H d. O-H e. Cl-H

Concept introduction:

Electronegativity is the comparative strength of an atom to pull electrons in a chemical bond. The character and type of a chemical bond can be predicted employing the electro negativity difference of the elements that bond.

Answer to Problem 120A

Increasing order of polarity is as follows: c. Si-H, a. C-H, b. N-H, e. Cl-H, d. O-H.

Explanation of Solution

Electronegativity value of

$\begin{array}{l}\text{O}=3.44\\ \text{C}=2.55\\ \text{Si}=1.90\\ \text{H}=2.2\end{array}$

$\begin{array}{l}\text{N}=3.04\\ \text{Cl}=3.16\end{array}$

From the above electro negativity value evaluate the electro negativity difference of each as follows:

Electronegativity difference of C-H is $2.2-2.55=-0.35$.

Electronegativity difference of N-H is $2.2-3.04=-0.84$

Electronegativity difference of Si-H is $2.2-1.90=-0.3$

Electronegativity difference of O-H is $2.2-3.44=-1.24$

Electronegativity difference of Cl-H is $2.2-3.16=-0.96$

The highest EN results more polar bond. Above calculation of EN indicates that the bond Si-H has highest EN among them. Hence, the increasing order of polarity is as follows: c. Si-H, a. C-H, b. N-H, e. Cl-H, d. O-H.

Conclusion

During a reaction the type of bond formation is related to each atom’s attraction for electrons.