Chapter 8, Problem 4STP

Interpretation Introduction

Interpretation:

The pairs of elements that would form an ionic bond are to be predicted.

Concept introduction:

The ability of an atom to attract electrons in a chemical bond is termed as electro negativity of that atom. The most electronegative element in the periodic table is fluorine (3.98) and the least electronegative element in the periodic table is francium (0.70).

Answer to Problem 4STP

The pair of elements having atomic number 8 and atomic number 12 would form an ionic bond. Therefore, option (D) is the correct option.

Explanation of Solution

Reason for correct option: Let’s assume a bond is formed between two elements A and B. For a bond to be ionic, the electro negativity difference (Δε) between A and B should be greater than 1.7.

The given graph is shown below.

From graph, the electronegativy value of element having atomic number 8 is 3.5 and the electronegativy value of element having atomic number 12 is 1.25. The corresponding graph is shown below.

The electro negativity difference (Δε) is calculated as shown below.

$\begin{array}{c}\Delta \epsilon ={\epsilon }_{\text{Atomic}\text{No}\text{.}\text{8}}-{\epsilon }_{\text{Atomic}\text{No}\text{.}\text{12}}\\ =3.5-1.25\\ \Delta \epsilon =2.25\end{array}$

The electro negativity difference (Δε) is 2.25 which is greater than 1.7. Therefore, the pair of elements having atomic number 8 and atomic number 12 would form an ionic bond. Therefore, option (D) is the correct option.

Conclusion

Reason for incorrect option: The electro negativity difference (Δε) between elements having atomic number 3 and 5 is 0.50. So, A is incorrect option.

The electro negativity difference (Δε) between elements having atomic number 7 and 8 is 0.50. So, B is incorrect option.

The electro negativity difference (Δε) between elements having atomic number 4 and 18 is 1.50. So, C is incorrect option.