Question

four atoms are arbitrarily labeled d, e, f, and g. their electronegativities are as follows: d 3.9 e 3.5 f 2.6 g 1.2 if the atoms of these elements form the molecules de, dg, eg, and df, how would you arrange these molecules in order of increasing covalent bond character? molecule covalent character de select eg select df select dg select

Explanation:

Step1: Recall electronegativity - covalent bond relation

The greater the electronegativity difference ($\Delta \chi$) between two atoms in a bond, the less covalent (more ionic) the bond. Smaller $\Delta \chi$ means more covalent character.

Step2: Calculate $\Delta \chi$ for DE

$\Delta \chi_{DE}=\vert3.9 - 3.5\vert= 0.4$

Step3: Calculate $\Delta \chi$ for DG

$\Delta \chi_{DG}=\vert3.9 - 1.2\vert = 2.7$

Step4: Calculate $\Delta \chi$ for EG

$\Delta \chi_{EG}=\vert3.5 - 1.2\vert=2.3$

Step5: Calculate $\Delta \chi$ for DF

$\Delta \chi_{DF}=\vert3.9 - 2.6\vert = 1.3$

Step6: Arrange in increasing covalent character (decreasing $\Delta \chi$)

The order of increasing $\Delta \chi$ (decreasing covalent character) is: DE < DF < EG < DG. So the order of increasing covalent - character is DG < EG < DF < DE.

Answer:

DG, EG, DF, DE