four atoms are arbitrarily labeled d, e, f, and g. their electronegativities are as follows: d 3.6 e 3.5 f…

four atoms are arbitrarily labeled d, e, f, and g. their electronegativities are as follows: d 3.6 e 3.5 f 2.9 g 1.5 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

four atoms are arbitrarily labeled d, e, f, and g. their electronegativities are as follows: d 3.6 e 3.5 f 2.9 g 1.5 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

Answer

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

$\Delta\chi_{DE}=\vert3.6 - 3.5\vert= 0.1$

Step3: Calculate $\Delta\chi$ for DG

$\Delta\chi_{DG}=\vert3.6 - 1.5\vert = 2.1$

Step4: Calculate $\Delta\chi$ for EG

$\Delta\chi_{EG}=\vert3.5 - 1.5\vert=2.0$

Step5: Calculate $\Delta\chi$ for DF

$\Delta\chi_{DF}=\vert3.6 - 2.9\vert = 0.7$

Step6: Arrange in increasing covalent character

Lower $\Delta\chi$ implies more covalent. So, the order is based on increasing $\Delta\chi$.

Answer:

DG < EG < DF < DE