Question

element\telectronegativity
h\t2.1
c\t2.5
s\t2.5
f\t4.0
cl\t3.0
si\t1.8
on the basis of the information above, which of the following arranges the binary compounds in order of increasing bond polarity?
a ch4 < sicl4 < sf4
b ch4 < sf4 < sicl4
c sf4 < ch4 < sicl4
d sicl4 < sf4 < ch4

Explanation:

Step1: Recall bond - polarity concept

Bond polarity is determined by the electronegativity difference ($\Delta\chi$) between the two atoms in a bond. The greater the electronegativity difference, the more polar the bond.

Step2: Calculate $\Delta\chi$ for $CH_4$

For $CH_4$, $\chi_C = 2.5$ and $\chi_H=2.1$. $\Delta\chi_{C - H}=|2.5 - 2.1| = 0.4$.

Step3: Calculate $\Delta\chi$ for $SiCl_4$

For $SiCl_4$, $\chi_{Si}=1.8$ and $\chi_{Cl}=3.0$. $\Delta\chi_{Si - Cl}=|3.0 - 1.8| = 1.2$.

Step4: Calculate $\Delta\chi$ for $SF_4$

For $SF_4$, $\chi_S = 2.5$ and $\chi_F=4.0$. $\Delta\chi_{S - F}=|4.0 - 2.5| = 1.5$.

Step5: Compare $\Delta\chi$ values

We have $\Delta\chi_{C - H}=0.4$, $\Delta\chi_{Si - Cl}=1.2$, $\Delta\chi_{S - F}=1.5$. So, the order of increasing bond - polarity is $CH_4

Answer:

A. $CH_4