Question

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

Explanation:

Step1: Calculate bond polarity for CH₄

For \(CH_4\), the electronegativity difference \(\Delta \chi\) between \(C\) (\(\chi = 2.5\)) and \(H\) (\(\chi=2.1\)) is \(\Delta \chi=2.5 - 2.1=0.4\)

Step2: Calculate bond polarity for SiCl₄

For \(SiCl_4\), the electronegativity difference \(\Delta \chi\) between \(Si\) (\(\chi = 1.8\)) and \(Cl\) (\(\chi = 3.0\)) is \(\Delta \chi=3.0 - 1.8 = 1.2\)

Step3: Calculate bond polarity for SF₄

For \(SF_4\), the electronegativity difference \(\Delta \chi\) between \(S\) (\(\chi = 2.5\)) and \(F\) (\(\chi = 4.0\)) is \(\Delta \chi=4.0 - 2.5=1.5\)

Answer:

A. \(CH_4