Question

The correct sequence of dipole moments among the chlorides of methane is

• A. $CHC{l}_3<C{H}_{2}C{l}_2>C{H}_{3}Cl>CC{l}_4$
• B. $C{H}_{2}C{l}_2>C{H}_{3}Cl>CHC{l}_3>CC{l}_4$
• C. $C{H}_{3}Cl>C{H}_{2}C{l}_2>CHC{l}_3>CC{l}_4$
• D. $C{H}_{2}C{l}_2>CHC{l}_3>C{H}_{3}Cl>CC{l}_4$

Answer 1

Answer: (C)

${\mu }_{CC{l}_4}=0$

The resultant dipole moment of three $C-H$ bonds $=0.4D$.
Net dipole moment $=1.9D$

The resultant of two $C-H$ bonds
$=\sqrt{{\left(0.4\right)}^2+{\left(0.4\right)}^2+2\times {\left(0.4\right)}^2\times cos109.5^{\circ }}$
$=\sqrt{0.16+0.16-0.2873}=0.18D$
The resultant of two $C-Cl$ bonds
$=\sqrt{{\left(1.5\right)}^2+{\left(1.5\right)}^2+2\times {\left(1.5\right)}^{2}cos109.5^{\circ }}$
$=\sqrt{2.25+2.25+2\times 2.25\times cos109.5^{\circ }}$
$=\sqrt{4.5-4.04}$
$=\sqrt{0.46}=0.678$
$\therefore {\mu }_{cal}=0.858D$ but experimentally for
$C{H}_{2}C{l}_2,\mu =1.59D$.
The resultant of three $C-Cl$ bonds $=1.5D$
Experimental dipole moment of $CHC{l}_3=1.15D$