19 Which of the following has been arranged in order of decreasing dipole moment (A) C*H_{3}*Cl > C*H_{3}*F > C*H_{3}*Br >CHyI (C) C*H_{3}*Cl > C*H_{3}*Br > C*H_{3}*I > C*H_{3}*F (B) C*H_{3}*F > C*H_{3}*Cl > C*H_{3}*Br > C*H_{3}*I (D) C*H_{3}*F > C*H_{3}*Cl > C*H_{3}*l > C*H_{3}*Br

Key Concepts Needed

1. Dipole Moment ($\mu$)

   $\mu = q \times r$

   where $q$ is the charge separation and $r$ is the bond length (distance between the partial charges).

2. Vector Nature

   In molecules like $CH_3X$, the geometry is tetrahedral. The three $C-H$ bond moments partially cancel each other; the net dipole is mainly along the $C-X$ bond but reduced by the opposite component of the three $C-H$ bonds.

3. Electronegativity vs. Bond Length Trade-off

   • Fluorine is the most electronegative ($\Delta EN$ largest) but has the shortest bond ($r$ smallest).
   • Going down the group (Cl, Br, I) electronegativity decreases, but bond length increases.
   • Therefore, the product $q \times r$ grows at first (F → Cl) because $r$ increases more than $q$ decreases, giving a larger $\mu$ for $CH_3Cl$.
   • After chlorine, $q$ falls off faster than $r$ grows, so $\mu$ drops for Br and even more for I.

Numerical Dipole Moments (experimental)

• $CH_3F$: ≈ 1.85 D
• $CH_3Cl$: ≈ 1.90 D
• $CH_3Br$: ≈ 1.82 D
• $CH_3I$: ≈ 1.62 D

Hence the decreasing order is:

$CH_3Cl > CH_3F > CH_3Br > CH_3I$

This matches Option (A).

Imagine a Tug-of-War!

Each molecule is like a four-way tug-of-war. One of the ropes is held by a big halogen atom (F, Cl, Br or I) and the other three ropes are held by tiny hydrogen atoms.

• The dipole moment tells us how strongly the big atom is pulling the electrons compared with the three little hydrogens.
• A stronger pull or a longer rope makes the ‘tug’ bigger.
• But the three hydrogens pull a little in the opposite direction, so sometimes the biggest, strongest atom doesn’t win as you would expect!

For $CH_3X$ (where $X$ is a halogen):

1. Fluorine: super-strong pull, but the rope (bond) is short.
2. Chlorine: good pull and a longer rope → ends up winning!
3. Bromine: weaker pull, longer rope than Cl, but not enough.
4. Iodine: weakest pull, longest rope, so the overall tug is the smallest.

So, the order of pull (dipole moment) is:

$CH_3Cl > CH_3F > CH_3Br > CH_3I$.

Step-by-Step Solution

1. Recall the Formula

   $\mu = q \times r$

2. Consider Electronegativity ($q$)

   Fluorine > Chlorine > Bromine > Iodine.

3. Consider Bond Length ($r$)

   C–F < C–Cl < C–Br < C–I.

4. Trade-off Analysis

   • Moving from F to Cl: $r$ increases more than $q$ decreases → $\mu$ increases.
   • Moving from Cl to Br: $r$ keeps increasing, but now $q$ drops faster → $\mu$ starts to fall.
   • Moving from Br to I: both low $q$ and long $r$ → $\mu$ smallest.

5. Experimental Confirmation

   $$\begin{aligned} CH_3F & : 1.85 \; D \\ CH_3Cl & : 1.90 \; D \\ CH_3Br & : 1.82 \; D \\ CH_3I & : 1.62 \; D \end{aligned}$$

6. Arrange in Decreasing Order

   $CH_3Cl > CH_3F > CH_3Br > CH_3I$

7. Match with Options

   Option (A) matches this order.

Answer: (A)