Which one of the following has the highest dipole moment?
(i)CH2Cl2
(ii)CHCl3
(iii)CCl4
CH2Cl2
CHCl3
CCl4
(i)
Dichlormethane(CH2Cl2) µ=1.60D
(ii) 
Chloroform(CHCl3) µ=1.08D
(iii) 
Carbon tetrachloride(CCl4)
µ=0D CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds cancel each other. Hence, its resultant dipole moment is zero. As shown in the above figure, in CHCl3, the resultant dipole moments of two C−Cl bonds are opposed by the resultant dipole moments of one C−H bond and one C−Cl bond. Since the resultant of one C−H bond and one C−Cl bond dipole moment is smaller than two C−Cl bonds, the opposition is to a small extent. As a result,CHCl3 has a small dipole moment of 1.08 D. the other hand, in the case of CH2Cl2, the resultant of the dipole moments of two C−Cl bonds is strengthened by the resultant of the dipole moments of two C−H bonds. As a result, CH2Cl2 has a higher dipole moment of 1.60 D than CHCl3 i.e.,CH2Cl2 has the highest dipole moment. Hence, the given compounds can be arranged in the increasing order of their dipole moments as CCl4<CHCl3<CH2Cl2.
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.
If vector \( \mathbf{a} = 3 \hat{i} + 2 \hat{j} - \hat{k} \) \text{ and } \( \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \), then which of the following is correct?