MCQ
The correct order of dipole moment is
- ✓$C{H_4} < N{F_3} < N{H_3} < {H_2}O$
- B$N{F_3} < C{H_4} < N{H_3} < {H_2}O$
- C$N{H_3} < N{F_3} < C{H_4} < {H_2}O$
- D${H_2}O < N{H_3} < N{F_3} < C{H_4}$
$N{F_3} = 0.2\,D,\,\,N{H_3} = 1.47\,D$
and ${H_2}O = 1.85\,D$.
Therefore the correct order of the dipole moment is
$C{H_4} < N{F_3} < N{H_3} < {H_2}O$.
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$Xe{F_6}\,\xrightarrow{{excess\,\,{H_2}O}}\,'X'\, + \,HF$
$Xe{F_6}\,\xrightarrow{{2\,moles\,{H_2}O}}\,'Y'\, + \,HF$

$I.$ There is no $P_{\pi } -P_{\pi }$ bonds present in the molecule
$II.$ There are eight lone pair of electrons
$III.$ Each $S$ atom is $sp^3$ hybridised