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
$C{H_4} < N{F_3} < N{H_3} < {H_2}O$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
because
$STATEMENT$-$2$: The small size of $\mathrm{B}^{3+}$ favours formation of covalent bond.
In the given reaction what will be the product
