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.
$(a)$ They might be configurational isomers $(b)$ They might be diastereomers
$(c)$ They might be constitutional isomers $(d)$ They might be tautomers
$(e)$ They might be conformational isomers $(f)$ They might be enantiomers
$(g)$ They might be positional isomers
ethyne $\xrightarrow[\begin{smallmatrix}
(2)\,\,excess\,I-\,C{{H}_{2}}\,-\,{{(C{{H}_{2}})}_{2}}-\,C{{H}_{3}} \\
(3)\,\,{{H}^{\oplus }}
\end{smallmatrix}]{(1)\,\,excess\,NaN{{H}_{2}}}$
| $A/mol\,L^{-1}$ | $0.2$ | $0.2$ | $0.4$ |
| $B/mol\,L^{-1}$ | $0.3$ | $0.1$ | $0.05$ |
| $r_0/mol^{-1}s^{-1}$ | $5.0\times 10^{-5}$ | $5.0\times 10^{-5}$ | $1.4\times 10^{-4}$ |
