MCQ
Electron affinity is the lowest for
- ANitrogen
- ✓Carbon
- COxygen
- DSulphur
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$(A)$ The acidity of compound $I$ is due to delocalization in the conjugate base.
$(B)$ The conjugate base of compound IV is aromatic.
$(C)$ Compound II becomes more acidic, when it has a $- NO _2$ substituent.
$(D)$ The acidity of compounds follows the order $I > IV > V > II > III$.
| List $I$ Element detected | List $II$ Reagent used/Product formed |
| $A$ Nitrogen | $I.$ $Na _2\left[ Fe ( CN )_5 NO \right]$ |
| $B$ Sulphur | $II.$ $AgNO _3$ |
| $C$ Phosphorous | $III.$ $Fe _4\left[ Fe ( CN )_6\right]_3$ |
| $D$ Halogen | $IV.$ $\left( NH _4\right)_2 MoO _4$ |
Choose the correct answer from the options given below:
$\begin{matrix}
C{{H}_{3}}\,\,\,\,\,\,\, \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
{{H}_{3}}C-C-CH=C{{H}_{2}} \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,\,\,\,\,\,\, \\
\end{matrix}$