MCQ
The reaction $C{H_2} = C{H_2} + {H_2}\mathop {\xrightarrow{{Ni}}}\limits_{250 - {{300}\,^o}C} C{H_3} - C{H_3}$ is called
- AWurtz's reaction
- BKolbe's reaction
- ✓Sabatier and Senderen's reaction
- DCarbylamine reaction
This is catalytic hydrogenation of unsaturated hydrocarbons. Methane cannot be obtained by this method.
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|
Column $I$ |
Column $II$ |
| $A$. Inert-gas elements | $1.\left( {n - 1} \right){d^{1 - 10}}\,n{s^{1 - 2}}$ |
| $B$. Transition elements | $2\,. ns^2\,np^6$ |
| $C$. Inner-transition elements | $3.\left( {n - 2} \right){f^{1 - 14}}\,\left( {n - 1} \right){s^2}{p^6}{d^{0 - 1}}\,n{s^2}$ |

| List$-I ($Compound$)$ | List$-II ($Colour$)$ | ||
| $A.$ | $Fe_4[Fe(CN)_6]_3.xH_2O$ | $I.$ | Violet |
| $B.$ | $[Fe(CN)_5NOS]^{4–}$ | $II.$ | Blood Red |
| $C.$ | $[Fe(SCN)]^{2+}$ | $III.$ | Prussian Blue |
| $D.$ | $(NH_4)_3PO_4.12MoO_3$ | $IV.$ | Yellow |
$2NO(g) + 2{H_2}(g) \to {N_2}(g) + 2{H_2}O(g)$ is :
Step $1$ : $2NO(g) + {H_2}(g)\xrightarrow{{slow}}{N_2} + {H_2}{O_2}$
Step : $2$ ${H_2}{O_2} + {H_2}\xrightarrow{{fast}}2{H_2}O$
Then the correct statement is
$(A)\,{{C}_{8}}{{H}_{10}}\xrightarrow{KMn{{O}_{4}}}(B){{C}_{8}}{{H}_{6}}{{O}_{4}}\xrightarrow[Fe]{B{{r}_{2}}}{{C}_{8}}{{H}_{5}}Br{{O}_{4}}(C)$ (one-product only)