MCQ
In the given compound, the number of $2^{\circ}$ carbon atom $/ \mathrm{s}$ is. . . . . .


- AThree
- ✓One
- CTwo
- DFour


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| List $-I$ (Molecule / Species) | List $-II$ |
| $(P)$ $NO_2$ (Unpaired electron in) | $(1)$ Vacant $p-$ orbital involved in hybridization |
| $(Q)$ $B_2H_6$ | $(2)$ $sp^2-$ orbital |
| $(R)$ $ClO_3$ (Unpaired electron in) | $(3)$ $sp^3-$orbital |
| $(S)$ $CH_3^+$ | $(4)$ $120^o$ Bond angle |
Reason : Graphite is a good conductor.
$KMn{O_4} + {H_2}S{O_4} + {H_2}{O_4} \to {K_2}S{O_4} + MnS{O_4} + {O_2} + {H_2}O$