MCQ
$NaOH$ is a strong base because
- ✓It gives $O{H^ - }$ion
- BIt can be oxidised
- CIt can be easily ionised
- DBoth $(a)$ and $(c)$
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Product $(A)$ of the reaction is

[Atomic Number: $\mathrm{Fe}=26, \mathrm{Mn}=25, \mathrm{Co}=27$ ]
| List-$I$ | List-$II$ |
| ($P$) $ t_{2 g}^6 e_g^0$ | ($1$)$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ |
| ($Q$) $t_{2 g}^3 e_g^2$ | ($2$) $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ |
| ($R$) $e^2 t_2^3$ | ($3$)$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ |
| ($S$) $t_{2 g}^{+} e_g^2$ | ($4$)$\left[\mathrm{FeCl}_4\right]^{-}$ |
| $\left[\mathrm{CoCl}_4\right]^{2-}$ |
$(I)\,\,Zn\,+\,Conc. HNO_3\,(hot) \to Zn(NO_3)_2\,+\,X\,+\,H_2O$
$(II)\,\,Zn\,+\,dil. HNO_3\,(Cold) \to Zn(NO_3)_2\,+\,Y\,+\,H_2O$
Compound $X$ and $Y$ are respectively
$A + H_2O + CO_2 \to B$
$B + NaCl \to C + NH_4Cl$
$C\xrightarrow{\Delta }D + {H_2}O + C{O_2}$
Incorrect statement is