- A$Na^+ > Mg^{2+} > Al^{3+} > Li^+ > Be^{2+}$
- ✓$Na^+ > Li^+ > Mg^{2+} > Al^{3+} > Be^{2+}$
- C$Na^+ > Mg^{2+} > Li^+ > Al^{3+} > Be^{2+}$
- D$Na^+ > Mg^{2+} > Li^+ > Be^{2+}$
In a period, on moving from left to right, the ionic radius decreases. Hence, $N a^{+}>M g^{2+}>A l^{3+}$ and $L i^{+}>B e^{2+}$
In a group, on moving down from top to bottom, the ionic radius increases. Hence, $N a^{+}>L i^{+}$ and $M g^{2+}>B e^{2+}$
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$Mn_{(S)} | Mn^{+2}_{(aq)} (0.4\,M) | | Sn^{+2}_{(aq)} (0.04\,M)| Sn_{(S)}$,
Calculate free energy change $\left( {\Delta G} \right)$ at $298\, K.$ .......... $\mathrm{kJ}$
Given : $E_{M{n^{ + 2}}/Mn}^o = - 1.18\,V ; \,E_{S{n^{ + 2}}/Sn}^o = - 0.14\,{\rm{volt}}\,\frac{{2.303RT}}{F}=0.06$
$2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g); K_1 = 2.1 \times 10^{-13}$
$2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g); K_2 = 1.4 \times 10^{-12}$
$K + {H_2}O + $Water $ \to KOH(aq) + \frac{1}{2}{H_2};\,\Delta H = - 48\,kcal$
$KOH + $Water $ \to KOH(aq);\,\Delta H = - 14\,kcal$
The heat of formation of $KOH$ is (in kcal)