- A$1$
- B$2$
- ✓$3$
- D$4$
$I \cdot E_2-I E_1=412-284=128 \cdot kJ / mol$
$I \cdot E_3-I \cdot E_2=654-412=242 kJ / mol$
$I \cdot E_4-I \cdot E_3=3210-656=2554 kJ / mol$.
The avalue of $2554\,kJ/mol$ is the differeace between $I-E_4$ and $I-E_3$ suggest that the element is not able to loose its electron After loosing the three election. i.e the element will become stable on loss of three electrons.
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$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$, the $\mathrm{K}_{\mathrm{p}}$ for the process is $\mathrm{x} \times 10^{-1} \cdot \mathrm{x}=$ $\qquad$
[Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
