- A$6$
- B$3$
- ✓$1$
- D$2$
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$C_{(graphite)} +O_2(g) \rightarrow CO_2(g)\,;$
$\Delta _rH^o=-395.5 \, kJ\,mol^{-1}$
$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O(l)\,;$ $\Delta _rH^o =-285.8\, kJ\, mol^{-1}$
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)\,;$
$\Delta _rH^o = + 890.3\, kJ\, mol^{-1}$
Based on the above thermochemical equations, the value of, $\Delta _rH^o $ at $298\, K$ for the reaction
$C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g) $ will be for $\Delta_{r} H^{\circ}$ ........... $kJ \,mol^{-1}$
$(I)$ the highest oxidation state of iron is $+ 6$ in $K_ 2FeO_4$
$(II)$ that the iron shows $+ 2$ oxidation state with six electrons in the $3d$ orbitals
$(III)$ the common oxidation state of iron is $+3$ with five unpaired electrons in $3d$ orbital
$(A)$ $\mathrm{O}$ $(B)$ $\mathrm{N}$ $(C)$ Be $(D)$ $\mathrm{F}$ $(E)$ $B$
Choose the correct answer from the options given below :