MCQ
The oxidation state of $\ce{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
- A$+4$
- B$+3$
- ✓$+6$
- D$+5$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $H - H$ bond energy | $:\, 431.37 \,kJ\, mol^{-1}$ |
| $C= C$ bond energy | $:\, 606.10\, kJ \,mol^{-1}$ |
| $C - C$ bond energy | $:\, 336.49\, kJ\, mol^{-1}$ |
| $C - H$ bond energy | $:\, 410.50\, kJ\, mol^{-1}$ |
Enthalpy for the reaction,
$\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{C = C} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\, + \,H - H\, \to \,\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{H - C - C - H} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\,$
will be .............. $\mathrm{kJ \,mol}^{-1}$