MCQ
The $H-C-H$ bond angle in $CH_4$ is $109.5^o$, due to lone pair repulsion, the $H - O - H$ angle in $H_2O$ will
- Aremain the same
- Bincrease
- ✓decrease
- Dbecome $180^o$
$1.$ Bonding pairs $2.$ Non-bonding or lone pairs
The combination of these determine the shape of molecule.
- single bond have big impact on shape and double bond has less.
- The bond angle decreases as non-bonding pair increases
- In $H _2 O$, there are two lone pair of $e^{-}$. Hence bond angle will decrease.
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$\mathrm{HF}, \mathrm{H}_2, \mathrm{H}_2 \mathrm{~S}, \mathrm{CO}_2, \mathrm{NH}_3, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{CHCl}_3, \mathrm{SiF}_4$, $\mathrm{H}_2 \mathrm{O}, \mathrm{BeF}_2$
$\begin{array}{*{20}{c}}
{C{H_3} - C{H_2} - C - CH - CHO} \\
{\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,CN}
\end{array}$
is :
Given : $\Delta H _{ f }^{\ominus}\left( Al _2 O _3\right)=-1700\,kJ\,mol ^{-1}$
$\Delta H _{ f }^\theta\left( Fe _2 O _3\right)=-840\,kJ\,mol ^{-1}$
Molar mass of $Fe , Al$ and $O$ are $56,27$ and $16\,g\,mol ^{-1}$ respectively.