- ✓Bent $T$ shape
- BSea saw
- CSquare pyramid
- DPyramidal
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$\mathop {\begin{array}{*{20}{c}}
{{\mkern 1mu} {\mkern 1mu} Ph}\\
|\\
{C{H_3} - C - OH}\\
|\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2} - C{H_3}{\mkern 1mu} {\mkern 1mu} }
\end{array}}\limits_{{\rm{'D'}}} $ $\xrightarrow{{HI}}$ products
If the configuration of substrate is $D$, then configuration of products will be

$(a)$ Number of $S-S$ bonds in $H_2S_nO_6$ are $(n +1)$
$(b)$ When $F_2$ reacts with $H_2O$ it forms $HF,$ $O_2$ & $O_3.$
$(c)$ $XeF_6$ on hydrolysis shows disproportionation reaction
$(d)$ $Al$ metal on reacting with dilute $NaOH$ gives a white precipitate of $Al (OH)_3$ as a final product

$C$ $ (graphite) $ $ + {O_2}(g) \to C{O_2}(g);\,\Delta H = - 393.5$
If the graphite is made of diamond then the above figure is $\Delta H$.......$kJ$
$Zn\,(s)\,\, + \,\,C{u^{2 + }}\,(aq)\, \rightleftharpoons \,Z{n^{2 + \,}}\,(aq)\, + Cu\,(s)$
At $300\,K$ is approximately $(R = 8 \,JK^{-1}\,mol^{-1},\, F = 96000\,C\,mol^{-1})$