- A$2,\, 0$ and $1$ lone pairs of central atom respectively
- B$1, \,0$ and $1$ lone pairs of central atom respectively
- C$0, \,0$ and $2$ lone pairs of central atom respectively
- ✓$1, \,0$ and $2$ lone pairs of central atom respectively
$CF_4$ : $sp^3$ hybridization
$XeF_4$ : $sp^3d^2$ hybridization
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$(A)$ $\Delta G$ is positive $(B)$ $\Delta S _{\text {system }}$ is positive
$(C)$ $\Delta S _{\text {surroundings }}=0$ $(D)$ $\Delta H =0$

$(I)$ $C{H_2} = CH\mathop C\limits^ + HC{H_3}$
$\begin{array}{*{20}{c}} {{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\ {(II)\,\,\,\,\,\,\,\,\,\,C{H_2} = C - \mathop {{\text{ }}C}\limits^ + {H_2}} \end{array}$
$(III)$ $C{H_3}CH = CH\mathop C\limits^ + {H_2}$
Statement $II :$ The lone pair of electrons on nitrogen in pyridine makes it basic.
Choose the CORRECT answer from the options given below: