The greater the $$s$$  character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an $$sp$$  hybrid orbital with $$50\% \,s$$  character is more electronegative than those possessing $$s{p^2}$$  or $$s{p^3}$$  hybridised orbitals .
The asterisk marked carbon in $$C{H_3} - C{H_2} - C \equiv \mathop C\limits^ * H$$     is $$sp$$  hybridised hence, it is most electronegative.

As $$F$$ is most electronegative and in this structure there are two $$F$$ atoms, therefore, dispersal of negative charge is maximum hence it is the most stable.

$$C$$  is molecule has non-zero dipole moment.

2-chlorobutane has the structure

It is optically active due to the presence of asymmetric carbon atom $$\left( {{C^*}} \right).$$

TIPS/FORMULAE :
$$ - N{O_2}, - Cl\,\,{\text{and}}\, - OH$$     are electron-attracting or withdrawing group due to $$ - M, - E\,\,\,{\text{and/or}}\, - I$$     effects where as $$C{H_3}$$  show, $$ + I$$  effect ( electron releasing ).
Because of the $$ + I$$  effect of the $$C{H_3}$$  group, toluene has the highest electron density in the  $$o - $$  and $$p - $$ positions and hence can be most readily sulphonated.

Due to similar charges on adjacent atoms, the structure (A) is least stable.

$$\eqalign{ & {\text{Increase in mass of }}U{\text{ - tube}} = 1.08\,g \cr & {\text{Mass of water formed}} = 1.08\,g \cr & 18\,g\,{H_2}O \equiv 2\,g\,H \cr & 1.08\,g\,{H_2}O = \frac{2}{{18}} \times 1.08 \cr} $$
$$\% \,{\text{ of hydrogen }} = \frac{2}{{18}} \times \frac{{1.08}}{{0.92}} \times 100$$        $$ = 13.04\% $$

The conjugation (delocalisation) of electrons in pyrrole can be visualised as

As resonating structures III and IV are more stable than II and V. Thus, maximum electron density will be found on carbon 2 and 5.

Carbon bonded with a triple bond $$\left( {i.e.\,\,{C_1}} \right)$$   is $$sp$$  hybridised. Carbon bonded with a double bond $$\left( {{C_2}} \right)\,$$ is $$s{p^2}$$ hybridised.