Product of the above given reaction is CH₃COOH + HOOC.CH₂CH₃.

Among all the given options, SF₄ would have permanent dipole moment.

Phenols are much more acidic than alcohols, due to the stabilisation of phenoxide ion by resonance.

Phenoxide ion is stabilized due to following resonating structures :

Ortho nitrophenol is most acidic because in it -NO₂ electron attracting group is attached on ortho position which helps in stabilising of negative charge on the oxygen of phenoxide ion.Hence, due to this reason acidic character of phenol is increased, while on attachment of -CH₃ group (electron donating group) acidic strength of phenol is decreased in cresol due to destabilisation of phenoxide ion.

Trans [CO (en)₂Cl₂] is able to exhibit the phenomena of optical isomerism because it can form a superimposable mirror image.

H₂S gas is passed in presence of HCl, in qualitative analysis of cation of second group, therefore,due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product (K_sp). Here, fourth group cations are not precipitated because they require for exceeding their ionic product to their solubility products and higher sulphide ions concentration due to their higher K_sp which is not obtained here due to common ion effect.

The most stable carbocation is t-alkyl carbocation because the order of stability of alkyl carbocation is

t-alkyl > s-alkyl > p-alkyl > CH${}_3^{+}$ carbocation

This stability order is described with the help of hyperconjugation and inductive effect.

Total vapour pressure of mixture

= (mole fraction of pentane x VP of pentane) + (mole fraction of hexane x VP of hexane)

= VP of pentane in mixture + VP of hexane in mixture

=$\left(\frac{1}{5}\times 440+\frac{4}{5}\times 120\right)$= 184 mm

Since,

VP of pentane in mixture = VP of mixture x mole fraction of pentane in vapour phase

88 = 184 xmole fraction of pentane in vapour phase

Therefore, Mole fraction of pentane in vapour phase =$\frac{88}{184}$= 0.478

Stability of alkene$\propto$$\frac{1}{\mathrm{Heat}\mathrm{of}\mathrm{hydrogenation}\mathrm{of}\mathrm{alkene}}$

Greater the number of alkyl groups attached to the doubly bonded carbon atoms, more stable isthe alkene. Hence, given alkene follow the following order of stability.

Hence, faster hydrogenation in

ClO₂ shows paramagnetic character due to presence of unpaired electron in its structure.

From second law of Faraday

$$\begin{gathered} \frac{{\mathrm{m}}_{\mathrm{Al}}}{{\mathrm{m}}_{\mathrm{H}}}=\frac{{\mathrm{E}}_{\mathrm{Al}}}{{\mathrm{E}}_{\mathrm{H}}} \\ \therefore \frac{4.5}{{\mathrm{m}}_{\mathrm{H}}}=\frac{{\displaystyle \raisebox{1ex}{$27$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{1} \\ {\mathrm{m}}_{\mathrm{H}}=0.5\mathrm{gm} \\ \because 2\mathrm{g}{\mathrm{H}}_2\mathrm{volume}\mathrm{at}\mathrm{STP}=22.4\mathrm{L} \\ \therefore 0.5\mathrm{gm}{\mathrm{H}}_2\mathrm{volume}\mathrm{at}\mathrm{STP}=\frac{22.4\times 0.5}{2}\mathrm{L}=5.6\mathrm{L} \end{gathered}$$