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Hybridisation of carbon $$ = s{p^{3 - {\text{number}}\,{\text{of}}\,\pi - bonds}}$$
In $$C{O_2},(\,O\mathop = \limits_{1\pi } C\mathop = \limits_{1\pi } O\,)$$    hybridisation of carbon $$ = s{p^{3 - 2}} = sp$$
In ethyne or acetylene, $${C_2}{H_2},(\,H - C\mathop \equiv \limits^{2\pi } C - H\,\,)$$      hybridisation of carbon $$ = s{p^{3 - 2}} = sp$$

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Compounds which dissociate below their boiling points or which have very high boiling points are distilled under reduced pressure.

$$C{l^ - }$$  Is the best leaving group among the given option.

The molecule in which all the carbon atoms are $$s{p^2}{\text{ - hybridised}}$$    will be coplanar. Thus, in option (A)
i.e.

all carbons are $$s{p^2}{\text{ - hybridised}}{\text{.}}$$

$$'Iso’$$  mean’s one $$Me$$  group is present in side chain. Hence, the structure of $$iso-butyl$$   group in an organic compound is

( $$'yl’$$ suffix is used to represent one $$-H$$  less than the parent hydrocarbon. )

Freeradicals are electrically neutral, unstable and very reactive on account of the presence of odd electrons.

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NOTE : Dipole moment is a vector quantity.
Methane molecule being symmetrical, has zero dipole moment. Replacement of one of the $$H - $$ atoms by $$Cl$$ atom increases the dipole moment. The increase in dipole moment is rather more than what can be expected because of the fact that the bond dipole moment of $$C - H$$   bond and that of $$C - Cl$$   bond reinforce one another. Replacement of another $$H$$ atom by $$Cl$$ increases the bond angle due to lone pair - lone pair repulsion between two $$Cl - $$ atoms thereby reducing the dipole moment of the molecule. Increase in angle is again caused by the the introduction of the third $$Cl - $$ atom. When the fourth $$Cl - $$ atom is introduced, the molecule $$\left( {CC{l_4}} \right)$$  again becomes symmetrical and dipole moment reduces to zero. So, $$C{H_3}Cl$$   will have the maximum dipole moment.