NEET Practice Questions (MCQs) with Answers & Solutions

Carbon shows tetravalency in its excited state. In its ground state, carbon has the electronic configuration $1s^2 2s^2 2p^2$. In the excited state, one of the 2s electrons is promoted to the 2p orbital, resulting in the configuration $1s^2 2s^1 2p^3$. This allows carbon to form four covalent bonds, utilizing one electron from the 2s orbital and three electrons from the 2p orbitals, making it tetravalent.

In organic chemistry, C-C bond lengths depend on the type of bond between the carbon atoms. In $C_2H_2$ (acetylene), the carbon atoms are connected by a triple bond, which is shorter than double bonds (found in $C_2H_4$) and single bonds (found in $C_2H_6$ and $C_4H_8$). Therefore, $C_2H_2$ has the shortest C-C distance.

In benzene ($C_6H_6$), each carbon is sp2 hybridized, leading to bond angles of $120^ ext{°}$. Similarly, in ethene ($C_2H_4$), the carbons are also sp2 hybridized, resulting in bond angles of $120^ ext{°}$ around each carbon atom.

The compound $ (CH_3)_3 COH $ (tert-butyl alcohol) involves carbon atoms that are all sp^3 hybridized. In this structure, each carbon forms four sigma bonds, which is characteristic of sp^3 hybridization. The central carbon is bonded to three methyl groups and one hydroxyl group, utilizing sp^3 hybrid orbitals for all its bonds.

When the hybridization state of carbon changes from $sp^3$ to $sp^2$ and finally to $sp$, the bond angles increase progressively. In $sp^3$ hybridization, the bond angles are approximately 109.5°. In $sp^2$ hybridization, the bond angles are 120°, and in $sp$ hybridization, the bond angles are 180°. This increase in bond angle is due to the increasing s-character in the hybrid orbitals.

In a homologous series, each successive member differs by a -CH2- group. The molecular weight of a -CH2- group is 14 amu (12 for carbon and 2 for hydrogen). Therefore, the difference in atomic mass units (amu) between consecutive members of a homologous series is 14.