Q.Which element exhibits maximum catenation and why?

Organic chemistry is built entirely on one unusual fact about carbon: its ground-state configuration is $1s^2 2s^2 2p^2$, four valence electrons short of a stable octet either way, and reaching a noble-gas configuration by simply gaining or losing four electrons (to form $C^{4-}$ or $C^{4+}$) costs far too much energy to ever happen in practice. So carbon essentially never forms simple ions -- instead, in almost every compound it makes, it forms four covalent bonds.

Those four bonds are explained by hybridisation. During bonding, one electron is promoted from the $2s$ orbital into the empty $2p_z$ orbital, and the resulting four orbitals mix and rearrange:

• sp3 hybridisation gives four equivalent sigma bonds arranged with tetrahedral geometry -- the default for a carbon making four single bonds.
• sp2 hybridisation leaves one unhybridised $p$ orbital free; the three sp2 orbitals give trigonal planar geometry, and sideways overlap of the leftover $p$ orbitals on two adjacent sp2 carbons forms one pi bond (a C=C double bond, one sigma + one pi).
• sp hybridisation leaves two unhybridised $p$ orbitals free; the two sp orbitals give linear geometry, and sideways overlap of both leftover $p$ orbitals forms two pi bonds (a C≡C triple bond, one sigma + two pi).

Counting pi bonds in a structure is therefore just a matter of counting double and triple bonds: a plain C=C contributes one pi bond; a C=O (as in an aldehyde or ketone carbonyl) also contributes one pi bond; a carboxylic acid's C=O contributes one pi bond (the C-OH bond is a plain sigma bond, not a second pi bond); a C≡C or C≡N contributes two pi bonds. Working out a molecule's hybridisation pattern carbon-by-carbon (sp3 wherever only single bonds meet that carbon, sp2 at each end of a double bond, sp at each end of a triple bond) is one of the most common early exam skills built on this idea. …