Explain the oxidation states of group $14$ elements.

(A) The group $14$ elements have four electrons in their outermost shell $(ns^{2}np^{2})$. The common oxidation states exhibited by these elements are $+4$ and $+2$.
Carbon also exhibits negative oxidation states. Since the sum of the first four ionization enthalpies is very high,compounds in the $+4$ oxidation state are generally covalent in nature.
In heavier members,the tendency to show the $+2$ oxidation state increases in the sequence $Ge < Sn < Pb$. This is due to the inert pair effect,which is the inability of $ns^{2}$ electrons of the valence shell to participate in bonding.
The relative stabilities of these two oxidation states vary down the group. Carbon and silicon mostly show the $+4$ oxidation state. Germanium forms stable compounds in the $+4$ state and only a few compounds in the $+2$ state.
Tin forms compounds in both oxidation states ($Sn^{2+}$ acts as a reducing agent). Lead compounds in the $+2$ state are stable,while those in the $+4$ state are strong oxidizing agents.
In the tetravalent state,the number of electrons around the central atom in a molecule (e.g.,carbon in $CCl_{4}$) is eight. Being electron-precise molecules,they are normally not expected to act as electron acceptors or electron donors.
Although carbon cannot exceed its covalence of $4$,other elements of the group can do so due to the presence of vacant $d$-orbitals. Consequently,their halides undergo hydrolysis and they have a tendency to form complexes by accepting electron pairs from donor species.
For example,species like $SiF_{6}^{2-}$,$[GeCl_{6}]^{2-}$,and $[Sn(OH)_{6}]^{2-}$ exist,where the hybridization of the central atom is $sp^{3}d^{2}$.