Question

Out of the elements C, S, Ge, Sn, and Pb, only +4 oxidation state is shown by how many elements?

Answer 1

Hint: The common oxidation state of group 14 is +4. Due to the inert pair effect the stability of the +4 oxidation state of group 14 changes on moving down the group.

Complete answer:
Carbon, silicon, germanium, tin, and lead are the elements of group 14. Their outer electronic configuration is \[n{{s}^{2}}n{{p}^{2}}\] where the n is the number of outermost principal shells.
Carbon and silicon show a +4 oxidation state. The remaining elements of this group, i.e., Ge (germanium), Sn (tin), and Pb (lead), however, shoe two oxidation states of +2 and +4 due to the inert pair effect which arises due to ineffective shielding of the valence s-electrons by the intervening d- and/or f-electrons. Evidently, as the number of d- and/or f-electrons increases down the group from Ge to Pb, the inert pair effect becomes more and more prominent. As a result, the stability of the +4 oxidation state decreases while that of the +2 oxidation state increases from Ge to Pb. In other words, the stability of the +2 oxidation state increases markedly in the sequence: $ Ge< Sn < Pb $ i.e. +2 oxidation state of Pb is more stable.
Hence all the elements of group 14 show a +4 oxidation state but stability decreases down the group.

Note: Further, in the tetravalent state, the number of electrons around the central atom in a molecule of group 14 is eight. Being electron-precise molecules, they neither act as electron-acceptor or electron-donors.