Oxidation Number And Oxidation States

The correct option is (C) : A-III, B-I, C-IV, D-II

To determine the oxidation state of the central atoms in the given chemical species, we need to use the rules for calculating oxidation numbers. Let's analyze each compound one by one:

Now that we have calculated the oxidation states, we can match List I with List II:

Therefore, the correct answer is: (A)- (IV), (B)- (III), (C)- (II), (D)- (I).

Fluorine's consistent -1 oxidation state can be attributed to its electronic configuration and the absence of d orbitals. Being the most electronegative element, fluorine seeks to gain an electron to fill its valence shell, achieving a stable octet configuration.

Fluorine has the electronic configuration: . Its outermost shell requires just one more electron to complete the p orbital. The absence of d orbitals in fluorine implies it cannot expand its valence shell, unlike other elements which can utilize d orbitals to form different oxidation states.

Thus, fluorine only shows an oxidation state of -1, as it cannot share or lose electrons to form higher positive oxidation states, a consequence of lacking d orbitals.