Q.Consider the elements : Cs, Ne, I and F

Concept: Oxidation states reflect an element's electron-gaining or electron-losing tendency. Highly electronegative elements gain electrons (negative states), electropositive elements lose electrons (positive states), and intermediate elements can do both.

Analysis by element:

1. Fluorine (F): Most electronegative element ($\chi = 4.0$). Always gains electrons, never loses them. Exhibits only $-1$ oxidation state in compounds.

2. Cesium (Cs): Highly electropositive alkali metal. Readily loses its single valence electron. Exhibits only $+1$ oxidation state.

3. Iodine (I): Moderately electronegative halogen. Can gain electrons (e.g., $-1$ in $\text{HI}$) or lose them to more electronegative atoms (e.g., $+1, +3, +5, +7$ in oxides and oxoacids). Exhibits both positive and negative states. …

Oxidation state is a bookkeeping tool that tracks electron distribution in compounds. It hinges on electronegativity: highly electronegative elements pull electrons toward themselves (negative oxidation states), while electropositive elements lose electrons (positive oxidation states). The key is to understand each element's position in the periodic table and its chemical behaviour.

Let's examine the four elements systematically.

Understanding each element

Fluorine (F): The most electronegative element in the entire periodic table (electronegativity ≈ 4.0). In any compound, fluorine always attracts electrons more strongly than its partner, meaning it always gains electrons and exhibits a negative oxidation state. There is no element more electronegative than fluorine, so it can never donate electrons to another atom.

Cesium (Cs): An alkali metal in Group 1, cesium is one of the most electropositive elements. It has a single valence electron in the 6s orbital that it readily loses to achieve a stable noble-gas configuration. Cesium is so eager to lose this electron that it never gains electrons from other atoms in chemical compounds.

Iodine (I): A halogen in Group 17, iodine sits in an interesting middle ground. While it's electronegative enough to gain electrons from metals (showing negative oxidation states), it's also less electronegative than oxygen, fluorine, and chlorine. When bonded to these more electronegative elements, iodine can lose electrons and exhibit positive oxidation states.

Neon (Ne): A noble gas with a complete octet ($1s^2 2s^2 2p^6$). Neon is chemically inert under ordinary conditions and forms essentially no stable compounds. Without forming bonds, it has no oxidation state to speak of.

Analysing oxidation states

1. Element exhibiting only negative oxidation state

   Fluorine's unmatched electronegativity means it always pulls electrons toward itself. In every known compound—whether $\text{HF}$, $\text{NaF}$, or $\text{OF}_2$—fluorine has an oxidation state of $-1$. No exceptions exist because no element can out-compete fluorine for electrons.

2. Element exhibiting only positive oxidation state

   Cesium's single valence electron is so loosely held that it invariably loses it in chemical reactions. In compounds like $\text{CsCl}$ or $\text{Cs}_2\text{O}$, cesium exhibits an oxidation state of $+1$. The idea of cesium gaining electrons is chemically unrealistic given its low electronegativity.

3. Element exhibiting both positive and negative oxidation states

   Iodine demonstrates remarkable versatility:

   • Negative oxidation states: In $\text{HI}$ or $\text{NaI}$, iodine is more electronegative than hydrogen or sodium, so it shows $-1$.
   • Positive oxidation states: In compounds with more electronegative elements—$\text{IF}_5$ (oxidation state $+5$), $\text{IF}_7$ ($+7$), $\text{I}_2\text{O}_5$ ($+5$), or $\text{HIO}_4$ ($+7$)—iodine loses electrons and exhibits positive oxidation states. …