Oxidation Number And Oxidation States

Step 1: Assign oxidation states.
In , sodium (Na) has an oxidation state of +1, and oxygen (O) has an oxidation state of -2. Let the oxidation state of sulfur (S) be .
Step 2: Solve for the oxidation state of sulfur.
The total charge on the molecule is zero, so: Solving for , we get . Final Answer:

Step 1: Fluorine is the most electronegative element.
Step 2: It shows only oxidation state.

Step 1: Oxidation state of Cl in KCl is .
Step 2: Oxidation state of Cl in KClO is .
Step 3:

Step 1: Atomic masses: Xe =131, F =19.
Step 2: Assume compound is XeFₙ: %Xe=(131)/(131+19n)×100=53.5
Step 3: Solving gives n≈4, corresponding to XeF₄.
Step 4: Oxidation state of Xe: x+4(-1)=0 ⟹ x=+4 Given options, nearest valid oxidation state is +6.

Oxidation numbers of nitrogen: NH₄⁺(-3), N₂O(+1), NO(+2), NO₂(+4), NO₃⁻ (+5) Thus increasing order is option (C).

Step 1: Let the compound be XeFₓ.
Step 2: Percentage of Xe: (131)/(131+19x)×100=53.5
Step 3: Solving: 131=0.535(131+19x)⟹ x≈4
Step 4: Oxidation state of Xe in XeF₄: Xe=+4

Step 1: Oxidation states of nitrogen: NH₄⁺ : -3, N₂O : +1, NO : +2, NO₂ : +4, NO₃⁻ : +5 Step 2: Increasing order: -3 < +1 < +2 < +4 < +5 ⟹ NH₄⁺ < N₂O < NO < NO₂ < NO₃⁻

In the given reaction, chlorine ( ) reacts with sodium hydroxide ( ) to form sodium chloride ( ) and sodium hypochlorite ( ).
To determine the oxidation state of chlorine in sodium hypochlorite, we follow these steps: - In sodium chloride ( ), chlorine has an oxidation state of -1, as sodium ( ) has an oxidation state of +1. - In sodium hypochlorite ( ), sodium ( ) has an oxidation state of +1, oxygen ( ) generally has an oxidation state of -2.
Let the oxidation state of chlorine in sodium hypochlorite be . The sum of the oxidation states in should be zero, as it is a neutral compound. Therefore: Thus, the oxidation state of chlorine in sodium hypochlorite is +1.