Oxidation Number

To determine which oxoacid of phosphorus has the highest number of oxygen atoms present in its chemical formula, let's examine each acid:

Pyrophosphorous acid (H₄P₂O₅): This acid has 5 oxygen atoms.

Hypophosphoric acid (H₄P₂O₆): This acid contains 6 oxygen atoms.

Phosphoric acid (H₃PO₄): This is the common phosphoric acid with 4 oxygen atoms.

Pyrophosphoric acid (H₄P₂O₇): This acid has 7 oxygen atoms.

From this analysis, we can see that Pyrophosphoric acid (H₄P₂O₇) contains the highest number of oxygen atoms, which is 7.

Thus, the correct answer is Pyrophosphoric acid.

Bromic acid is . Let the oxidation state of bromine be . In , the sum of oxidation states is:

Thus, the oxidation state of bromine in bromic acid is +5.

Perbromic acid is . Let the oxidation state of bromine be . In , the sum of oxidation states is:

Thus, the oxidation state of bromine in perbromic acid is +7.

Sum of oxidation states:

The reaction involves the addition of to an alkene. The major product is determined by Markovnikov's Rule, which states that the hydrogen atom adds to the carbon of the double bond that has more hydrogen atoms, and the bromine atom adds to the carbon that has fewer hydrogen atoms.

Step-by-Step Mechanism:

1. The starting compound is cyclobutyl methyl alkene ( ).
2. When is added, the double bond undergoes electrophilic addition: - The proton ( ) adds to the less substituted carbon of the double bond (the terminal carbon), creating a carbocation on the more substituted carbon. - This follows Markovnikov's rule, resulting in a tertiary carbocation.
3. The bromide ion ( ) then attacks the carbocation, leading to the formation of the major product.

Conclusion: The correct answer is Option (4): Cyclopentyl bromide with at C2 and at C1.

Step 1: Understanding the oxidation states of Xenon in XeF and its hydrolysis product.
1. In XeF , Xenon is bonded to 4 fluorine atoms.
The oxidation state of xenon in XeF can be calculated using the fact that the oxidation state of fluorine is -1. Therefore, the oxidation state of Xe in XeF is +4. 2. When XeF undergoes complete hydrolysis with water, the products formed are Xenon (Xe), Xenon trioxide (XeO ), oxygen (O ), and hydrofluoric acid (HF): In the product XeO , Xenon is bonded to three oxygen atoms. The oxidation state of oxygen is -2 in most compounds, so the oxidation state of Xenon can be determined as follows: Therefore, the oxidation state of Xenon in XeO is +6.
Step 2: Calculating the difference in oxidation state of Xe.
The oxidation state of Xenon in XeF is +4, and in XeO it is +6. The difference in oxidation state is: Thus, the difference in oxidation state of Xe between XeF and its oxidized product XeO is 2.

Reaction in acidic medium:

Mn⁺⁷ → Mn⁺², Change in Oxidation State = 7 − 2 = 5.

• Manganese reduces from +7 in KMnO₄ to +2 in Mn²⁺.
• Total change in oxidation state per manganese atom is 5.

In the compound , the oxidation state of manganese (Mn) is +7. When is reduced to , the oxidation state of manganese decreases to +2.

This is a reduction process, as the manganese atom gains electrons, reducing its oxidation state.