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For each of the following balanced equations, write theoxidation number above the symbol of each atom thatchanges the oxidation state in the course of the reactions.

(a) 2PF2I(l)+2Hg(l)P2F4(g)+Hg2I2(s)

(b)2KClO3(s)2KCl(s)+3O2(g)

(c)4NH3(g)+5O2(g)4NO(g)+6H2O(g)

(d)2As(s)+6NaOH(l)2Na3AsO3(s)+3H2(g)

Short Answer

Expert verified

The oxidation number is the property of an atom that shows the number of electrons lost or gained by a respective atom.

Step by step solution

01

Step 1:Oxidation and reduction.

a.2PF2I(l)+2Hg(l)P2F4g+Hg2I2s

The oxidation state of the atom of the reactant.

  1. P: +3
  2. F: -1
  3. I: -1
  4. Hg: 0

The oxidation state of the atom of the product.

  1. P: +2
  2. F: -1
  3. I: -1
  4. Hg: +1

b.2KClO3s2KCls+3O2g

The oxidation state of the atom of the reactant.

  1. K: +1
  2. Cl: +5
  3. O: -2

The oxidation state of the atom of the product.

  1. K: +1
  2. Cl: -1
  3. O: 0
02

Ammonia and Arsenic.

c.4NH3g+5O2g4NOg+6H2Og

The oxidation state of the atom of the reactant.

  1. N: -3
  2. H: +3
  3. O: 0

The oxidation state of the atom of the product.

  1. N: +2
  2. H: 0
  3. O: -2

d.2Ass+6NaOH(l)2Na3AsO3s+3H2g

The oxidation state of the atom of the reactant.

  1. As: 0
  2. Na: +1
  3. O: -2
  4. H: +1

The oxidation state of the atom of the reactant.

  1. As: +3
  2. Na: +3
  3. O: -6
  4. H: 0

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