Verified Answer :

The oxidation number of Cl may vary in each compound. In a, this is a neutral compound which means the overall charge is zero. oxygen has a total of -6, k has +1, so Cl has to be +5. Cl in b has a zero charge. In c, oxygen has -16, Ca has +2 and Cl has to be +7. In d, oxygen has -2 so Cl is +1.

Answer : The oxidation number of chlorine in \(KClO_3,Cl_2,Ca(ClO_4)_2\text{ and }Cl_2O\) are, \(+5,0,+7\text{ and }+1\) respectively.

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with sign (+ and -) first and then the magnitude.

Rules for Oxidation Numbers :

  • The oxidation number of a free element is always zero.
  • The oxidation number of a monatomic ion equals the charge of the ion.
  • The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.
  • The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.
  • The oxidation number of a Group 1 element in a compound is +1.
  • The oxidation number of a Group 2 element in a compound is +2.
  • The oxidation number of a Group 17 element in a binary compound is -1.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to calculate the oxidation number of chlorine (Cl) in each of the following substances.

(a) \(KClO_3\)

Let the oxidation state of Cl be, 'x'

\((+1)+x+3(-2)=0\\\\x=+5\)

Thus, the oxidation state of Cl is, (+5)

(b) \(Cl_2\)

The oxidation number of a free element is always zero.

Thus, the oxidation state of Cl is, (0)

(c) \(Ca(ClO_4)_2\)

Let the oxidation state of Cl be, 'x'

\((+2)+2[x+4(-2)]=0\\\\x=+7\)

Thus, the oxidation state of Cl is, (+7)

(d) \(Cl_2O\)

Let the oxidation state of Cl be, 'x'

\(2x+(-2)=0\\\\x=+1\)

Thus, the oxidation state of Cl is, (+1)

Therefore, the oxidation number of chlorine in \(KClO_3,Cl_2,Ca(ClO_4)_2\text{ and }Cl_2O\) are, \(+5,0,+7\text{ and }+1\) respectively.

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