Chemistry, 27.09.2019 16:30 josuemartinez1030

Determine the oxidation number of chlorine in each of the following substances:
a. kclo3
b. cl2
c. ca(clo4)2
d. cl2o

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Answer from: plewse

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)

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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