Question

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. (a) H2SO4 (b) Ca(OH)2 (c) BrOH (d) ClNO2 (e) TiCl4 (f) NaH'

Answer 1

Answer :

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

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

Rules for Oxidation Numbers are :

• 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.
• 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 determine the oxidation state of the elements in the compound.

(a) $H_2SO_4$

Let the oxidation state of 'S' be, 'x'

$2(+1)+x+4(-2)=0\\\\x=+6$

Hence, the oxidation state of 'S' is, (+6)

(b) $Ca(OH)_2$

Let the oxidation state of 'Ca' be, 'x'

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

Hence, the oxidation state of 'Ca' is, (+2)

(c) $BrOH$

Let the oxidation state of 'Br' be, 'x'

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

Hence, the oxidation state of 'Br' is, (+1)

(d) $ClNO_2$

Let the oxidation state of 'N' be, 'x'

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

Hence, the oxidation state of 'N' is, (+5)

(e) $TiCl_4$

Let the oxidation state of 'Ti' be, 'x'

$x+4(-1)=0\\\\x=+4$

Hence, the oxidation state of 'Ti' is, (+4)

(f) $NaH$

Let the oxidation state of 'Na' be, 'x'

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

Hence, the oxidation state of 'Na' is, (+1)