Question

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

Answer 1

Explanation:

a) $H_2SO_4$

oxidation state of H = +1

Oxidation state of O = -2

Let the oxidation no. of S be x.

$2\times1+ x+2\times -4 =0$

$2 +x-8 = 0\\x=+6$

Oxidation state of S in $H_2SO_4$ = +6

b) $Ca(OH)_2$

oxidation state of H = +1

Oxidation state of O = -2

Let the oxidation no. of Ca be x.

$x+2\times-2+2\times+1 =0\\x=4 -2\\x=+2$

Oxidation state of Ca in $Ca(OH)_2$ = +2

c) BrOH

oxidation state of H = +1

Oxidation state of O = -2

Let the oxidation no. of Br be x.

$x+1\times-2+1\times+1 =+1$

Oxidation no. of Br in BrOH is +1

d) $ClNO_2$

oxidation state of Cl = -1

Oxidation state of O = -2

Let the oxidation no. of N be x.

$-1+x+2\times-2\\x = 4+1 =+5$

Oxidation state of N in $ClNO_2$ = +3

e) $TiCl_4$

oxidation state of Cl = -1

Let the oxidation no. of Ti be x.

$x+4\times -1 = 0\\x=+4$

Oxidation state of Ti in $TiCl_4$ = +4

f) NaH

Na is more electropositive than H.

Therefore, oxidation state of Na = +1

Oxidation state of H = -1