Question

Determine the oxidation states of the elements in the following compounds
(a) NaI
(b)GdCl3
(c)LiNO3
(d)H2Se
(e)Mg2Si
(f)RbO2, rubidium superoxide
(g)HF

Answer 1

Explanation:

There are certain rules for determination of oxidation states of elements:

• Oxidation states of elements in their elemental state is zero.
• The elements of the group 1 and group 2 of the periodic table (metals) have oxidation state of +1 and +2 respectively.
• Hydrogen has generally an oxidation number of +1 except when hydrogen is bonded to metals (say Na, K, etc) where it has an oxidation state of -1.
• Oxygen has an oxidation state of -2 generally. but in peroxides, its oxidation state is -1.
• Florine has an oxidation of -1 always as it is the most electronegative atom.
• Oxidation state of halogens is generally -1 if not bonded to more electronegative atom.

a) NaI

Na is a group 1 element, therefore its oxidation state is +1

Let the oxidation no. of I be x.

+1 + x = 0

x = -1

Oxidation state of Na in NaI = +1

Oxidation state of I in NaI = -1

b) $GdCl_3$

Oxidation state of Cl is -1 as it is more electronegative than Gd.

Let the oxidation no. of Gs be x.

x + -1×3 = 0

x = +3

Oxidation state of Gd in $GdCl_3$ = +3

Oxidation state of Cl in  $GdCl_3$ = -1

c) $LiNO_3$

Li is a group 1 element, therefore its oxidation state is +1.

It is an oxide, so oxidation state of O is -2.

Let the oxidation no. of N be x.

+1 + x + -2×3 = 0

x = +5

Oxidation state of Li in = +1

Oxidation state of O = -2

Oxidation state of N = +5

d) $H_2Se$

As Se is more electronegative than H, so oxidation state of H is +1.

Let the oxidation no. of Se be x.

+1×2 + x = 0

x = -2

Oxidation state of H in $H_2Se$ = +1

Oxidation state of Se in $H_2Se$ = -2

e) $Mg_2Si$

Mg is a group 2 elements, so its oxidation state is +2

Let the oxidation state of Si be x.

+2×2 + x= 0

x = -4

Oxidation state of Mg in $Mg_2Si$ = +2

Oxidation state of Si in tex]Mg_2Si[/tex] = -4

f) $RbO_2$

In oxides, oxidation state of O = -2

Let the oxidation state of Rb be x

x  -2×2 += 0

x = +4

Oxidation state of Rb in $RbO_2$ = +4

Oxidation state of O in $RbO_2$ = -2

g) HF

F is the most electronegative element, so its oxidation state is -1

Oxidation state of H = +1