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

Step-by-step 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