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

Explanation:

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

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)

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 = +3

Oxidation state of Cl in   = -1

c)

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)

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 = +1

Oxidation state of Se in = -2

e)

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 = +2

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

f)

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 = +4

Oxidation state of O in = -2

g) HF

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

Oxidation state of H = +1

it would be d. henry moseley

hope this !

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