Determine the oxidation number for the indicated element in each of the following compounds: (a) Co in LiCoO2, (b) Al in NaAlH4, (c) C in CH3OH (methanol), (d) N in GaN, (e) Cl in HClO2, (f) Cr in BaCrO4.

(a) Co = + 3

(b) Al = + 3

(c) C = - 2

(d) N = - 3

(e) Cl = - 3

(f) Cr = + 6

Explanation: Oxidation number: Oxidation number of an element in any particular chemical reaction is the defined as the electrical charge.

(a) The oxidation number of Co in LiCoO₂

Where the oxidation numbers of

Li = + 1, O = - 2 and let the oxidation number of Co = x

Therefore,

x + (+1) + (-2*2) = 0

x + 1 - 4 = 0

x - 3 = 0

x = + 3.

Therefore the oxidation number of Co = + 3

(b) Al in NaAlH₄

Where Oxidation of Na = + 1, H = - 1, and let the oxidation number of Al = y

+1 + (y) + (-1*4) = 0

y - 4 + 1 = 0

y - 3 = 0

y = + 3.

Therefore the oxidation number of Al = + 3

(c) C in CH₃OH

Where the oxidation number of

H = + 1, O = - 2, let the oxidation number of C = z

Therefore,

z + (+1*3) + (-2) + (+1) = 0

z + 3 - 2 + 1 = 0

z + 2 = 0

z = - 2.

Therefore the oxidation number of C = - 2.

(d) N in GaN

Where Ga = + 3 and let the oxidation number of N = a.

Therefore,

a + (+3) = 0

a + 3 = 0

a = - 3.

Therefore the oxidation number of N = - 3

(e) Cl in HClO₂

Where H = + 1, O = - 2 and let The oxidation number of Cl = b

+1 + b + (-2*2) = 0

1+b-4 = 0

b + 3 = 0

b = - 3

Therefore the oxidation number of Cl = - 3

(f) Cr in BaCrO₄

Where Ba = + 2, O = - 2 and let the oxidation number of Cr = c

+2 + c + (-2*4) = 0

+2+c-8 = 0

c - 6 = 0

c = 6

Therefore the oxidation number of Cr = + 6