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7 September, 21:43

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.

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  1. 7 September, 22:10
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    (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
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