Answer:
(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
