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.

LiCoO2.

Li is +1.
O is -2; there are two of them for a total of -4. To make the compound zero (all compounds are zero) Co must be +3.
NaAlH4.
Na is +1. Al is +3. That makes H a total of -4 or -1 for each H. (H is a hydride in this compound)
The others are done the same way.

(a) Co in LiCoO2:

To determine the oxidation number of Co in LiCoO2, we need to assign oxidation numbers to the other elements in the compound.

The oxidation number of Li is +1 since it is a Group 1 element.

The oxidation number of O is -2 since it is typically -2 in compounds, and there are two O atoms, giving a total charge of -4.

Since the overall charge of the compound is zero, we can calculate the oxidation number of Co as follows:

(+1) + Co + (-4) = 0
Co - 3 = 0
Co = +3

Therefore, the oxidation number of Co in LiCoO2 is +3.

(b) Al in NaAlH4:
The oxidation number of Na is +1 since it is a Group 1 element.

The oxidation number of H is +1 since it is typically +1 in compounds.

The overall charge of the compound is zero. Let's assume the oxidation number of Al is x.

(+1) + x + 4(+1) = 0
x + 4 = 0
x = -4

Therefore, the oxidation number of Al in NaAlH4 is -4.

(c) C in CH3OH (methanol):
The oxidation number of H is +1 since it is typically +1 in compounds.

The overall charge of the compound is zero. Let's assume the oxidation number of C is x.

(+1) + 4(+1) + x + (-2) = 0
x + 3 = 0
x = -3

Therefore, the oxidation number of C in CH3OH (methanol) is -3.

(d) N in GaN:
The overall charge of the compound is zero. Let's assume the oxidation number of N is x.

x + (-3) = 0
x = +3

Therefore, the oxidation number of N in GaN is +3.

(e) Cl in HClO2:
The oxidation number of H is +1 since it is typically +1 in compounds.

The overall charge of the compound is zero. Let's assume the oxidation number of Cl is x.

(+1) + x + (-2) + 2(-2) = 0
x - 3 = 0
x = +3

Therefore, the oxidation number of Cl in HClO2 is +3.

(f) Cr in BaCrO4:
The oxidation number of Ba is +2 since it is a Group 2 element.

The oxidation number of O is -2 since it is typically -2 in compounds.

The overall charge of the compound is zero. Let's assume the oxidation number of Cr is x.

+2 + x + 4(-2) = 0
x - 6 = 0
x = +6

Therefore, the oxidation number of Cr in BaCrO4 is +6.

To determine the oxidation number of an element in a compound, we need to assign a hypothetical charge to the element. Here's how you can determine the oxidation numbers for each of the indicated elements:

(a) Co in LiCoO2:
In LiCoO2, since Li is a Group 1 element, its oxidation number would be +1. Oxygen is usually assigned an oxidation number of -2.
Using this information, we can find the oxidation number of Co:
(+1) + x + 2(-2) = 0
Simplifying this equation, we get:
x - 4 = 0
Therefore, the oxidation number of Co is +4 in LiCoO2.

(b) Al in NaAlH4:
In NaAlH4, since Na is a Group 1 element, its oxidation number would be +1. Hydrogen is typically assigned an oxidation number of +1.
Using this information, we can find the oxidation number of Al:
+1 + x + 4(1) = 0
Simplifying this equation, we get:
x + 4 = 0
Therefore, the oxidation number of Al is -4 in NaAlH4.

(c) C in CH3OH (methanol):
Hydrogen is typically assigned an oxidation number of +1. In this case, oxygen is assigned an oxidation number of -2.
Using this information, we can find the oxidation number of C:
+1 + 3(1) + x + (-2) = 0
Simplifying this equation, we get:
x + 2 = 0
Therefore, the oxidation number of C is -2 in CH3OH.

(d) N in GaN:
In this case, since GaN is an ionic compound, we know that the overall charge of GaN is zero. Since nitrogen is less electronegative than gallium, we assume that N takes the negative oxidation number. Therefore, the oxidation number of N in GaN is -3.

(e) Cl in HClO2:
In this case, hydrogen is usually assigned an oxidation number of +1 and oxygen is typically assigned an oxidation number of -2.
Using this information, we can find the oxidation number of Cl:
+1 + x + 2(-2) = 0
Simplifying this equation, we get:
x - 3 = 0
Therefore, the oxidation number of Cl is +3 in HClO2.

(f) Cr in BaCrO4:
In this case, since BaCrO4 is an ionic compound, we know that the overall charge of BaCrO4 is zero. Oxygen is usually assigned an oxidation number of -2.
Using this information, we can find the oxidation number of Cr:
+2 + x + 4(-2) = 0
Simplifying this equation, we get:
x - 6 = 0
Therefore, the oxidation number of Cr is +6 in BaCrO4.

THank YOu Dr.BoB222