Assign oxidation states to all atoms in each of the following. explain the order in which you determine the oxidation states for each element if applicable.
a)O2
b)HSO4-
c)Na2HPO4
d)CrCl3
a) For O2, the oxidation state of each oxygen atom can be determined as follows:
Since oxygen is a highly electronegative element, it tends to have an oxidation state of -2 in most compounds. However, in the case of O2, it is a diatomic molecule consisting of two oxygen atoms bonded together. Since the overall charge of O2 is neutral, each oxygen atom must have the same oxidation state.
Let's assume the oxidation state of each oxygen atom is x.
Total oxidation state of O2 = 2x
Since O2 is a neutral molecule, the total oxidation state must be 0.
Therefore, 2x = 0
x = 0
So, for O2, the oxidation state of each oxygen atom is 0.
b) For HSO4-, the oxidation state of each element can be determined as follows:
Let's assume the oxidation state of sulfur (S) is x.
Since hydrogen (H) is less electronegative than sulfur, it has an oxidation state of +1 in most compounds.
Since the overall charge of HSO4- is -1, the sum of oxidation states must equal -1.
So, the oxidation state of hydrogen (H) = +1.
Now, considering the overall charge of HSO4-, we can determine the oxidation state of sulfur (S).
Total oxidation state of HSO4- = (oxidation state of H) + (oxidation state of S) + (4 * oxidation state of O)
-1 = +1 + x + (4 * (-2))
Simplifying the equation, we get:
-1 = +1 + x - 8
x - 7 = -1
x = +6
So, for HSO4-, the oxidation state of sulfur (S) is +6.
c) For Na2HPO4, the oxidation state of each element can be determined as follows:
Let's assume the oxidation state of sodium (Na) is x.
Since sodium is a group 1 element, it typically has an oxidation state of +1 in its compounds.
Since the overall charge of Na2HPO4 is neutral (0), we need to determine the oxidation states of the remaining elements: hydrogen (H), phosphorus (P), and oxygen (O).
Considering the overall charge of Na2HPO4, we can determine the oxidation state of hydrogen (H) and oxygen (O).
Total oxidation state of Na2HPO4 = (2 * oxidation state of Na) + (oxidation state of H) + (oxidation state of P) + (4 * oxidation state of O)
Since Na2HPO4 is neutral, the total oxidation state must be 0.
So, the equation becomes:
0 = (2 * 1) + (oxidation state of H) + (oxidation state of P) + (4 * oxidation state of O)
Simplifying the equation, we get:
0 = 2 + (oxidation state of H) + (oxidation state of P) + (4 * oxidation state of O)
We know that oxidation state of sodium (Na) is +1, so we substitute it in the equation:
0 = 2 + (oxidation state of H) + (oxidation state of P) + (4 * oxidation state of O)
Simplifying further, we get:
0 = 2 + (oxidation state of H) + (oxidation state of P) + 4(oxidation state of O)
Now, considering that hydrogen (H) typically has an oxidation state of +1, we get:
0 = 2 + 1 + (oxidation state of P) + 4(oxidation state of O)
Simplifying once again, we get:
0 = 3 + (oxidation state of P) + 4(oxidation state of O)
Considering the oxidation state of oxygen (O) is typically -2, we can substitute this value and solve for the oxidation state of phosphorus (P):
0 = 3 + (oxidation state of P) + 4(-2)
Simplifying the equation:
0 = 3 + (oxidation state of P) - 8
(oxidation state of P) - 5 = 0
(oxidation state of P) = +5
So, for Na2HPO4, the oxidation states are as follows:
Sodium (Na) = +1
Hydrogen (H) = +1
Phosphorus (P) = +5
Oxygen (O) = -2
d) For CrCl3, the oxidation state of each element can be determined as follows:
Let's assume the oxidation state of chromium (Cr) is x.
Chlorine (Cl) is a highly electronegative element and typically has an oxidation state of -1 in compounds.
The overall charge of CrCl3 is neutral (0), so the sum of the oxidation states must be 0.
Total oxidation state of CrCl3 = (oxidation state of Cr) + (3 * oxidation state of Cl)
0 = x + (3 * (-1))
Simplifying the equation, we get:
0 = x - 3
x = +3
So, for CrCl3, the oxidation state of chromium (Cr) is +3, and the oxidation state of chlorine (Cl) is -1.
a) O2:
To determine the oxidation state of the atoms in O2, we need to consider the following rules:
1. The sum of oxidation states of atoms in a neutral molecule is always zero.
2. Oxygen (O) usually has an oxidation state of -2, unless it is in a peroxide where it is -1.
3. The sum of oxidation states in an ion is equal to the charge of the ion.
Since O2 is a neutral molecule, the sum of the oxidation states must be zero. Let x be the oxidation state of oxygen. Given that there are two oxygen atoms, the equation becomes:
2x + 0 = 0
Solving this equation, we find that x = 0. Therefore, the oxidation state of both oxygen atoms in O2 is 0.
The order in which we determine the oxidation states for each element is as follows:
1. Start by assigning oxidation states to the atoms with known oxidation states or rules (if applicable).
2. Then, assign oxidation states to the remaining atoms based on the sum of oxidation states and the known rules.
b) HSO4-:
In HSO4-, we need to determine the oxidation states of hydrogen (H), sulfur (S), and oxygen (O). Following the same rules, we have:
H usually has an oxidation state of +1 when bonded to nonmetals and -1 when bonded to metals.
Oxygen usually has an oxidation state of -2.
The sum of oxidation states in an ion is equal to the charge of the ion.
Since the charge of HSO4- is -1, the sum of oxidation states must be -1. Let x be the oxidation state of sulfur. Given that there is one sulfur atom, the equation becomes:
(+1) + x + 4(-2) = -1
Simplifying this equation, we find that x - 7 = -1, and thus x = +6. Therefore, the oxidation states are as follows:
Hydrogen (H): +1
Sulfur (S): +6
Oxygen (O): -2
The order in which we determined the oxidation states in this case was by starting with the known oxidation states of hydrogen and oxygen and then assigning the oxidation state of sulfur based on the sum of oxidation states.
c) Na2HPO4:
In Na2HPO4, we need to determine the oxidation states of sodium (Na), hydrogen (H), phosphorus (P), and oxygen (O). Following the same rules, we have:
Sodium usually has an oxidation state of +1.
Hydrogen usually has an oxidation state of +1 when bonded to nonmetals and -1 when bonded to metals.
Oxygen usually has an oxidation state of -2, unless it is in a peroxide where it is -1.
The sum of oxidation states in an ion is equal to the charge of the ion.
Since Na2HPO4 is a neutral molecule, the sum of the oxidation states must be zero. Let x be the oxidation state of phosphorus. Given that there is one phosphorus atom, the equation becomes:
2(+1) + 4x + 12(-2) = 0
Simplifying this equation, we find that 2 + 4x - 24 = 0, and thus 4x - 22 = 0. Solving for x, we get x = +5. Therefore, the oxidation states are as follows:
Sodium (Na): +1
Hydrogen (H): +1
Phosphorus (P): +5
Oxygen (O): -2
The order in which we determine the oxidation states in this case was by starting with the known oxidation states of sodium, hydrogen, and oxygen, and then assigning the oxidation state of phosphorus based on the sum of oxidation states.
d) CrCl3:
In CrCl3, we need to determine the oxidation states of chromium (Cr) and chlorine (Cl). Following the same rules, we have:
Chromium usually has multiple oxidation states, so we don't know it yet.
Chlorine usually has an oxidation state of -1.
The sum of oxidation states in a neutral molecule is zero.
Since CrCl3 is a neutral molecule, the sum of the oxidation states must be zero. Let x be the oxidation state of chromium. Given that there is one chromium atom and three chlorine atoms, the equation becomes:
x + 3(-1) = 0
Simplifying this equation, we find that x - 3 = 0, and thus x = +3. Therefore, the oxidation states are as follows:
Chromium (Cr): +3
Chlorine (Cl): -1
In this case, we determined the oxidation state of chlorine first, as it has a known oxidation state. Then, we assigned the oxidation state of chromium based on the sum of oxidation states.