I have this question:
Consider the dichloropropane molecule, C3H6Cl2. Despite being a molecular compound, oxidation states can be assigned to the elements in butane.
+1Calculate the oxidation state of the carbon atoms. The oxidation state of the Cl atoms is at a value of -1 here. Show your work. Also fill in the oxidation state of the H atoms.
I have:
C3H6Cl2
Carbon 4x+6*1=0
4x+6=0
-6 -6
-1.5
4x = -6 , x=-6/4
/4 /4
x= -3/2 or -1.5
Oxidation state of c= -1.5
For H = 1+
Is this correct? Also what is strange about the oxidation state of carbon in dichloropropane?
C3H6Cl2
I would do it this way. Cl = 2*-1 = 2-
H = 6*+1 = 6+
3C + 6 + (-2) = 0
3C = -4
C = -4/3 each
C =
C = -4/3
Your calculation for the oxidation state of carbon in dichloropropane (C3H6Cl2) is incorrect. Let's go through it step by step:
1. Start with the molecular formula: C3H6Cl2
2. We know that the oxidation state of chlorine (Cl) is -1.
3. Since this is a neutral molecule, the sum of the oxidation states of each element must equal zero.
4. Let's assign the oxidation state of carbon (C) as x. Since there are three carbon atoms, the total oxidation state for carbon is 3x.
5. The total oxidation state for hydrogen (H) is 6*1 = 6.
6. The total oxidation state for chlorine is 2*(-1) = -2.
7. Now we can set up the equation: 3x + 6 - 2 = 0
8. Simplify the equation: 3x + 4 = 0
9. Subtract 4 from both sides: 3x = -4
10. Divide both sides by 3: x = -4/3
Therefore, the oxidation state of carbon in dichloropropane is -4/3 or approximately -1.33.
Now, to answer your second question, what is strange about the oxidation state of carbon in dichloropropane is that oxidation states are generally considered to be integers or whole numbers. It is rare to come across fractional oxidation states such as -1.33. This indicates an unusual distribution of electrons in the molecule.
To determine the oxidation state of carbon in dichloropropane, you correctly set up an equation using the concept that the sum of the oxidation states of all the atoms in a neutral molecule is equal to zero.
Let's break down the calculation step by step:
1. The formula for dichloropropane is C3H6Cl2, which means it contains three carbon atoms, six hydrogen atoms, and two chlorine atoms.
2. Assign a variable, let's say x, to the oxidation state of carbon. Since there are three carbon atoms, we have 3x oxidation states from carbon.
3. Hydrogen is always assigned an oxidation state of +1, so we have 6*(+1) = +6 from hydrogen.
4. Chlorine in this molecule is assigned an oxidation state of -1. Therefore, we have 2*(-1) = -2 from chlorine.
Now, setting up the equation, taking into account that the sum of the oxidation states of all the atoms in a neutral molecule is equal to zero:
3x + (+6) + (-2) = 0
Simplifying the equation:
3x + 6 - 2 = 0
3x + 4 = 0
Subtracting 4 from both sides:
3x = -4
Dividing both sides by 3:
x = -4/3 = -1.33
Therefore, the oxidation state of carbon in dichloropropane is approximately -1.33. This value is not a whole number, which is what makes it strange. Oxidation states are typically whole numbers or integers, but in this case, due to the presence of different elements and their electronegativities, the oxidation state of carbon does not come out as a whole number.
Additionally, you mentioned that the oxidation state of chlorine in dichloropropane is -1, which is correct since chlorine is more electronegative than carbon and tends to take electrons away.
Also, you correctly stated that the oxidation state of hydrogen is +1, as it usually is in most compounds.