What is the oxidation number of Si in (CH3)2SiCl2
well, Dimethyldichlorosilane , you have the formula wrong, consider
Si(CH3)2Cl2
as you can see from this Si has four things attached to it, two methyl groups, and two Cl- ions. Si must be +4
CH3
CH3-Si=Cl
Cl
disregard that diagram, it came out way wrong.
To determine the oxidation number of an element in a compound, you need to assign oxidation numbers (also known as oxidation states) to the other elements in the compound and use the known rules for assigning oxidation numbers.
In the compound (CH3)2SiCl2, let's start by assigning oxidation numbers to the other elements involved.
The oxidation number of hydrogen (H) is almost always +1 in compounds. Since there are three hydrogen atoms in each CH3 group and two CH3 groups in the compound, the total oxidation number for hydrogen (H) is +1 * 3 * 2 = +6.
The oxidation number of chlorine (Cl) is typically -1 in compounds. Since there are two chlorine atoms in the compound, the total oxidation number for chlorine (Cl) is -1 * 2 = -2.
To find the oxidation number of silicon (Si), we will make use of the fact that the sum of the oxidation numbers in a neutral compound is zero. In this case, we know that the sum of the oxidation numbers of hydrogen (H), chlorine (Cl), and silicon (Si) must be zero.
So, let's set up an equation:
(+6) + (-2) + (oxidation number of Si) = 0
Simplifying the equation, we have:
+6 - 2 + (oxidation number of Si) = 0
Rearranging and solving for the oxidation number of silicon (Si), we get:
(oxidation number of Si) = 0 - 6 + 2 = -4
Therefore, the oxidation number of silicon (Si) in (CH3)2SiCl2 is -4.