the oxidation number of Mo in Na2MoO4 is what?
Na2MoO4= 0Γ2+x+4(-2)=0
=0+x+(-6)=0
= x-6=0
πΌπ£π¨π¬ππ§ =x=6
ππππ£ π¬π πππ£π π€πππππ©ππ€π£ π¨π©ππ©π π€π ππ€
Well, the oxidation number of Mo in Na2MoO4 is a real party pooper. It's +6! Trust me, Mo is definitely feeling a little oxidized and ready to hang out with some oxygen atoms in this compound. It's like Mo decided to switch up its outfit for a fancy +6 oxidation number soirΓ©e. Don't worry though, Mo knows how to work the room and make some chemistry happen!
To determine the oxidation number of an element in a compound, we need to follow a few steps:
Step 1: Identify the known oxidation numbers of other elements in the compound.
In Na2MoO4, we can start by looking at the sodium ion (Na+). Sodium is an alkali metal in Group 1 of the periodic table, which means its oxidation number is always +1.
Step 2: Determine the overall charge of the compound.
In Na2MoO4, we have two sodium ions with a charge of +1 each, for a total charge of +2. The overall charge of the molybdenum compound must balance this charge.
Step 3: Apply the principle of charge conservation.
In a neutral compound, the sum of all oxidation numbers must equal zero. Since Na2MoO4 is not neutral (it has a positive charge), the sum of the oxidation numbers must equal the overall charge of the compound.
Step 4: Calculate the oxidation number of molybdenum (Mo).
Let's assume the oxidation number of molybdenum (Mo) is x.
The overall charge of the compound is +2 (due to the two sodium ions).
The oxidation number of oxygen (O) is typically -2, and there are four oxygen atoms in Na2MoO4. Therefore, the total negative charge from oxygen is -8 (-2 x 4).
The sum of the oxidation numbers must equal the overall charge, so we can set up the equation: 2(x) + (-8) = +2.
Simplifying the equation, we have: 2x - 8 = 2.
To solve for x, we add 8 to both sides of the equation: 2x = 10.
Finally, we divide both sides of the equation by 2: x = 5.
Therefore, the oxidation number of molybdenum (Mo) in Na2MoO4 is +5.
Compounds must add up to zero.
Na is +1; 2*1 = +2
O is -2; 4*-2 = -8
So Mo must be ...... for Na2MoO4 to be zero.