Draw the Lewis structure for the sulfur dioxide molecule. Be sure to include all resonance structures that satisfy the octet rul
e.
1 answer:
Answer:
See explanation
Explanation:
We have to remember that the formula of <u>sulfur dioxide</u> is . So, the "S" atom is the central one and the "O" atoms are in the left and right of S. Additionally, "S" and "O" atom has <u>6 valence electrons each</u>. With this in mind, we can start with a structure in which the left oxygen has 2 lone pairs and a double bond, the S atom can have a double bond a single bond and a lone pair and the right oxygen with 3 lone pairs with a single bond <u>(structure A)</u>. Now, if we move the negative charge in the right (resonance) to the middle and the double bond to the left oxygen we will obtain <u>structure B</u>. Finally, we can move the negative charge in the left to the middle and we will obtain structure C. (See figure 1).
You might be interested in
Here we have to calculate the amount of ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.