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vichka [17]
1 year ago
10

Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? Arrange the substances with polar

covalent bonds in order of increasing bond polarity:
(d) SO₂
Chemistry
1 answer:
KengaRu [80]1 year ago
3 0

Covalent bonds can be classified as nonpolar and polar covalent given the electronegativity difference between two atoms (ΔEN).

Nonpolar covalent bond electrons are shared equally between two atoms, polar covalent bond electrons are shared unequally, atoms have partial charges, ionic bond electrons are completely transferred to one atom, full charges present. Therefore, the greater the electronegativity difference, the greater the bond polarity. Let's determine the types of bonds present in the compounds and arrange the ones with polar covalent in order of increasing ΔEN. Sulfur and oxygen are both nonmetals so the substance is covalent. Sulfur has EN = 2.5 and oxygen has EN = 3.5. Since there is an electronegativity difference, the S−O bonds in the substance can be classified as polar covalent bonds.

Learn more about polar covalent bond here:

brainly.com/question/25150590

#SPJ4

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5 0
3 years ago
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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
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Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

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First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

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