Acidic solution in acidic solution, the iodate ion can be used to react with a number of metal ions. one such reaction is io3−(aq)+sn2+(aq)→i−(aq)+sn4+(aq) since this reaction takes place in acidic solution, h2o(l) and h+(aq) will be involved in the reaction. places for these species are indicated by the blanks in the following restatement of the equation: io3−(aq)+sn2+(aq)+ −−−→i−(aq)+sn4+(aq)+ −−−

Explanation:

IO₃⁻ + Sn²⁺ ⟶ I⁻ + Sn⁴⁺

1: Separate the equation into two half-reactions.

IO₃⁻ ⟶ I⁻

Sn²⁺ ⟶ Sn⁴⁺

2: Balance all atoms other than H and O.

Done

3: Balance O.

IO₃⁻ ⟶ I⁻ + 3H₂O

Sn²⁺ ⟶ Sn⁴⁺

4: Balance H

IO₃⁻ + 6H ⟶ I⁻ + 3H₂O  

Sn²⁺ ⟶ Sn⁴⁺

5: Balance charge.

IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O  

Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻

6: Equalize electrons transferred.

1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]

3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]

7: Add the two half-reactions.

1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]

3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]                                  

      IO₃⁻ + 3Sn²⁺ + 6H⁺ ⟶ I⁻ + 3Sn⁴⁺ + 3H₂O

8: Check mass balance.

 On the left: 1 I, 3 O, 3 Sn, 6 H

On the right: 1 I, 3 O, 3 Sn, 6 H

Step 9: Check charge balance.

  On the left: 1- + 12+ = 11+

On the right: 1-  + 12+ = 11+

The equation is balanced.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

the answer is option a.