When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be . I–(aq) + NO3–(aq) → NO(g) + I2(s) (acidic solution)

When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be __6__.

Explanation:

From the redox equation, we can see that NO₃⁻ is reduced to NO (from oxidation state +5 to +2), whereas I⁻ is oxidized to I₂ (from oxidation state -1 to 0). The half reactions are balanced with H⁺ (acidic solution), as follows:

Reduction :        2 x (NO₃⁻(aq) + 3 e-  + 4 H⁺ → NO(g) + 2 H₂O)

Oxidation :                                   3 x (2 I⁻(aq) → I₂(s) + 2 e-)

Total equation: 6 I⁻(aq) + 2 NO₃⁻(aq)+ 8 H⁺ → 3 I₂(s) + 2 NO(g) + 4 H₂O

That is the redox equation with the smallest whole number coefficients.

Accordin to this, the coefficient for the iodide ion (I⁻) is: 6.

"the periodic table is arranged in order of atomic number."