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 positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be