Answer: N (the Nitrogen)
Explanation:
Reduction refers to a decrease in oxidation number/state due to the gaining of electrons. As such the species that is being reduced will show a decrease in oxidation state.
Based on the redox rules,
Zn(s) has oxidation number of 0 [<em>rule 1: the oxidation number of an element in its free (uncombined) state is zero</em>]
Zn²⁺ has oxidation number of +2 [<em>rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion</em>]
Now, since Nitrogen is enbedded in a polyatomic ion in both cases, you have to do a bit a calculation to obtain the oxidation state.
For NO₃⁻ : N + (-2 × 3) = -1
N - 6 = -1
N = 5
<em>[Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 6: The oxidation state of hydrogen in a compound is usually +1]</em>
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For NH₄⁺ :
N + (4 x 1) = 1
N + 4 = 1
N = -3
[<em>Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 5: The oxidation number of oxygen in a compound is usually –2]</em>
Therefore, Zn moves from oxidation state of 0 to +2 (oxidation), while N moves from +5 to -3 (reduction).
In an ionic compound the atoms are linked via ionic bonds. These are formed by the transfer of electrons from one atom to the other. The atom that loses electrons gains a positive charge whereas the atom that accepts electrons gains a negative. This happens in accordance with the octet rule wherein each atom is surrounded by 8 electrons
In the given example:
The valence electron configuration of Iodine (I) = 5s²5p⁵
It needs only one electron to complete its octet.
In the given options:
K = 4s¹
C = 2s²2p²
Cl = 3s²3p⁵
P = 3s²3p³
Thus K can donate its valence electron to Iodine. As a result K, will gain a stable noble gas configuration of argon while iodine would gain an octet. This would also balance the charges as K⁺I⁻ creating a neutral molecule.
Ans: Potassium (K)
Answer:
The answer is Sodium chloride.
Na is sodium and Cl is chlorine.