My question may seem too simple. But I couldn't find why this is not a redox.

$$\ce{Hg2(NO3)2 + 2KBr -> Hg2Br2 + 2KNO3}$$

In $\ce{Hg2(NO3)2}$ compound atoms have following states:

O: -2
N: -3
Hg: +9

In $\ce{KBr}$ compound atoms have following states:

K: +1
Br: -1

In $\ce{Hg2Br2}$ compound atoms have following states:

Hg: +1
Br: -1

In $\ce{KNO3}$ compound atoms have following states:

K: +1
N: +5
O: -2

Hg changed from +9 to +1, so reduced. N changed from -3 to +5, so oxidized.

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    $\begingroup$ No atom ever can be +9. $\endgroup$ Sep 23 '19 at 13:17
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    $\begingroup$ @IvanNeretin Tetroxoiridium(IX): *cough cough* $\endgroup$
    – andselisk
    Sep 23 '19 at 14:51
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    $\begingroup$ Well, at least no mercury atom can hit +9 in ordinary Y chemistry (I've read that maybe there us +4, on the outside edge). Nitrogen in nitrate ion is +5, not -3. $\endgroup$ Sep 23 '19 at 18:32

You got the oxidation numbers in the $\ce{NO3^-}$ anion wrong. Remember that the sum of the oxidation numbers in a species must be equal to the total charge. You claimed oxygen has a $-2$ oxidation number and nitrogen has $-3$, this would mean that you would not have an $\ce{NO3^-}$ ion but an $\ce{NO3^9-}$, which is clearly not the case. Try fixing the oxidation numbers here and you would hopefully be able to understand the problem.

  • $\begingroup$ So if a compound has more than two atoms, firstly I must split it into two. One or each of them is a polyatomic ion. Each of them has suitable ions. Then find the oxidation states of each atom in the polyatomic ion. My mistake was trying to find oxidation states of each atom directly. Am I right? $\endgroup$ Sep 23 '19 at 13:27
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    $\begingroup$ If talking about oxidation numbers then yes, you will need to find them for each atom. For ionic compounds, it is often the most helpful to do this for each ion in the system, because as I have said, the charge needs to add up for each ionic species too, and this can help in many cases. Also note that you had the same anion in the KNO3 compound and there you got it right. It is exactly the same anion there, no reason to have different oxidation numbers than before. $\endgroup$
    – user23638
    Sep 23 '19 at 13:31

You got oxidation numbers mixed up. As the comments pointed out, the +9 state is possible only in very few elements and mercury is not one of those.

To my understanding in general chemistry, this problem can easily solved using Solubility Rules:

  1. Salts containing Group I elements ($\ce{Li+, Na+, K+, Cs+, Rb+}$) and salts containing ammonium ion ($\ce{NH4+}$) are soluble in water. There are few exceptions to this rule.
  2. Salts containing nitrate ion ($\ce{NO3-}$) are generally soluble.
  3. Salts containing $\ce{Cl-, Br-, or I-}$ are generally soluble. Important exceptions to this rule are halide salts of $\ce{Ag+, Pb2+,}$ and $\ce{Hg2^2+}$. For example, $\ce{AgCl, PbBr2,}$ or $\ce{Hg2Cl2}$ are insoluble.
  4. Most silver salts are insoluble. $\ce{AgNO3}$ and $\ce{Ag(C2H3O2)}$ are common soluble salts of silver; virtually all others are insoluble.
  5. Most sulfate salts are soluble. Important exceptions to this rule include $\ce{CaSO4, BaSO4, PbSO4, Ag2SO4}$ and $\ce{SrSO4}$.
  6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements ($\ce{Ca, Sr,}$ and $\ce{Ba}$) are slightly soluble. Hydroxide salts of transition metals and $\ce{Al^3+}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not soluble.
  7. Most sulfides of transition metals are highly insoluble, including $\ce{CdS, FeS, ZnS,}$ and $\ce{Ag2S}$. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
  8. Carbonates are frequently insoluble. Group II carbonates ($\ce{CaCO3, SrCO3,}$ and $\ce{BaCO​3}$) are insoluble, as are $\ce{FeCO3}$ and $\ce{PbCO3}$.
  9. Chromates are frequently insoluble. Examples include $\ce{PbCrO4}$ and $\ce{BaCrO4}$.
  10. Phosphates such as $\ce{Ca3(PO4)2}$ and $\ce{Ag3PO4}$ are frequently insoluble.
  11. Fluorides such as $\ce{BaF2, MgF2,}$ and $\ce{PbF2}$ are frequently insoluble.

Thus, according to Rules 1, 2, and 3, $\ce{KNO3}$ (Rules 1 & 2), $\ce{Hg2(NO3)2}$ (Rules 2), and $\ce{KBr}$ (Rules 1 & 3) are soluble in water: Thus, they tends to dissociate in water and exist in their ionic forms such as $\ce{K+ (aq), Br- (aq), Hg2^2+ (aq)}$ and $\ce{NO3- (aq)}$ until $\ce{Hg2^2+ (aq)}$ reacts with $\ce{Br- (aq)}$ to give $\ce{Hg2Br2}$.

According to Rule 3, $\ce{Hg2Br2}$ is insoluble in water so that it stay in solid form.

Now, look at your reaction: $$\ce{Hg2(NO3)2 (aq) + 2 KBr(aq) -> Hg2Br2(s) + 2 KNO3(aq)}$$

Thus, in total ionic form:

$$\ce{Hg2^2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + 2 Br- (aq) -> Hg2Br2(s) + 2 K+ (aq) + 2 NO3- (aq)}$$

When remove the common ions, you got the net ionic equation:

$$\ce{Hg2^2+ (aq) + 2 Br- (aq) -> Hg2Br2(s) }$$

Can you say it is a redox reaction now? It is a simple precipitation.

Reference sited by the source:

Ralph H. Petrucci, F. Geoffrey Herring, Jeffrey D. Madura, and Carey Bissonnette, In General Chemistry: Principles and Modern Applications; 10th ed.; Pearson Education: Upper Saddle River, New Jersey, 2011.

  • 1
    $\begingroup$ @Oscar Lanzi: Nice edit. I appreciate your good edit to make this answer better. $\endgroup$ Sep 24 '19 at 14:55

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