Method of electronic balance in an accessible presentation. Nitrogen compounds The oxidation state of hno3 is

Compounds with an oxidation state of –3. Nitrogen compounds in the -3 oxidation state are represented by ammonia and metal nitrides.

Ammonia- NH 3 is a colorless gas with a characteristic pungent odor. The ammonia molecule has the geometry of a trigonal pyramid with a nitrogen atom at the top. The atomic orbitals of nitrogen are in sp 3- hybrid state. Three orbitals are involved in the formation of nitrogen-hydrogen bonds, and the fourth orbital contains an unshared electron pair, the molecule has a pyramidal shape. The repulsive action of the lone pair of electrons causes the bond angle to decrease from the expected 109.5 to 107.3°.

At -33.4 °C, ammonia condenses to form a liquid with a very high heat of vaporization, which allows it to be used as a refrigerant in industrial refrigeration systems.

The presence of an unshared electron pair at the nitrogen atom allows it to form another covalent bond according to the donor-acceptor mechanism. Thus, in an acidic environment, the formation of the molecular ammonium cation - NH 4 + occurs. The formation of a fourth covalent bond leads to alignment of bond angles (109.5°) due to the uniform repulsion of hydrogen atoms.

Liquid ammonia is a good self-ionizing solvent:

2NH 3 NH 4 + + NH 2 -

amide anion

It dissolves alkali and alkaline earth metals, forming colored conductive solutions. In the presence of a catalyst (FeCl 3), the dissolved metal reacts with ammonia to release hydrogen and form an amide, for example:

2Na + 2NH 3 \u003d 2NaNH 2 + H 2

sodium amide

Ammonia is very soluble in water (at 20 °C, about 700 volumes of ammonia dissolve in one volume of water). In aqueous solutions, it exhibits the properties of a weak base.

NH 3 + H 2 O ® NH 3 × H 2 O NH 4 + + OH -

= 1.85 10 -5

In an oxygen atmosphere, ammonia burns with the formation of nitrogen; on a platinum catalyst, ammonia is oxidized to nitric oxide (II):

4NH 3 + 3O 2 = 2N 2 + 6H 2 O; 4NH 3 + 5O 2 \u003d 4NO + 6H 2 O

As a base, ammonia reacts with acids to form salts of the ammonium cation, for example:

NH 3 + HCl = NH 4 Cl

Ammonium salts are highly soluble in water and slightly hydrolyzed. In the crystalline state, they are thermally unstable. The composition of thermolysis products depends on the properties of the acid forming the salt:

NH 4 Cl ® NH 3 + HCl; (NH 4) 2 SO 4 ® NH 3 + (NH 4) HSO 4

(NH 4) 2 Cr 2 O 7 ® N 2 + Cr 2 O 3 + 4H 2 O

Under the action of alkalis on aqueous solutions of ammonium salts, ammonia is released during heating, which makes it possible to use this reaction as a qualitative one for ammonium salts and as a laboratory method for obtaining ammonia.

NH 4 Cl + NaOH \u003d NaCl + NH 3 + H 2 O

In industry, ammonia is obtained by direct synthesis.

N 2 + 3H 2 2NH 3

Since the reaction is highly reversible, the synthesis is carried out at elevated pressure (up to 100 MPa). To speed up the process, the process is carried out in the presence of a catalyst (spongy iron promoted by additives) and at a temperature of about 500°C.

Nitride are formed as a result of the reactions of many metals and non-metals with nitrogen. The properties of nitrides naturally change in a period. For example, for elements of the third period:

Nitrides of s-elements of groups I and II are crystalline salt-like substances that are easily decomposed by water to form ammonia.

Li 3 N + 3H 2 O \u003d 3LiOH + NH 3

Of the halogen nitrides in the free state, only Cl 3 N was isolated, the acid character manifests itself in the reaction with water:

Cl 3 N + 3H 2 O \u003d 3HClO + NH 3

The interaction of nitrides of different nature leads to the formation of mixed nitrides:

Li 3 N + AlN \u003d Li 3 AlN 2; 5Li 3 N + Ge 3 N 4 = 3Li 5 GeN 3

lithium nitridoaluminate nitridogermanate(IV) lithium

BN, AlN, Si 3 N 4, Ge 3 N 4 nitrides are solid polymeric substances with high melting points (2000-3000 ° C), they are semiconductors or dielectrics. Nitrides of d-metals - crystalline compounds of variable composition (bertolides), very hard, refractory and chemically stable, exhibit metallic properties: metallic luster, electrical conductivity.

Compounds with an oxidation state of –2. Hydrazine - N 2 H 4 - the most important inorganic nitrogen compound in the -2 oxidation state.

Hydrazine is a colorless liquid with a boiling point of 113.5 °C, fuming in air. Hydrazine vapors are extremely toxic and form explosive mixtures with air. Hydrazine is obtained by oxidizing ammonia with sodium hypochlorite:

2N -3 H 3 + NaCl +1 O \u003d N 2 -2 H 4 + NaCl -1 + H 2 O

Hydrazine is miscible with water in any ratio and behaves in solution as a weak diacid base, forming two series of salts.

N 2 H 4 + H 2 O N 2 H 5 + + OH - , K b = 9.3×10 -7 ;

hydrosonium cation

N 2 H 5 + + H 2 O N 2 H 6 2+ + OH - , K b = 8.5×10 -15;

dihydrosonium cation

N 2 H 4 + HCl N 2 H 5 Cl; N 2 H 5 Cl + HCl N 2 H 6 Cl 2

hydrosonium chloride dihydrosonium dichloride

Hydrazine is the strongest reducing agent:

4KMn +7 O 4 + 5N 2 -2 H 4 + 6H 2 SO 4 \u003d 5N 2 0 + 4Mn +2 SO 4 + 2K 2 SO 4 + 16H 2 O

Unsymmetrical dimethylhydrazine (heptyl) is widely used as a rocket fuel.

Compounds with an oxidation state of –1. Hydroxylamine - NH 2 OH - the main inorganic nitrogen compound in the oxidation state -1.

Hydroxylamine is obtained by reducing nitric acid with hydrogen at the time of isolation during electrolysis:

HNO 3 + 6H \u003d NH 2 OH + 2H 2 O

This is a colorless crystalline substance (mp. 33 ° C), highly soluble in water, in which it exhibits the properties of a weak base. With acids it gives hydroxylammonium salts - stable colorless substances soluble in water.

NH 2 OH + H 2 O + + OH - , K b = 2×10 -8

hydroxylammonium ion

The nitrogen atom in the NH 2 OH molecule exhibits an intermediate oxidation state (between -3 and +5), so hydroxylamine can act both as a reducing agent and as an oxidizing agent:

2N -1 H 2 OH + I 2 + 2KOH = N 0 2 + 2KI + 4H 2 O;

reducing agent

2N -1 H 2 OH + 4FeSO 4 + 3H 2 SO 4 = 2Fe 2 (SO 4) 3 + (N -3 H 4) 2 SO 4 + 2H 2 O

oxidizing agent

NH 2 OH easily decomposes when heated, undergoing disproportionation:

3N -1 H 2 OH \u003d N 0 2 + N -3 H 3 + 3H 2 O;

Compounds with an oxidation state of +1. Nitric oxide (I) - N 2 O (nitrous oxide, laughing gas). The structure of its molecule can be conveyed by the resonance of two valence schemes, which show that this compound can be considered as nitric oxide (I) only formally, in reality it is nitrogen (V) oxynitride - ON +5 N -3.

N 2 O is a colorless gas with a slight pleasant smell. In small concentrations it causes bouts of unbridled joy, in large doses it has a general anesthetic effect. A mixture of nitrous oxide (80%) and oxygen (20%) was used in medicine for anesthesia.

Under laboratory conditions, nitric oxide (I) can be obtained by decomposition of ammonium nitrate. N 2 O obtained by this method contains impurities of higher nitrogen oxides, which are extremely toxic!

NH 4 NO 3 ¾® N 2 O + 2H 2 O

According to its chemical properties, nitric oxide (I) is a typical non-salt-forming oxide; it does not react with water, acids and alkalis. When heated, it decomposes to form oxygen and nitrogen. For this reason, N 2 O can act as an oxidizing agent, for example:

N 2 O + H 2 \u003d N 2 + H 2 O

Compounds with an oxidation state of +2. Nitric oxide (II) - NO - colorless gas, extremely toxic. In air, it is rapidly oxidized by oxygen to form no less toxic nitric oxide (IV). In industry, NO is produced by the oxidation of ammonia on a platinum catalyst or by passing air through an electric arc (3000-4000 °C).

4NH 3 + 5O 2 \u003d 4NO + 6H 2 O; N 2 + O 2 \u003d 2NO

A laboratory method for obtaining nitric oxide (II) is the interaction of copper with dilute nitric acid.

3Cu + 8HNO 3 (diff.) = 3Cu(NO 3) 2 + 2NO + 4H 2 O

Nitric oxide (II) is a non-salt-forming oxide, a strong reducing agent, easily reacts with oxygen and halogens.

2NO + O 2 \u003d 2NO 2; 2NO + Cl 2 = 2NOCl

nitrosyl chloride

At the same time, when interacting with strong reducing agents, NO acts as an oxidizing agent:

2NO + 2H 2 = N 2 + 2H 2 O; 10NO + 4Р = 5N 2 + 2Р 2 O 5

Compounds with an oxidation state of +3. Nitric oxide (III) - N 2 O 3 - an intensely blue liquid (t.cr. -100 ° C). Stable only in liquid and solid state at low temperatures. It appears to exist in two forms:

Nitric oxide(III) is obtained by co-condensation of NO and NO 2 vapors. Dissociates in liquids and vapors.

NO 2 + NO N 2 O 3

The properties are typical acidic oxide. It reacts with water, forming nitrous acid, with alkalis forms salts - nitrites.

N 2 O 3 + H 2 O \u003d 2HNO 2; N 2 O 3 + 2NaOH \u003d 2NaNO 2 + H 2 O

Nitrous acid- medium strength acid (K a = 1×10 -4). It has not been isolated in its pure form, in solutions it exists in two tautomeric forms (tautomers are isomers that are in dynamic equilibrium).

nitrite form nitro form

Salts of nitrous acid are stable. The nitrite anion exhibits a pronounced redox duality. Depending on the conditions, it can perform both the function of an oxidizing agent and the function of a reducing agent, for example:

2NaNO 2 + 2KI + 2H 2 SO 4 = I 2 + 2NO + K 2 SO 4 + Na 2 SO 4 + 2H 2 O

oxidizing agent

KMnO 4 + 5NaNO 2 + 3H 2 SO 4 = 2MnSO 4 + 5NaNO 3 + K 2 SO 4 + 3H 2 O

reducing agent

Nitrous acid and nitrites are prone to disproportionation:

3HN +3 O 2 \u003d HN +5 O 3 + 2N +2 O + H 2 O

Compounds with an oxidation state of +4. Nitric oxide (IV) - NO 2 - brown gas, with a sharp unpleasant odor. Extremely toxic! In industry, NO 2 is produced by the oxidation of NO. The laboratory method for obtaining NO 2 is the interaction of copper with concentrated nitric acid, as well as the thermal decomposition of lead nitrate.

Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O;

2Pb(NO 3) 2 \u003d 2PbO + 4NO 2 + O 2

The NO 2 molecule has one unpaired electron and is a stable free radical, so nitric oxide easily dimerizes.

The dimerization process is reversible and very sensitive to temperature:

paramagnetic, diamagnetic,

brown colorless

Nitrogen dioxide is an acidic oxide that reacts with water to form a mixture of nitric and nitrous acid (mixed anhydride).

2NO 2 + H 2 O \u003d HNO 2 + HNO 3; 2NO 2 + 2NaOH \u003d NaNO 3 + NaNO 2 + H 2 O

Compounds with an oxidation state of +5. Nitric oxide (V) - N 2 O 5 - white crystalline substance. Obtained by dehydration of nitric acid or oxidation of nitric oxide (IV) with ozone:

2HNO 3 + P 2 O 5 \u003d N 2 O 5 + 2HPO 3; 2NO 2 + O 3 \u003d N 2 O 5 + O 2

In the crystalline state, N 2 O 5 has a salt-like structure - + -, in vapor (t. vozg. 33 ° C) - molecular.

N 2 O 5 - acid oxide - nitric acid anhydride:

N 2 O 5 + H 2 O \u003d 2HNO 3

Nitric acid- HNO 3 - a colorless liquid with a boiling point of 84.1 ° C, decomposes when heated and in the light.

4HNO 3 \u003d 4NO 2 + O 2 + 2H 2 O

Nitrogen dioxide impurities give concentrated nitric acid a yellow-brown color. Nitric acid is miscible with water in any ratio and is one of the strongest mineral acids; it completely dissociates in solution.

The structure of the nitric acid molecule is described by the following structural formulas:

Difficulties with writing the structural formula of HNO 3 are caused by the fact that, showing in this compound the oxidation state +5, nitrogen, as an element of the second period, can form only four covalent bonds.

Nitric acid is one of the strongest oxidizing agents. The depth of its recovery depends on many factors: concentration, temperature, reducing agent. Usually, when oxidized with nitric acid, a mixture of reduction products is formed:

HN +5 O 3 ® N +4 O 2 ® N +2 O ® N +1 2 O ® N 0 2 ® +

The predominant product of the oxidation of non-metals and inactive metals with concentrated nitric acid is nitric oxide (IV):

I 2 + 10HNO 3 (conc) = 2HIO 3 + 10NO 2 + 4H 2 O;

Pb + 4HNO 3 (conc) = Pb (NO 3) 2 + 2NO 2 + 2H 2 O

Concentrated nitric acid passivates iron and aluminum. Aluminum is passivated even with dilute nitric acid. Nitric acid of any concentration does not affect gold, platinum, tantalum, rhodium and iridium. Gold and platinum are dissolved in aqua regia - a mixture of concentrated nitric and hydrochloric acids in a ratio of 1: 3.

Au + HNO 3 + 4HCl \u003d H + NO + 2H 2 O

The strong oxidizing effect of aqua regia is due to the formation of atomic chlorine during the decomposition of nitrosyl chloride, a product of the interaction of nitric acid with hydrogen chloride.

HNO 3 + 3HCl \u003d Cl 2 + NOCl + 2H 2 O;

NOCl = NO + Cl×

An effective solvent for low-active metals is a mixture of concentrated nitric and hydrofluoric acids.

3Ta + 5HNO 3 + 21HF = 3H 2 + 5NO + 10H 2 O

Diluted nitric acid, when interacting with non-metals and low-active metals, is reduced mainly to nitric oxide (II), for example:

3P + 5HNO 3 (razb) + 2H 2 O \u003d 3H 3 PO 4 + 5NO;

3Pb + 8HNO 3 (razb) \u003d 3Pb (NO 3) 2 + 2NO + 4H 2 O

Active metals reduce dilute nitric acid to N 2 O, N 2 or NH 4 NO 3, for example,

4Zn + 10HNO 3 (razb) \u003d 4Zn (NO 3) 2 + NH 4 NO 3 + 3H 2 O

The bulk of nitric acid goes to the production of fertilizers and explosives.

Nitric acid is produced industrially by the contact or arc method, which differ in the first stage - the production of nitric oxide (II). The arc method is based on the production of NO by passing air through an electric arc. In the contact process, NO is produced by the oxidation of ammonia with oxygen over a platinum catalyst. Next, nitric oxide (II) is oxidized to nitric oxide (IV) by atmospheric oxygen. By dissolving NO 2 in water in the presence of oxygen, nitric acid is obtained with a concentration of 60-65%.

4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3

If necessary, nitric acid is concentrated by distillation with concentrated sulfuric acid. In the laboratory, 100% nitric acid can be obtained by the action of concentrated sulfuric acid on crystalline sodium nitrate when heated.

NaNO 3 (cr) + H 2 SO 4 (conc) = HNO 3 + NaHSO 4

Salts of nitric acid- nitrates - highly soluble in water, thermally unstable. The decomposition of nitrates of active metals (excluding lithium), which are in the series of standard electrode potentials to the left of magnesium, leads to the formation of nitrites. For example:

2KNO 3 \u003d 2KNO 2 + O 2

During the decomposition of lithium and magnesium nitrates, as well as metal nitrates located in the series of standard electrode potentials to the right of magnesium, up to copper, a mixture of nitric oxide (IV) and oxygen is released. For example:

2Cu(NO 3) 2 \u003d 2CuO + 4NO 2 + O 2

Nitrates of metals located at the end of the activity series decompose to free metal:

2AgNO 3 \u003d 2Ag + 2NO 2 + O 2

Sodium, potassium and ammonium nitrates are widely used for the production of gunpowder and explosives, as well as nitrogen fertilizers (saltpeter). Ammonium sulfate, ammonia water and carbamide (urea) - full carbonic acid amide are also used as fertilizers:

Hydrogen azide(dinitridonitrate) - HN 3 (HNN 2) - a colorless volatile liquid (mp -80 ° C, bp 37 ° C) with a pungent odor. The central nitrogen atom is in sp hybridization, the oxidation state is +5, the atoms adjacent to it have an oxidation state of –3. Molecule structure:

An aqueous solution of HN 3 - hydronitrous acid is close in strength to acetic acid, K a \u003d 2.6 × 10 -5. Stable in dilute solutions. It is obtained by the interaction of hydrazine and nitrous acid:

N 2 H 4 + HNO 2 \u003d HN 3 + 2H 2 O

In terms of oxidizing properties, HN 3 (HN +5 N 2) resembles nitric acid. So, if the interaction of a metal with nitric acid produces nitric oxide (II) and water, then with hydrazoic acid - nitrogen and ammonia. For example,

Cu + 3HN +5 N 2 = Cu(N 3) 2 + N 2 0 + NH 3

A mixture of HN 3 and HCl behaves like aqua regia. Salts of hydronitrous acid - azides. Only alkali metal azides are relatively stable; at temperatures > 300 °C they are destroyed without explosion. The rest disintegrate with an explosion on impact or heating. Lead azide is used in the production of detonators:

Pb(N 3) 2 = Pb + 3N 2 0

The starting product for the production of azides is NaN 3, which is formed as a result of the reaction of sodium amide and nitric oxide (I):

NaNH 2 + N 2 O \u003d NaN 3 + H 2 O

4.2 Phosphorus

Phosphorus is represented in nature by one isotope - 31 P, the clarke of phosphorus is 0.05 mol.%. It occurs in the form of phosphate minerals: Ca 3 (PO 4) 2 - phosphorite, Ca 5 (PO 4) 3 X (X \u003d F, Cl, OH) - apatites. It is part of the bones and teeth of animals and humans, as well as the composition of nucleic acids (DNA and RNA) and adenosine phosphoric acids (ATP, ADP and AMP).

Phosphorus is obtained by reduction of phosphorite with coke in the presence of silicon dioxide.

Ca 3 (PO 4) 2 + 3SiO 2 + 5C = 3CaSiO 3 + 2P + 5CO

A simple substance - phosphorus - forms several allotropic modifications, of which the main ones are white, red and black phosphorus. White phosphorus is formed during the condensation of phosphorus vapor and is a white wax-like substance (mp 44 ° C), insoluble in water, soluble in some organic solvents. White phosphorus has a molecular structure and consists of tetrahedral molecules P 4 .

The bond strength (valence angle P-P-P is only 60 °) determines the high reactivity and toxicity of white phosphorus (lethal dose is about 0.1 g). Since white phosphorus is highly soluble in fats, milk cannot be used as an antidote for poisoning. In air, white phosphorus spontaneously ignites, so it is stored in a hermetically sealed chemical container under a layer of water.

Red phosphorus has a polymer structure. It is obtained by heating white phosphorus or irradiating it with light. Unlike white phosphorus, it is slightly reactive and non-toxic. However, residual amounts of white phosphorus can make red phosphorus toxic!

Black phosphorus is obtained by heating white phosphorus under a pressure of 120 thousand atm. It has a polymer structure, has semiconductor properties, is chemically stable and non-toxic.

Chemical properties. White phosphorus is spontaneously oxidized by atmospheric oxygen at room temperature (oxidation of red and black phosphorus occurs when heated). The reaction proceeds in two stages and is accompanied by luminescence (chemiluminescence).

2P + 3O 2 \u003d 2P 2 O 3; P 2 O 3 + O 2 \u003d P 2 O 5

Phosphorus also reacts stepwise with sulfur and halogens.

2P + 3Cl 2 \u003d 2PCl 3; PCl 3 + Cl 2 = PCl 5

When interacting with active metals, phosphorus acts as an oxidizing agent, forming phosphides - phosphorus compounds in the -3 oxidation state.

3Ca + 2P = Ca 3 P 2

Oxidizing acids (nitric and concentrated sulfuric acids) oxidize phosphorus to phosphoric acid.

P + 5HNO 3 (conc) = H 3 PO 4 + 5NO 2 + H 2 O

When boiling with alkali solutions, white phosphorus disproportionates:

4P 0 + 3KOH + 3H 2 O = P -3 H 3 + 3KH 2 P +1 O 2

phosphine potassium hypophosphite

Under normal conditions, nitric acid is a colorless liquid (density 1.52 g/cm 3 ), boiling at 82.6 o C, and at a temperature (-41.6 o C) solidifying into a transparent crystalline mass. Gross formula - HNO 3 . Molar mass - 93 g/mol. The structure of the nitric acid molecule is shown in fig. one.

Nitric acid is miscible with water in any ratio. It is a strong electrolyte, i.e. in aqueous solution almost completely dissociates into ions. In OVR, it acts as an oxidizing agent.

Rice. 1. The structure of the nitric acid molecule, indicating the bond angles between bonds and the lengths of chemical bonds.

HNO3, oxidation states of elements in it

To determine the oxidation states of the elements that make up nitric acid, you first need to figure out for which elements this value is exactly known.

The oxidation states of hydrogen and oxygen in the composition of inorganic acids are always equal to (+1) and (-2), respectively. To find the oxidation state of nitrogen, let's take its value as "x" and determine it using the electroneutrality equation:

(+1) + x + 3×(-2) = 0;

1 + x - 6 = 0;

So the degree of oxidation of nitrogen in nitric acid is (+5):

H +1 N +5 O -2 3 .

Examples of problem solving

EXAMPLE 1

Chemicals can be divided into typical oxidizers, typical reducing agents, and substances that may exhibit both oxidizing and reducing properties. Some substances practically do not show redox activity.

To typical oxidizers include:

• simple substances - non-metals with the strongest oxidizing properties (fluorine F 2, oxygen O 2, chlorine Cl 2);
• ionsmetals or non-metals with high positive (usually higher) oxidation states : acids (HN +5 O 3, HCl +7 O 4), salts (KN +5 O 3, KMn +7 O 4), oxides (S +6 O 3, Cr +6 O 3)
• compounds containing some metal cations having high oxidation states: Pb 4+ , ​​Fe 3+ , Au 3+ etc.

Typical reducing agents is usually:

• simple substances - metals(reducing abilities of metals are determined by a series of electrochemical activity);
• complex substances that contain atoms or ions of non-metals with a negative (usually lower) oxidation state: binary hydrogen compounds (H 2 S, HBr), salts of oxygen-free acids (K 2 S, NaI);
• some compounds containing cations with the lowest positive oxidation state(Sn 2+, Fe 2+, Cr 2+), which, donating electrons, can increase their oxidation state;
• compounds containing complex ions, consisting of non-metals with an intermediate positive oxidation state(S +4 O 3) 2–, (НР +3 O 3) 2– , in which elements can, by donating electrons, increase its positive oxidation state.

Most other substances can show both oxidizing and reducing properties.

Typical oxidizing and reducing agents are shown in the table.

In laboratory practice the most commonly used are the following oxidizers :

potassium permanganate (KMnO 4);

potassium dichromate (K 2 Cr 2 O 7);

nitric acid (HNO 3);

concentrated sulfuric acid (H 2 SO 4);

hydrogen peroxide (H 2 O 2);

oxides of manganese (IV) and lead (IV) (MnO 2 , PbO 2);

molten potassium nitrate (KNO 3) and melts of some other nitrates.

To reducers that apply in laboratory practice relate:

• magnesium (Mg), aluminum (Al), zinc (Zn) and other active metals;
• hydrogen (H 2) and carbon (C);
• potassium iodide (KI);
• sodium sulfide (Na 2 S) and hydrogen sulfide (H 2 S);
• sodium sulfite (Na 2 SO 3);
• tin chloride (SnCl 2).

Classification of redox reactions

Redox reactions are usually divided into four types: intermolecular, intramolecular, disproportionation reactions (self-oxidation-self-reduction), and counter-disproportionation reactions.

Intermolecular reactions proceed with a change in the degree of oxidation different elements from different reagents. At the same time, they form various products of oxidation and reduction .

2Al0 + Fe +3 2 O 3 → Al +3 2 O 3 + 2Fe 0,

C 0 + 4HN +5 O 3 (conc) = C +4 O 2 + 4N +4 O 2 + 2H 2 O.

Intramolecular reactions are reactions in which different elements from one reagent move into different products such as:

(N -3 H 4) 2 Cr+6 2 O 7 → N 2 0 + Cr +3 2 O 3 + 4 H 2 O,

2 NaN +5 O -2 3 → 2 NaN +3 O 2 + O 0 2 .

Disproportionation reactions (self-oxidation-self-healing) - these are reactions in which the oxidizing agent and reducing agent - the same element of the same reagent, which goes into different products:

3Br 2 + 6 KOH → 5KBr + KBrO 3 + 3 H 2 O,

Reproportionation (proportionation, counterdisproportionation ) are reactions in which an oxidizing agent and a reducing agent are the same element, Which one of different reagents goes into one product. Reaction inverse to disproportionation.

2H 2 S -2 + S + 4 O 2 \u003d 3S + 2H 2 O

Basic rules for compiling redox reactions

Redox reactions are accompanied by oxidation and reduction processes:

Oxidation is the process of donating electrons by a reducing agent.

Recovery is the process of adding electrons to an oxidizing agent.

Oxidizing agent recovering, and the reducing agent oxidized .

In redox reactions, the electronic balance: The number of electrons that the reducing agent donates is equal to the number of electrons that the oxidizing agent receives. If the balance is drawn up incorrectly, you will not be able to draw up complex OVRs.

Several methods for compiling redox reactions (ORRs) are used: the electron balance method, the electron-ion balance method (half-reaction method), and others.

Consider in detail electronic balance method .

It is quite easy to “recognize” the OVR - it is enough to arrange the oxidation states in all compounds and determine that the atoms change the oxidation state:

K + 2 S -2 + 2K + Mn +7 O -2 4 = 2K + 2 Mn +6 O -2 4 + S 0

We write out separately the atoms of the elements that change the oxidation state, in the state BEFORE the reaction and AFTER the reaction.

The oxidation state is changed by manganese and sulfur atoms:

S -2 -2e = S 0

Mn +7 + 1e = Mn +6

Manganese absorbs 1 electron, sulfur donates 2 electrons. At the same time, it is necessary to comply electronic balance. Therefore, it is necessary to double the number of manganese atoms, and leave the number of sulfur atoms unchanged. We indicate the balance coefficients both before the reagents and before the products!

Scheme for compiling OVR equations using the electronic balance method:

Attention! There can be several oxidizing or reducing agents in a reaction. The balance must be drawn up so that the TOTAL number of given and received electrons is the same.

General patterns of redox reactions

The products of redox reactions often depend on process conditions. Consider the main factors affecting the course of redox reactions.

The most obvious determining factor is reaction solution medium - . As a rule (but not necessarily), the substance defining the medium is listed among the reagents. The following options are possible:

• oxidative activity intensifies in a more acidic environment and the oxidant is reduced more deeply(for example, potassium permanganate, KMnO 4, where Mn +7 is reduced to Mn +2 in an acidic environment, and to Mn +6 in an alkaline environment);
• oxidative activity intensifies in a more alkaline environment, and the oxidizing agent is reduced more deeply (for example, potassium nitrate KNO 3, where N +5, when interacting with a reducing agent in an alkaline medium, is reduced to N -3);
• or the oxidizing agent is practically not subject to changes in the environment.

The reaction medium makes it possible to determine the composition and form of existence of the remaining OVR products. The basic principle is that products are formed that do not interact with reagents!

Note! E If the solution medium is acidic, then bases and basic oxides cannot be present among the reaction products, because they interact with acid. Conversely, in an alkaline medium, the formation of acid and acid oxide is excluded. This is one of the most common, and most gross mistakes.

Also, the direction of OVR flow is affected by the nature of the reactants. for example, during the interaction of nitric acid HNO 3 with reducing agents, a pattern is observed - the greater the activity of the reducing agent, the more nitrogen N + 5 is reduced.

With an increase temperature most OVRs tend to be more intense and deeper.

In heterogeneous reactions, the composition of products is often influenced by fineness of the solid . For example, powdered zinc with nitric acid forms one product, while granular zinc forms completely different products. The greater the degree of grinding of the reagent, the greater its activity, usually.

Consider the most typical laboratory oxidizers.

Basic schemes of redox reactions

Scheme for the recovery of permanganates

Permanganates contain a powerful oxidizing agent - manganese in oxidation state +7. Salts of manganese +7 color the solution in Violet Colour.

Permanganates, depending on the medium of the reaction solution, are reduced in different ways.

AT acidic environment recovery is deeper Mn2+. Manganese oxide in the +2 oxidation state exhibits basic properties, therefore, in acidic environment salt is formed. Salts of manganese +2 colorless. AT neutral solution manganese is recovered to the degree of oxidation +4 , with education amphoteric oxide MnO 2 — brown sediment insoluble in acids and alkalis. AT alkaline environment, manganese is reduced minimally - to the nearest oxidation states +6 . Manganese compounds +6 exhibit acidic properties, in an alkaline medium they form salts - manganates. Manganates give the solution green coloring .

Consider the interaction of potassium permanganate KMnO 4 with potassium sulfide in acidic, neutral and alkaline media. In these reactions, the oxidation product of the sulfide ion is S 0 .

5 K 2 S + 2 KMnO 4 + 8 H 2 SO 4 \u003d 5 S + 2 MnSO 4 + 6 K 2 SO 4 + 8 H 2 O,

3 K 2 S + 2 KMnO 4 + 4 H 2 O = 2 MnO 2 ↓ + 3 S↓ + 8 KOH,

A common mistake in this reaction is an indication of the interaction of sulfur and alkali in the reaction products. However, sulfur interacts with alkali under rather harsh conditions (elevated temperature), which does not correspond to the conditions for this reaction. Under normal conditions, it will be correct to indicate exactly molecular sulfur and alkali separately, and not the products of their interaction.

K 2 S + 2 KMnO 4 - (KOH) \u003d 2 K 2 MnO 4 + S ↓

Difficulties also arise when compiling this reaction. The fact is that in this case, writing the molecule of the medium (KOH or other alkali) in the reagents is not required to equalize the reaction. Alkali takes part in the reaction and determines the product of the reduction of potassium permanganate, but the reactants and products are equalized even without its participation. This seemingly paradox can be easily resolved if we remember that a chemical reaction is just a conditional notation that does not indicate every ongoing process, but is just a reflection of the sum of all processes. How to determine it yourself? If you act according to the classical scheme - balance-balance coefficients - metal equalization, then you will see that metals are equalized by balance coefficients, and the presence of alkali on the left side of the reaction equation will be superfluous.

Permanganates oxidize:

• nonmetals with a negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5 ;
• nonmetals with an intermediate oxidation state to the highest degree of oxidation;
• active metals stable positive the degree of oxidation of the metal.

KMnO 4 + NeMe (lowest d.d.) = NeMe 0 + other products

KMnO 4 + NeMe (intermediate s.o.) = NeMe (higher s.o.) + other products

KMnO 4 + Me 0 = Me (stable s.d.) + other products

KMnO 4 + P -3, As -3 = P +5, As +5 + other products

Chromate/Bichromate Recovery Scheme

A feature of chromium with valence VI is that it forms 2 types of salts in aqueous solutions: chromates and bichromates, depending on the solution medium. Active metal chromates (for example, K 2 CrO 4) are salts that are stable in alkaline environment. Dichromates (bichromates) of active metals (for example, K 2 Cr 2 O 7) - salts, stable in an acidic environment .

Chromium(VI) compounds are reduced to chromium(III) compounds . Chromium compounds Cr +3 are amphoteric, and depending on the medium of the solution, they exist in solution in various forms: in an acidic medium in the form salts(amphoteric compounds form salts when interacting with acids), in a neutral medium - insoluble amphoteric chromium (III) hydroxide Cr(OH) 3 , and in an alkaline environment, chromium (III) compounds form complex salt, For example, potassium hexahydroxochromate (III) K 3 .

Chromium VI compounds oxidize:

• nonmetals in a negative oxidation state to simple substances (with oxidation state 0), exceptions — phosphorus, arsenic - up to +5;
• nonmetals in an intermediate oxidation state to the highest degree of oxidation;
• active metals from simple substances (oxidation point 0) to compounds with stable positive the degree of oxidation of the metal.

Chromate/bichromate + neMe (negative d.d.) = neMe 0 + other products

Chromate/bichromate + NeMe (intermediate positive s.d.) = NeMe (highest s.d.) + other products

Chromate / bichromate + Me 0 \u003d Me (stable s.d.) + other products

Chromate/dichromate + P, As (negative d.d.) = P, As +5 + other products

Decomposition of nitrates

Nitrate salts contain nitrogen in oxidation state +5 - strong oxidizing agent. Such nitrogen can oxidize oxygen (O -2). This happens when nitrates are heated. In this case, in most cases, oxygen is oxidized to the oxidation state 0, i.e. before molecular oxygen O2 .

Depending on the type of metal that forms the salt, various products are formed during the thermal (temperature) decomposition of nitrates: if metal active(in the series of electrochemical activity are to magnesium), then nitrogen is reduced to an oxidation state of +3, and upon decomposition nitrite salts and molecular oxygen are formed .

for example:

2NaNO 3 → 2NaNO 2 + O 2 .

Active metals occur in nature in the form of salts (KCl, NaCl).

If a metal is in the electrochemical activity series to the right of magnesium and to the left of copper (including magnesium and copper) , then the decomposition produces metal oxide in a stable oxidation state, nitric oxide (IV)(brown gas) and oxygen. Metal oxide also forms during decomposition lithium nitrate .

for example, decomposition zinc nitrate:

2Zn(NO 3) 2 → 2ZnO + 4NO 2 + O 2.

Metals of medium activity are most often found in nature in the form of oxides (Fe 2 O 3, Al 2 O 3, etc.).

ions metals, located in the series of electrochemical activity to the right of copper are strong oxidizing agents. At decomposition of nitrates they, like N +5, participate in the oxidation of oxygen, and are reduced to simple substances, i.e. metal is formed and gases are released nitric oxide (IV) and oxygen .

for example, decomposition silver nitrate:

2AgNO 3 → 2Ag + 2NO 2 + O 2 .

Inactive metals occur in nature in the form of simple substances.

Some exceptions!

Decomposition ammonium nitrate :

In the ammonium nitrate molecule there is both an oxidizing agent and a reducing agent: nitrogen in the -3 oxidation state exhibits only reducing properties, nitrogen in the +5 oxidation state only oxidizing.

When heated, ammonium nitrate decomposing. At temperatures up to 270 o C, nitric oxide (I)("laughing gas") and water:

NH 4 NO 3 → N 2 O + 2H 2 O

This is an example of a reaction counterdisproportionation .

The resulting oxidation state of nitrogen is the arithmetic mean of the oxidation state of nitrogen atoms in the original molecule.

At a higher temperature, nitric oxide (I) decomposes into simple substances - nitrogen and oxygen:

2NH 4 NO 3 → 2N 2 + O 2 + 4H 2 O

At decomposition ammonium nitrite NH4NO2 counter-disproportionation also occurs.

The resulting oxidation state of nitrogen is also equal to the arithmetic mean of the oxidation states of the initial nitrogen atoms - the oxidizing agent N +3 and the reducing agent N -3

NH 4 NO 2 → N 2 + 2H 2 O

Thermal decomposition manganese(II) nitrate accompanied by metal oxidation:

Mn(NO 3) 2 \u003d MnO 2 + 2NO 2

Iron(II) nitrate at low temperatures it decomposes to iron oxide (II), when heated, iron is oxidized to an oxidation state of +3:

2Fe(NO 3) 2 → 2FeO + 4NO 2 + O 2 at 60°C
4Fe(NO 3) 2 → 2Fe 2 O 3 + 8NO 2 + O 2 at >60°C

Nickel(II) nitrate decomposes to nitrite when heated.

Oxidizing properties of nitric acid

Nitric acid HNO 3 when interacting with metals is practically never forms hydrogen , unlike most mineral acids.

This is due to the fact that the acid contains a very strong oxidizing agent - nitrogen in the +5 oxidation state. When interacting with reducing agents - metals, various products of nitrogen reduction are formed.

Nitric acid + metal \u003d metal salt + nitrogen reduction product + H 2 O

Nitric acid can be converted into nitric oxide (IV) NO 2 (N +4); nitric oxide (II) NO (N +2); nitric oxide (I) N 2 O ("laughing gas"); molecular nitrogen N 2 ; ammonium nitrate NH 4 NO 3. As a rule, a mixture of products is formed with a predominance of one of them. Nitrogen is reduced in this case to oxidation states from +4 to −3. The depth of recovery depends primarily by nature reducing agent and from the concentration of nitric acid . This is how the rule works: the lower the concentration of the acid and the higher the activity of the metal, the more electrons nitrogen receives, and the more reduced products are formed.

Some patterns will allow you to correctly determine the main product of nitric acid reduction by metals in the reaction:

• under action very dilute nitric acid on the metals usually formed ammonium nitrate NH 4 NO 3 ;

for example, interaction of zinc with very dilute nitric acid:

4Zn + 10HNO 3 = 4Zn(NO 3) 2 + NH 4 NO 3 + 3H 2 O

• concentrated nitric acid in the cold passivates some metals - chromium Cr, aluminum Al and iron Fe . When the solution is heated or diluted, the reaction proceeds;

metal passivation - this is the transfer of the metal surface to an inactive state due to the formation of thin layers of inert compounds on the metal surface, in this case mainly metal oxides, which do not react with concentrated nitric acid

• Nitric acid does not react with platinum subgroup metals — gold Au, platinum Pt, and palladium Pd;

• when interacting concentrated acid with inactive metals and metals of medium activity nitric acid is reduced to nitric oxide (IV) NO 2 ;

for example, oxidation of copper with concentrated nitric acid:

Cu + 4HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O

• when interacting concentrated nitric acid with active metals formed nitrogen oxide (I) N 2 O ;

for example, oxidation sodium concentrated nitric acid:

Na + 10HNO 3 \u003d 8NaNO 3 + N 2 O + 5H 2 O

• when interacting dilute nitric acid with inactive metals (in the activity series to the right of hydrogen) the acid is reduced to nitric oxide (II) NO ;
• when interacting dilute nitric acid with intermediate activity metals either nitric oxide (II) NO, or nitric oxide N 2 O, or molecular nitrogen N 2 - depending on additional factors (metal activity, metal grinding degree, acid dilution degree, temperature).
• when interacting dilute nitric acid with active metals formed molecular nitrogen N 2 .

For an approximate determination of the products of the reduction of nitric acid in the interaction with different metals, I propose to use the pendulum principle. The main factors that shift the position of the pendulum are the concentration of the acid and the activity of the metal. To simplify, we use 3 types of acid concentrations: concentrated (more than 30%), dilute (30% or less), very dilute (less than 5%). We divide metals by activity into active (before aluminum), medium activity (from aluminum to hydrogen) and inactive (after hydrogen). The products of the reduction of nitric acid are arranged in descending order of the degree of oxidation:

NO2; NO; N 2 O; N 2 ; NH4NO3

The more active the metal, the more we move to the right. The greater the concentration or the lower the dilution of the acid, the more we shift to the left.

for example , concentrated acid and inactive metal copper Cu interact. Therefore, we shift to the extreme left position, nitric oxide (IV), copper nitrate and water are formed.

The interaction of metals with sulfuric acid

Dilute sulfuric acid interacts with metals like a normal mineral acid. Those. interacts with metals that are located in a series of electrochemical voltages up to hydrogen. The oxidizing agent here is H + ions, which are reduced to molecular hydrogen H 2. In this case, metals are oxidized, as a rule, to minimum degree of oxidation.

for example:

Fe + H 2 SO 4 (razb) \u003d FeSO 4 + H 2

interacts with metals standing in a series of voltages both before and after hydrogen.

H 2 SO 4 (conc) + metal \u003d metal salt + sulfur reduction product (SO 2, S, H 2 S) + water

When concentrated sulfuric acid interacts with metals, a metal salt (in a stable oxidation state), water and a sulfur reduction product are formed - sulfur dioxide S +4 O 2, molecular sulfur S or hydrogen sulfide H 2 S -2, depending on the degree of concentration, activity of the metal, its degree of grinding, temperature, etc. When concentrated sulfuric acid interacts with metals, molecular hydrogen is not formed!

The basic principles of the interaction of concentrated sulfuric acid with metals:

1. concentrated sulfuric acid passivates aluminium, chrome, iron at room temperature, or in the cold;

2. concentrated sulfuric acid does not interact with gold, platinum and palladium ;

3. With inactive metals concentrated sulfuric acid recovers to sulfur oxide (IV).

for example, copper is oxidized with concentrated sulfuric acid:

Cu 0 + 2H 2 S +6 O 4 (conc) = Cu +2 SO 4 + S +4 O 2 + 2H 2 O

4. When interacting with active metals and zinc concentrated sulfuric acid formssulfur S or hydrogen sulfide H 2 S 2- (depending on temperature, degree of grinding and activity of the metal).

for example , interaction of concentrated sulfuric acid with zinc:

8Na 0 + 5H 2 S +6 O 4(conc) → 4Na 2 + SO 4 + H 2 S — 2 + 4H2O

Hydrogen peroxide

Hydrogen peroxide H 2 O 2 contains oxygen in the -1 oxidation state. Such oxygen can both increase and decrease the oxidation state. Thus, hydrogen peroxide exhibits both oxidizing and reducing properties.

When interacting with reducing agents, hydrogen peroxide exhibits the properties of an oxidizing agent, and is reduced to an oxidation state of -2. As a rule, the product of reduction of hydrogen peroxide is water or hydroxide ion, depending on the reaction conditions. For example:

S +4 O 2 + H 2 O 2 -1 → H 2 S +6 O 4 -2

When interacting with oxidizing agents, peroxide is oxidized to molecular oxygen (oxidation state 0): O 2 . for example :

2KMn +7 O 4 + 5H 2 O 2 -1 + 3H 2 SO 4 → 5O 2 0 + 2Mn +2 SO 4 + K 2 SO 4 + 8H 2 O

A. H2S B.SO3 C.H2SO3

2. The oxidation state of carbon in calcium carbonate is:
A. -4 B.+2 C.+4

3. A substance in which the oxidation state of phosphorus is zero:
A. P4 B.PH3 C.P2O5

4. Redox is a reaction whose equation is:
A.2Al(OH)3=Al2O3+3H2O B.H2+Cl2=2HCl C.NaOH+HNO3=NaNO3+H2O

5. An oxidizing agent in a chemical reaction, CuO+H2=Cu+H2O is:
A.H20 B.Cu2+ C.O2- D.Cu0

6. The oxidation state of chlorine decreases in the series:
A.Cl2– HCl–HClO B.NaCl–Cl2–KClO3 C.HClO4–NaClO2–BaCl2

7. The transition process, the scheme of which N-3→N+2 is:
A. Recovery
B. Oxidation
B. Not a redox process.

8. In the reaction equation S + O2 → SO2, the number of electrons donated by the oxidizing agent is:
A.2 B.4 C.6

9. Phosphorus in oxidation state 0 can be:
A.Only reductant
B. Oxidizer only
B. Oxidizing and reducing agent

10. A simple substance is a non-metal with the strongest oxidizing properties:
A. Br2
B. Cl2
B. F2

Part B.
11. Write the formulas for nitric oxide (III) and nitric oxide (V)

12. Arrange the coefficients in the reaction scheme using the electronic balance method:
Ca + O2 = CaO
Name the processes of oxidation and reduction, and indicate the oxidizing agent and reducing agent.

13. Arrange the formulas of chemical compounds: CH4, CO2, CO - in order of decreasing oxidation states of carbon atoms.

14. According to the scheme Cu + 2 + 2ē → Cu0, make up an equation for a chemical reaction and consider it from the point of view of the OVR.

15. Complete the phrase: "Restoration is ..."

1) add the reaction equations, indicate the oxidation states of the elements and arrange the coefficients using the electronic balance method: Ca + O2 ->, N2 + H2 ->. 2)

determine the oxidation state of each element, arrange the coefficients using the electronic balance method: KCIO3 + S -> KCI + SO2. 3) Please determine the oxidation state of sulfur in the following compounds: H2SO4, SO2, H2S, SO2, H2SO3. 4 towards the atoms of which chemical element do common electron pairs shift in the molecules of the following compounds: H2O, HI, PCI3, H3N, H2S, CO2? please give a valid answer! 5) tell me, do the oxidation states of atoms change when water is formed from hydrogen and oxygen? 6) write the equations of electrolytic dissociation: copper nitrate, hydrochloric acid, aluminum sulfate, barium hydroxide, zinc sulfate. 7) please write the molecular and ionic equations of reactions between solutions: lithium hydroxide and nitric acid, copper nitrate and sodium hydroxide, potassium carbonate and phosphoric acid. 8) in the interaction of solutions of which substances, one of the reaction products is water? K2CO3 and HCI: Ca(OH)2 and HNO3: NaOH and H2SO4: NaNO3 and H2SO4? Please write the reaction equations in molecular and ionic formulas. 9) which of the following salts undergo hydrolysis when dissolved in water: aluminum chloride, potassium sulfide, sodium chloride? Write equations for hydrolysis.