Chromium exhibits the same degree of oxidation in compounds. Oxidation state of chromium. Role in biology

Chromium oxide(II) and chromium(II) hydroxide are basic in nature

Cr(OH)+2HCl→CrCl+2HO

Chromium(II) compounds are strong reducing agents; transform into a chromium(III) compound under the influence of atmospheric oxygen.

2CrCl+ 2HCl → 2CrCl+ H

4Cr(OH)+O+ 2HO→4Cr(OH)

Chromium oxide(III) CrO is a green, water-insoluble powder. Can be obtained by calcination of chromium(III) hydroxide or potassium and ammonium dichromates:

2Cr(OH)-→CrO+ 3HO

4KCrO-→ 2CrO + 4KCrO + 3O

(NH)CrO-→ CrO+ N+ HO

It is difficult to interact with concentrated solutions of acids and alkalis:

Cr 2 O 3 + 6 KOH + 3H 2 O = 2K 3 [Cr(OH) 6 ]

Cr 2 O 3 + 6HCl = 2CrCl 3 + 3H 2 O

Chromium (III) hydroxide Cr(OH) 3 is obtained by the action of alkalis on solutions of chromium (III) salts:

CrCl 3 + 3KOH = Cr(OH) 3 ↓ + 3KCl

Chromium (III) hydroxide is a gray-green precipitate, upon receipt of which the alkali must be taken in deficiency. The chromium (III) hydroxide obtained in this way, in contrast to the corresponding oxide, easily interacts with acids and alkalis, i.e. exhibits amphoteric properties:

Cr(OH) 3 + 3HNO 3 = Cr(NO 3) 3 + 3H 2 O

Cr(OH) 3 + 3KOH = K 3 [Cr(OH)6] (hexahydroxochromite K)

When Cr(OH) 3 is fused with alkalis, metachromites and orthochromites are obtained:

Cr(OH) 3 + KOH = KCrO 2 (metachromite K)+ 2H 2 O

Cr(OH) 3 + KOH = K 3 CrO 3 (orthochromite K)+ 3H 2 O

Chromium compounds(VI).

Chromium oxide (VI) - CrO 3 – dark red crystalline substance, highly soluble in water – a typical acidic oxide. This oxide corresponds to two acids:

CrO 3 + H 2 O = H 2 CrO 4 (chromic acid – formed when there is excess water)

CrO 3 + H 2 O =H 2 Cr 2 O 7 (dichromic acid - formed at a high concentration of chromium oxide (3)).

Chromium oxide (6) is a very strong oxidizing agent, therefore it energetically interacts with organic substances:

C 2 H 5 OH + 4CrO 3 = 2CO 2 + 2Cr 2 O 3 + 3H 2 O

Also oxidizes iodine, sulfur, phosphorus, coal:

3S + 4CrO 3 = 3SO 2 + 2Cr 2 O 3

When heated to 250 0 C, chromium oxide (6) decomposes:

4CrO3 = 2Cr2O3 + 3O2

Chromium oxide (6) can be obtained by the action of concentrated sulfuric acid on solid chromates and dichromates:

K 2 Cr 2 O 7 + H 2 SO 4 = K 2 SO 4 + 2CrO 3 + H 2 O

Chromic and dichromic acids.

Chromic and dichromic acids exist only in aqueous solutions and form stable salts, chromates and dichromates, respectively. Chromates and their solutions are yellow in color, dichromates are orange.

Chromate - CrO 4 2- ions and dichromate - Cr2O 7 2- ions easily transform into each other when the solution environment changes

In an acidic solution, chromates transform into dichromates:

2K 2 CrO 4 + H 2 SO 4 = K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O

In an alkaline environment, dichromates turn into chromates:

K 2 Cr 2 O 7 + 2 KOH = 2 K 2 CrO 4 + H 2 O

When diluted, dichromic acid turns into chromic acid:

H 2 Cr 2 O 7 + H 2 O = 2H 2 CrO 4

Dependence of the properties of chromium compounds on the degree of oxidation.

Redox properties of chromium compounds.

Reactions in an acidic environment.

In an acidic environment, Cr +6 compounds transform into Cr +3 compounds under the action of reducing agents: H 2 S, SO 2, FeSO 4

K 2 Cr 2 O 7 + 3H 2 S + 4H 2 SO 4 = 3S + Cr 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O

S -2 – 2e → S 0

2Cr +6 + 6e → 2Cr +3

Reactions in an alkaline environment.

In an alkaline environment, chromium compounds Cr +3 transform into compounds Cr +6 under the action of oxidizing agents: J2, Br2, Cl2, Ag2O, KClO3, H2O2, KMnO4:

2KCrO 2 +3 Br2 +8NaOH =2Na 2 CrO 4 + 2KBr +4NaBr + 4H 2 O

Cr +3 - 3e → Cr +6

DEFINITION

Chromium located in the fourth period of group VI of the secondary (B) subgroup periodic table. Designation – Cr. In the form of a simple substance - a grayish-white shiny metal.

Chrome has a body-centered cubic lattice structure. Density - 7.2 g/cm3. The melting and boiling points are 1890 o C and 2680 o C, respectively.

Oxidation state of chromium in compounds

Chromium can exist in the form of a simple substance - a metal, and the oxidation state of metals in the elemental state is equal to zero, since the distribution of electron density in them is uniform.

Oxidation states (+2) And (+3) chromium appears in oxides (Cr +2 O, Cr +3 2 O 3), hydroxides (Cr +2 (OH) 2, Cr +3 (OH) 3), halides (Cr +2 Cl 2, Cr +3 Cl 3 ), sulfates (Cr +2 SO 4, Cr +3 2 (SO 4) 3) and other compounds.

Chromium is also characterized by its oxidation state (+6) : Cr +6 O 3, H 2 Cr +6 O 4, H 2 Cr +6 2 O 7, K 2 Cr +6 2 O 7, etc.

Examples of problem solving

EXAMPLE 1

EXAMPLE 2

Exercise	Phosphorus has the same oxidation state in the following compounds: a) Ca 3 P 2 and H 3 PO 3; b) KH 2 PO 4 and KPO 3; c) P 4 O 6 and P 4 O 10; d) H 3 PO 4 and H 3 PO 3.
Solution	In order to give the correct answer to the question posed, we will alternately determine the degree of oxidation of phosphorus in each pair of proposed compounds. a) The oxidation state of calcium is (+2), oxygen and hydrogen - (-2) and (+1), respectively. Let us take the value of the oxidation state of phosphorus as “x” and “y” in the proposed compounds: 3 ×2 + x ×2 = 0; 3 + y + 3×(-2) = 0; The answer is incorrect. b) The oxidation state of potassium is (+1), oxygen and hydrogen are (-2) and (+1), respectively. Let us take the value of the oxidation state of chlorine as “x” and “y” in the proposed compounds: 1 + 2×1 +x + (-2)×4 = 0; 1 + y + (-2)×3 = 0; The answer is correct.
Answer	Option (b).

Introduction

Oxidation state (CO) - this is a symbol in chemistry that serves to determine the charge of an atom of a chemical element (or group of elements). Without oxidation states, not a single problem can be solved, not a single equation can be compiled, but most importantly, without them we cannot clearly determine the properties of an element and what role it will play in various compounds.

It is significant that the periodic system (PS) of D.I. Mendeleev is grouped in the most ingenious way: all elements are divided into periods, groups, subgroups, their serial numbers also correspond to certain indicators. Thanks to this, we do not have to learn the qualities of each chemical element (CE) by heart, because we can easily find it in the table and determine everything that is required. However, even in this case, some people, forgetting school knowledge in a chemistry course (or having neglected them once), we are forced to return to studying this topic in more detail.

So, first you need to form the correct objective ideas about chromium ( Cr ), understand its position in the PS, and then you can proceed to the most important part - practice.
Chromium - Cr, position in the periodic table, physical and chemical properties
Chromium - it is a solid, metallic, shiny, silvery-white (or bluish) color. It is quite brittle, but at the same time it has an incomparable advantage compared to many other metals - resistance to corrosion; This is why it is an important component in the production of stainless steel, and is also used to coat the surface of other metals that are more prone to corrosion. Chrome has poor thermal and electrical conductivity.

ChE is located in group VI, period 4, has serial number 24 and has atomic mass equal to 52 g/mol. Thanks to passivation, chromium does not interact with sulfur ( H2SO4) and nitrogen ( HNO3) acids, exhibits stability in air.

This amphoteric metal - This means that it can dissolve in both acids and alkalis. The element dissolves in strong dilute acids (for example, hydrochloric acid HCl), under normal conditions (n.s.) interacts only with fluorine ( F). When heated, chromium can interact with elements of group VII (halogens), oxygen O 2, boron B, nitrogen N 2, gray S 2, silicon Si. If you heat it up Cr, then it is able to react with water vapor.

Now let's talk directly about what oxidation states a given CE has: it can acquire CO +4, +6, as well as +2 in an airless space, +3 in a space with air. Chromium, like any other metal, is a strong reducing agent.

Substances with different oxidation states

• +2. When Cr acquires CO+2, the substance demonstrates basic and very strong restorative properties. For example, chromium (II) oxide - CrO, chromium hydroxide - Cr(OH)2, lots of salts. Compounds of this element with fluorine are synthesized( CrF 2), chlorine( CrCl2) and so on.
• +3. These substances have amphoteric properties and can be of different colors (but mostly green H2O). For example, let's take the oxide Cr2O3(this is a greenish powder that does not dissolve in), Cr(OH)3, chromites NaCrO2.
• +4. Such compounds are very rare: they do not form salts or acids, and almost no work is done with them. But from known substances there are oxide CrO2, tetrahalide CrF 4, CrCl 4.
• +6. Chrome s CO+6, forming salts, is acidic in nature, very poisonous, hydroscopic, and also has strong oxidizing properties. Examples: CrO3(looks like red crystals), K2CrO4, H2CrO4, H2Cr2O7. The element is capable of forming two types of hydroxides (already listed).

How to determine CO in complex substances

You are probably already familiar with the “criss-cross” rule. What if the connection has, for example, as many as three elements?

In this case, we look at the last element of the substance, determine its oxidation state and multiply by the coefficient on the right (of course, if it exists). We mentally separate the last element (with an already certain oxidation state) from the other two elements. We require that CO the first two and last elements added up to zero.

Let's look at an example:

• PbCrO4 - lead (II) chromate, which looks like a red salt. At the end of the formula is oxygen, the oxidation state of which will always (except in some cases) be -2. -2*4=-8. Pb (lead) has CO+2. Further actions will be similar to an algebraic equation, but to be honest, when a person is already well versed in determining oxidation states and knows how to use the solubility table, it is quite possible to avoid such calculations. So, we denote an element with an unknown oxidation state (chromium) as a letter variable. 2+x-8=0;x=8-2;x=6. The variable is 6, therefore the oxidation state of chromium becomes +6.

Oxidation states in following formulas try to arrange it yourself:

1. Na2CrO4;
2. BaCrO4;
3. Fe(CrO 2) 2;
4. Cr2O7;
5. H2CrO4.

Chromium -one of the most interesting chemical elements, connections with which are a complex thing, but necessary to understand. It would be great if these examples help to understand such a painstaking topic.

Editorial "site"

Chromium is an element of the side subgroup of the 6th group of the 4th period of the periodic system of chemical elements of D.I. Mendeleev, with atomic number 24. It is designated by the symbol Cr (lat. Chromium). The simple substance chromium is a hard metal of a bluish-white color.

Chemical properties of chromium

Under normal conditions, chromium reacts only with fluorine. At high temperatures (above 600°C) it interacts with oxygen, halogens, nitrogen, silicon, boron, sulfur, phosphorus.

4Cr + 3O 2 – t° →2Cr 2 O 3

2Cr + 3Cl 2 – t° → 2CrCl 3

2Cr + N 2 – t° → 2CrN

2Cr + 3S – t° → Cr 2 S 3

When heated, it reacts with water vapor:

2Cr + 3H 2 O → Cr 2 O 3 + 3H 2

Chromium dissolves in dilute strong acids(HCl, H2SO4)

In the absence of air, Cr 2+ salts are formed, and in air, Cr 3+ salts are formed.

Cr + 2HCl → CrCl 2 + H 2

2Cr + 6HCl + O 2 → 2CrCl 3 + 2H 2 O + H 2

The presence of a protective oxide film on the surface of the metal explains its passivity in relation to concentrated solutions of acids - oxidizers.

Chromium compounds

Chromium(II) oxide and chromium(II) hydroxide are basic in nature.

Cr(OH) 2 + 2HCl → CrCl 2 + 2H 2 O

Chromium (II) compounds are strong reducing agents; transform into chromium (III) compounds under the influence of atmospheric oxygen.

2CrCl 2 + 2HCl → 2CrCl 3 + H 2

4Cr(OH) 2 + O 2 + 2H 2 O → 4Cr(OH) 3

Chromium oxide (III) Cr 2 O 3 is a green, water-insoluble powder. Can be obtained by calcination of chromium(III) hydroxide or potassium and ammonium dichromates:

2Cr(OH) 3 – t° → Cr 2 O 3 + 3H 2 O

4K 2 Cr 2 O 7 – t° → 2Cr 2 O 3 + 4K 2 CrO 4 + 3O 2

(NH 4) 2 Cr 2 O 7 – t° → Cr 2 O 3 + N 2 + 4H 2 O (volcano reaction)

Amphoteric oxide. When Cr 2 O 3 is fused with alkalis, soda and acid salts, chromium compounds with an oxidation state of (+3) are obtained:

Cr 2 O 3 + 2NaOH → 2NaCrO 2 + H 2 O

Cr 2 O 3 + Na 2 CO 3 → 2NaCrO 2 + CO 2

When fused with a mixture of alkali and oxidizing agent, chromium compounds are obtained in the oxidation state (+6):

Cr 2 O 3 + 4KOH + KClO 3 → 2K 2 CrO 4 + KCl + 2H 2 O

Chromium (III) hydroxide C r (OH) 3 . Amphoteric hydroxide. Gray-green, decomposes when heated, losing water and forming green metahydroxide CrO(OH). Does not dissolve in water. Precipitates from solution as a gray-blue and bluish-green hydrate. Reacts with acids and alkalis, does not interact with ammonia hydrate.

It has amphoteric properties - it dissolves in both acids and alkalis:

2Cr(OH) 3 + 3H 2 SO 4 → Cr 2 (SO 4) 3 + 6H 2 O Cr(OH) 3 + ZN + = Cr 3+ + 3H 2 O

Cr(OH) 3 + KOH → K, Cr(OH) 3 + ZON - (conc.) = [Cr(OH) 6 ] 3-

Cr(OH) 3 + KOH → KCrO 2 + 2H 2 O Cr(OH) 3 + MOH = MSrO 2 (green) + 2H 2 O (300-400 °C, M = Li, Na)

Cr(OH) 3 →(120 o C – H 2 O) CrO(OH) →(430-1000 0 C –H 2 O) Cr2O3

2Cr(OH) 3 + 4NaOH (conc.) + ZN 2 O 2 (conc.) = 2Na 2 CrO 4 + 8H 2 0

Receipt: precipitation with ammonia hydrate from a solution of chromium(III) salts:

Cr 3+ + 3(NH 3 H 2 O) = WITHr(OH) 3 ↓+ ЗNН 4+

Cr 2 (SO 4) 3 + 6NaOH → 2Cr(OH) 3 ↓+ 3Na 2 SO 4 (in excess alkali - the precipitate dissolves)

Chromium (III) salts have a purple or dark green color. Their chemical properties resemble colorless aluminum salts.

Cr(III) compounds can exhibit both oxidizing and reducing properties:

Zn + 2Cr +3 Cl 3 → 2Cr +2 Cl 2 + ZnCl 2

2Cr +3 Cl 3 + 16NaOH + 3Br 2 → 6NaBr + 6NaCl + 8H 2 O + 2Na 2 Cr +6 O 4

Hexavalent chromium compounds

Chromium(VI) oxide CrO 3 - bright red crystals, soluble in water.

Obtained from potassium chromate (or dichromate) and H 2 SO 4 (conc.).

K 2 CrO 4 + H 2 SO 4 → CrO 3 + K 2 SO 4 + H 2 O

K 2 Cr 2 O 7 + H 2 SO 4 → 2CrO 3 + K 2 SO 4 + H 2 O

CrO 3 is an acidic oxide, with alkalis it forms yellow chromates CrO 4 2-:

CrO 3 + 2KOH → K 2 CrO 4 + H 2 O

In an acidic environment, chromates turn into orange dichromates Cr 2 O 7 2-:

2K 2 CrO 4 + H 2 SO 4 → K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O

In an alkaline environment, this reaction proceeds in the opposite direction:

K 2 Cr 2 O 7 + 2KOH → 2K 2 CrO 4 + H 2 O

Potassium dichromate is an oxidizing agent in an acidic environment:

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3Na 2 SO 3 = Cr 2 (SO 4) 3 + 3Na 2 SO 4 + K 2 SO 4 + 4H 2 O

K 2 Cr 2 O 7 + 4H 2 SO 4 + 3NaNO 2 = Cr 2 (SO 4) 3 + 3NaNO 3 + K 2 SO 4 + 4H 2 O

K 2 Cr 2 O 7 + 7H 2 SO 4 + 6KI = Cr 2 (SO 4) 3 + 3I 2 + 4K 2 SO 4 + 7H 2 O

K 2 Cr 2 O 7 + 7H 2 SO 4 + 6FeSO 4 = Cr 2 (SO 4) 3 + 3Fe 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O

Potassium chromate K 2 Cr O 4 . Oxosol. Yellow, non-hygroscopic. Melts without decomposition, thermally stable. Very soluble in water ( yellow the color of the solution corresponds to the CrO 4 2- ion), slightly hydrolyzes the anion. In an acidic environment it turns into K 2 Cr 2 O 7 . Oxidizing agent (weaker than K 2 Cr 2 O 7). Enters into ion exchange reactions.

Qualitative reaction on the CrO 4 2- ion - the precipitation of a yellow precipitate of barium chromate, which decomposes in a strongly acidic environment. It is used as a mordant for dyeing fabrics, a leather tanning agent, a selective oxidizing agent, a reagent in analytical chemistry.

Equations of the most important reactions:

2K 2 CrO 4 +H 2 SO 4(30%)= K 2 Cr 2 O 7 +K 2 SO 4 +H 2 O

2K 2 CrO 4 (t) +16HCl (concentration, horizon) = 2CrCl 3 +3Cl 2 +8H 2 O+4KCl

2K 2 CrO 4 +2H 2 O+3H 2 S=2Cr(OH) 3 ↓+3S↓+4KOH

2K 2 CrO 4 +8H 2 O+3K 2 S=2K[Cr(OH) 6 ]+3S↓+4KOH

2K 2 CrO 4 +2AgNO 3 =KNO 3 +Ag 2 CrO 4(red) ↓

Qualitative reaction:

K 2 CrO 4 + BaCl 2 = 2KCl + BaCrO 4 ↓

2BaCrO 4 (t) + 2HCl (dil.) = BaCr 2 O 7 (p) + BaC1 2 + H 2 O

Receipt: sintering of chromite with potash in air:

4(Cr 2 Fe ‖‖)O 4 + 8K 2 CO 3 + 7O 2 = 8K 2 CrO 4 + 2Fe 2 O 3 + 8СO 2 (1000 °C)

Potassium dichromate K 2 Cr 2 O 7 . Oxosol. Technical name chrome peak. Orange-red, non-hygroscopic. Melts without decomposition, and decomposes upon further heating. Very soluble in water ( orange The color of the solution corresponds to the Cr 2 O 7 2- ion. In an alkaline environment it forms K 2 CrO 4 . A typical oxidizing agent in solution and during fusion. Enters into ion exchange reactions.

Qualitative reactions - blue color of an ethereal solution in the presence of H 2 O 2, blue color of an aqueous solution under the action of atomic hydrogen.

It is used as a leather tanning agent, a mordant for dyeing fabrics, a component of pyrotechnic compositions, a reagent in analytical chemistry, a metal corrosion inhibitor, in a mixture with H 2 SO 4 (conc.) - for washing chemical dishes.

Equations of the most important reactions:

4K 2 Cr 2 O 7 =4K 2 CrO 4 +2Cr 2 O 3 +3O 2 (500-600 o C)

K 2 Cr 2 O 7 (t) +14HCl (conc) = 2CrCl 3 +3Cl 2 +7H 2 O+2KCl (boiling)

K 2 Cr 2 O 7 (t) +2H 2 SO 4(96%) ⇌2KHSO 4 +2CrO 3 +H 2 O (“chromium mixture”)

K 2 Cr 2 O 7 +KOH (conc) =H 2 O+2K 2 CrO 4

Cr 2 O 7 2- +14H + +6I - =2Cr 3+ +3I 2 ↓+7H 2 O

Cr 2 O 7 2- +2H + +3SO 2 (g) = 2Cr 3+ +3SO 4 2- +H 2 O

Cr 2 O 7 2- +H 2 O +3H 2 S (g) =3S↓+2OH - +2Cr 2 (OH) 3 ↓

Cr 2 O 7 2- (conc.) +2Ag + (dil.) =Ag 2 Cr 2 O 7 (red) ↓

Cr 2 O 7 2- (dil.) +H 2 O +Pb 2+ =2H + + 2PbCrO 4 (red) ↓

K 2 Cr 2 O 7(t) +6HCl+8H 0 (Zn)=2CrCl 2(syn) +7H 2 O+2KCl

Receipt: treatment of K 2 CrO 4 with sulfuric acid:

2K 2 CrO 4 + H 2 SO 4 (30%) = K 2Cr 2 O 7 + K 2 SO 4 + H 2 O



Target: deepen students' knowledge on the topic of the lesson.

Tasks:

• characterize chromium as a simple substance;
• introduce students to chromium compounds of different oxidation states;
• show the dependence of the properties of compounds on the degree of oxidation;
• show the redox properties of chromium compounds;
• continue to develop students’ skills in writing down equations of chemical reactions in molecular and ionic form and creating an electronic balance;
• continue to develop the skills to observe a chemical experiment.

Lesson form: lecture with elements independent work students and observing a chemical experiment.

Progress of the lesson

I. Repetition of material from the previous lesson.

1. Answer questions and complete tasks:

What elements belong to the chromium subgroup?

Write electronic formulas of atoms

What type of elements are they?

What oxidation states do the compounds exhibit?

How does the atomic radius and ionization energy change from chromium to tungsten?

You can ask students to complete the table using the tabulated values ​​of atomic radii, ionization energies and draw conclusions.

Sample table:

2. Listen to a student’s report on the topic “Elements of the chromium subgroup in nature, preparation and application.”

II. Lecture.

Lecture outline:

1. Chromium.
2. Chromium compounds. (2)
• Chromium oxide; (2)
• Chromium hydroxide. (2)
1. Chromium compounds. (3)
• Chromium oxide; (3)
• Chromium hydroxide. (3)
1. Chromium compounds (6)
• Chromium oxide; (6)
• Chromic and dichromic acids.
1. Dependence of the properties of chromium compounds on the degree of oxidation.
2. Redox properties of chromium compounds.

1. Chrome.

Chrome is a white, shiny metal with a bluish tint, very hard (density 7.2 g/cm3), melting point 1890˚C.

Chemical properties: Chromium is an inactive metal under normal conditions. This is explained by the fact that its surface is covered with an oxide film (Cr 2 O 3). When heated, the oxide film is destroyed, and chromium reacts with simple substances at high temperatures:

• 4Сr +3О 2 = 2Сr 2 О 3
• 2Сr + 3S = Сr 2 S 3
• 2Сr + 3Cl 2 = 2СrСl 3

Exercise: draw up equations for the reactions of chromium with nitrogen, phosphorus, carbon and silicon; Compose an electronic balance for one of the equations, indicate the oxidizing agent and the reducing agent.

Interaction of chromium with complex substances:

At very high temperatures, chromium reacts with water:

• 2Сr + 3Н2О = Сr2О3 + 3Н2

Exercise:

Chromium reacts with dilute sulfuric and hydrochloric acids:

• Cr + H 2 SO 4 = CrSO 4 + H 2
• Cr + 2HCl = CrCl 2 + H 2

Exercise: draw up an electronic balance, indicate the oxidizing agent and reducing agent.

Concentrated sulfuric hydrochloric acid and nitric acid passivate chrome.

2. Chromium compounds. (2)

1. Chromium oxide (2)- CrO is a solid, bright red substance, a typical basic oxide (it corresponds to chromium (2) hydroxide - Cr(OH) 2), does not dissolve in water, but dissolves in acids:

• CrO + 2HCl = CrCl 2 + H 2 O

Exercise: draw up a reaction equation in molecular and ionic form for the interaction of chromium oxide (2) with sulfuric acid.

Chromium oxide (2) is easily oxidized in air:

• 4CrO+ O 2 = 2Cr 2 O 3

Exercise: draw up an electronic balance, indicate the oxidizing agent and reducing agent.

Chromium oxide (2) is formed by the oxidation of chromium amalgam with atmospheric oxygen:

2Сr (amalgam) + O 2 = 2СrО

2. Chromium hydroxide (2)- Cr(OH) 2 is a yellow substance, poorly soluble in water, with a pronounced basic character, therefore it interacts with acids:

• Cr(OH) 2 + H 2 SO 4 = CrSO 4 + 2H 2 O

Exercise: draw up reaction equations in molecular and ionic form for the interaction of chromium oxide (2) with hydrochloric acid.

Like chromium(2) oxide, chromium(2) hydroxide is oxidized:

• 4 Cr(OH) 2 + O 2 + 2H 2 O = 4Cr(OH) 3

Exercise: draw up an electronic balance, indicate the oxidizing agent and reducing agent.

Chromium hydroxide (2) can be obtained by the action of alkalis on chromium salts (2):

• CrCl 2 + 2KOH = Cr(OH) 2 ↓ + 2KCl

Exercise: write ionic equations.

3. Chromium compounds. (3)

1. Chromium oxide (3)- Cr 2 O 3 – dark green powder, insoluble in water, refractory, close in hardness to corundum (chromium hydroxide (3) – Cr(OH) 3) corresponds to it. Chromium oxide (3) is amphoteric in nature, but is poorly soluble in acids and alkalis. Reactions with alkalis occur during fusion:

• Cr 2 O 3 + 2KOH = 2KSrO 2 (chromite K)+ H 2 O

Exercise: draw up a reaction equation in molecular and ionic form for the interaction of chromium oxide (3) with lithium hydroxide.

It is difficult to interact with concentrated solutions of acids and alkalis:

• Cr 2 O 3 + 6 KOH + 3H 2 O = 2K 3 [Cr(OH) 6 ]
• Cr 2 O 3 + 6HCl = 2CrCl 3 + 3H 2 O

Exercise: draw up reaction equations in molecular and ionic form for the interaction of chromium oxide (3) with concentrated sulfuric acid and a concentrated solution of sodium hydroxide.

Chromium oxide (3) can be obtained from the decomposition of ammonium dichromate:

• (NН 4)2Сr 2 О 7 = N 2 + Сr 2 О 3 +4Н 2 О

2. Chromium hydroxide (3) Cr(OH) 3 is obtained by the action of alkalis on solutions of chromium salts (3):

• CrCl 3 + 3KOH = Cr(OH) 3 ↓ + 3KCl

Exercise: write ionic equations

Chromium hydroxide (3) is a gray-green precipitate, upon receipt of which the alkali must be taken in deficiency. The chromium hydroxide (3) obtained in this way, in contrast to the corresponding oxide, easily interacts with acids and alkalis, i.e. exhibits amphoteric properties:

• Cr(OH) 3 + 3HNO 3 = Cr(NO 3) 3 + 3H 2 O
• Cr(OH) 3 + 3KOH = K 3 [Cr(OH)6] (hexahydroxochromite K)

Exercise: draw up reaction equations in molecular and ionic form for the interaction of chromium hydroxide (3) with hydrochloric acid and sodium hydroxide.

When Cr(OH) 3 is fused with alkalis, metachromites and orthochromites are obtained:

• Cr(OH) 3 + KOH = KCrO 2 (metachromite K)+ 2H 2 O
• Cr(OH) 3 + KOH = K 3 CrO 3 (orthochromite K)+ 3H 2 O

4. Chromium compounds. (6)

1. Chromium oxide (6)- CrO 3 – dark red crystalline substance, highly soluble in water – a typical acidic oxide. This oxide corresponds to two acids:

• CrO 3 + H 2 O = H 2 CrO 4 (chromic acid – formed when there is excess water)
• CrO 3 + H 2 O =H 2 Cr 2 O 7 (dichromic acid - formed at a high concentration of chromium oxide (3)).

Chromium oxide (6) is a very strong oxidizing agent, therefore it energetically interacts with organic substances:

• C 2 H 5 OH + 4CrO 3 = 2CO 2 + 2Cr 2 O 3 + 3H 2 O

Also oxidizes iodine, sulfur, phosphorus, coal:

• 3S + 4CrO 3 = 3SO 2 + 2Cr 2 O 3

Exercise: write equations chemical reactions chromium oxide (6) with iodine, phosphorus, coal; create an electronic balance for one of the equations, indicate the oxidizing agent and reducing agent

When heated to 250 0 C, chromium oxide (6) decomposes:

• 4CrO3 = 2Cr2O3 + 3O2

Chromium oxide (6) can be obtained by the action of concentrated sulfuric acid on solid chromates and dichromates:

• K 2 Cr 2 O 7 + H 2 SO 4 = K 2 SO 4 + 2CrO 3 + H 2 O

2. Chromic and dichromic acids.

Chromic and dichromic acids exist only in aqueous solutions and form stable salts, chromates and dichromates, respectively. Chromates and their solutions are yellow in color, dichromates are orange.

Chromate - CrO 4 2- ions and dichromate - Cr 2O 7 2- ions easily transform into each other when the solution environment changes

In an acidic solution, chromates transform into dichromates:

• 2K 2 CrO 4 + H 2 SO 4 = K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O

In an alkaline environment, dichromates turn into chromates:

• K 2 Cr 2 O 7 + 2 KOH = 2 K 2 CrO 4 + H 2 O

When diluted, dichromic acid turns into chromic acid:

• H 2 Cr 2 O 7 + H 2 O = 2H 2 CrO 4

5. Dependence of the properties of chromium compounds on the degree of oxidation.

Oxidation state	+2	+3	+6
Oxide	CrO	Cr 2 O 3	СrО 3
Character of the oxide	basic	amphoteric	acid
Hydroxide	Cr(OH) 2	Cr(OH) 3 – H 3 CrO 3	H 2 CrO 4
Nature of the hydroxide	basic	amphoteric	acid
→ weakening of basic properties and strengthening of acidic properties→

6. Redox properties of chromium compounds.

Reactions in an acidic environment.

In an acidic environment, Cr +6 compounds transform into Cr +3 compounds under the action of reducing agents: H 2 S, SO 2, FeSO 4

• K 2 Cr 2 O 7 + 3H 2 S + 4H 2 SO 4 = 3S + Cr 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O
• S -2 – 2e → S 0
• 2Cr +6 + 6e → 2Cr +3

Exercise:

1. Equalize the reaction equation using the electronic balance method, indicate the oxidizing agent and reducing agent:

• Na 2 CrO 4 + K 2 S + H 2 SO 4 = S + Cr 2 (SO 4) 3 + K 2 SO 4 + Na 2 SO 4 + H 2 O

2. Add the reaction products, equalize the equation using the electronic balance method, indicate the oxidizing agent and reducing agent:

• K 2 Cr 2 O 7 + SO 2 + H 2 SO 4 =? +? +H 2 O

Reactions in an alkaline environment.

In an alkaline environment, chromium compounds Cr +3 transform into compounds Cr +6 under the action of oxidizing agents: J2, Br2, Cl2, Ag2O, KClO3, H2O2, KMnO4:

• 2KCrO 2 +3 Br 2 +8NaOH =2Na 2 CrO 4 + 2KBr +4NaBr + 4H 2 O
• Cr +3 - 3e → Cr +6
• Br2 0 +2e → 2Br -

Exercise:

Equalize the reaction equation using the electronic balance method, indicate the oxidizing agent and reducing agent:

• NaCrO 2 + J 2 + NaOH = Na 2 CrO 4 + NaJ + H 2 O

Add the reaction products, equalize the equation using the electronic balance method, indicate the oxidizing agent and reducing agent:

• Cr(OH) 3 + Ag 2 O + NaOH = Ag + ? + ?

Thus, the oxidizing properties consistently increase with a change in oxidation states in the series: Cr +2 → Cr +3 → Cr +6. Chromium compounds (2) are strong reducing agents and are easily oxidized, turning into chromium compounds (3). Chromium compounds (6) are strong oxidizing agents and are easily reduced to chromium compounds (3). Chromium compounds (3) when interacting with strong reducing agents exhibit oxidizing properties, turning into chromium compounds (2), and when interacting with strong oxidizing agents they exhibit reducing properties, turning into chromium compounds (6)

To the lecture methodology:

1. To enhance students’ cognitive activity and maintain interest, it is advisable to conduct a demonstration experiment during the lecture. Depending on capabilities educational laboratory You can demonstrate the following experiments to students:
• obtaining chromium oxide (2) and chromium hydroxide (2), proof of their basic properties;
• obtaining chromium oxide (3) and chromium hydroxide (3), proving their amphoteric properties;
• obtaining chromium oxide (6) and dissolving it in water (preparation of chromic and dichromic acids);
• transition of chromates to dichromates, dichromates to chromates.
1. Independent work tasks can be differentiated taking into account the real learning capabilities of students.
2. You can complete the lecture by completing the following tasks: write equations of chemical reactions that can be used to carry out the following transformations:

.III. Homework: improve the lecture (add the equations of chemical reactions)

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