Such a school curriculum subject as chemistry causes numerous difficulties for most modern schoolchildren; few can determine the degree of oxidation in compounds. The greatest difficulties are experienced by schoolchildren who study, that is, primary school students (grades 8-9). Misunderstanding of the subject leads to the emergence of hostility among schoolchildren towards this subject.
Teachers identify a number of reasons for this “dislike” of middle and high school students for chemistry: reluctance to understand complex chemical terms, inability to use algorithms to consider a specific process, problems with mathematical knowledge. The Ministry of Education of the Russian Federation has made serious changes to the content of the subject. In addition, the number of hours for teaching chemistry was also “cut.” This had a negative impact on the quality of knowledge in the subject and decreased interest in studying the discipline.
What chemistry course topics are most difficult for schoolchildren?
According to the new program, the course of the basic school discipline “Chemistry” includes several serious topics: D.I. Mendeleev’s periodic table of elements, classes of inorganic substances, ion exchange. The most difficult thing for eighth graders is determining the degree of oxidation of oxides.
Arrangement rules
First of all, students should know that oxides are complex two-element compounds that include oxygen. A prerequisite for a binary compound to belong to the class of oxides is the location of oxygen second in this compound.
Algorithm for acid oxides
To begin with, let us note that degrees are numerical expressions of the valency of elements. Acidic oxides are formed by non-metals or metals with a valence of four to seven, the second in such oxides is always oxygen.
In oxides, the valence of oxygen always corresponds to two; it can be determined from the periodic table of elements by D.I. Mendeleev. A typical nonmetal like oxygen, being in group 6 of the main subgroup of the periodic table, accepts two electrons to completely complete its outer energy level. Nonmetals in compounds with oxygen most often exhibit a higher valence, which corresponds to the number of the group itself. It is important to remember that the oxidation state of chemical elements is an indicator that assumes a positive (negative) number.
The nonmetal at the beginning of the formula has a positive oxidation state. The nonmetal oxygen in oxides is stable, its index is -2. In order to check the reliability of the arrangement of values in acid oxides, you will have to multiply all the numbers you entered by the indices of a specific element. Calculations are considered reliable if the total sum of all the pros and cons of the given degrees is 0.
Compiling two-element formulas
The oxidation state of the atoms of elements gives the chance to create and write compounds from two elements. When creating a formula, firstly, both symbols are written side by side, and oxygen is always placed second. Above each of the recorded signs, the values of the oxidation states are written down, then between the found numbers there is a number that will be divisible by both numbers without any remainder. This indicator must be divided separately by the numerical value of the oxidation state, obtaining indices for the first and second components of the two-element substance. The highest oxidation state is numerically equal to the value of the highest valence of a typical non-metal and is identical to the number of the group where the non-metal is located in the PS.
Algorithm for setting numerical values in basic oxides
Oxides of typical metals are considered such compounds. In all compounds they have an oxidation state index of no more than +1 or +2. In order to understand what oxidation state a metal will have, you can use the periodic table. For metals of the main subgroups of the first group, this parameter is always constant, it is similar to the group number, that is, +1.
The metals of the main subgroup of the second group are also characterized by a stable oxidation state, in digital terms +2. The total oxidation states of oxides, taking into account their indices (numbers), should give zero, since the chemical molecule is considered a neutral particle, devoid of charge.
Arrangement of oxidation states in oxygen-containing acids
Acids are complex substances consisting of one or more hydrogen atoms that are bonded to some kind of acidic moiety. Given that oxidation states are numbers, calculating them will require some math skills. This indicator for hydrogen (proton) in acids is always stable and is +1. Next, you can indicate the oxidation state for the negative oxygen ion; it is also stable, -2.
Only after these steps can the oxidation state of the central component of the formula be calculated. As a specific example, consider determining the oxidation state of elements in sulfuric acid H2SO4. Considering that the molecule of this complex substance contains two hydrogen protons and 4 oxygen atoms, we obtain an expression of the form +2+X-8=0. In order for the sum to form zero, sulfur will have an oxidation state of +6
Arrangement of oxidation states in salts
Salts are complex compounds consisting of metal ions and one or more acidic residues. The method for determining the oxidation states of each of the constituent parts in a complex salt is the same as in oxygen-containing acids. Considering that the oxidation state of elements is a digital indicator, it is important to correctly indicate the oxidation state of the metal.
If the metal forming the salt is located in the main subgroup, its oxidation state will be stable, corresponds to the group number, and is a positive value. If the salt contains a metal of a similar PS subgroup, the different metals can be revealed by the acid residue. After the oxidation state of the metal is established, set (-2), then calculate the oxidation state of the central element using a chemical equation.
As an example, consider the determination of oxidation states of elements in (average salt). NaNO3. The salt is formed by a metal of the main subgroup of group 1, therefore, the oxidation state of sodium will be +1. Oxygen in nitrates has an oxidation state of -2. To determine the numerical value of the oxidation state, the equation is +1+X-6=0. Solving this equation, we find that X should be +5, this is
Basic terms in OVR
There are special terms for the oxidation and reduction processes that schoolchildren must learn.
The oxidation state of an atom is its direct ability to attach to itself (donate to others) electrons from some ions or atoms.
An oxidizing agent is considered to be neutral atoms or charged ions that gain electrons during a chemical reaction.
The reducing agent will be uncharged atoms or charged ions that lose their own electrons in the process of chemical interaction.
Oxidation is thought of as a procedure of donating electrons.
Reduction involves the acceptance of additional electrons by an uncharged atom or ion.
The redox process is characterized by a reaction during which the oxidation state of an atom necessarily changes. This definition provides insight into how one can determine whether a reaction is ODD.
Rules for parsing OVR
Using this algorithm, you can arrange the coefficients in any chemical reaction.
Knowledge and ability to determine the oxidation state of elements in molecules allows one to solve very complex reaction equations and, accordingly, correctly calculate the amounts of selected substances for reactions, experiments and technological processes. The oxidation state is one of the most important, key concepts in chemistry. This table helps in determining the oxidation state of elements, exceptions to the rule are also indicated, and an algorithm for performing tasks of this type is given
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RULES FOR DETERMINING THE DEGREE OF OXIDATION. |
Rule #1 | Rule № 2 | Rule № 3 | Rule № 4 | Rule № 5 | Rule № 6 | Rule № 7 | Rule № 8 |
Isolated atoms of chemical elements have an oxidation state of 0. | Simple substances have an oxidation state of 0. | Hydrogen has Oxidation state | Oxygen has an oxidation state of -2. | Fluorine in compounds has an oxidation state of -1. | Alkali metals (chief subgroup I group) have an oxidation state, +1 | Alkaline earth metals (chief subgroup II group, Ca-Ra) and Mg have an oxidation state+2. | Aluminum has an oxidation state of +3 in compounds. |
Examples. | Examples. | Examples. | Examples. | Examples. | Examples. | Examples. | Examples. |
H2O | Na2S | CaF2 | Al2O3 |
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H3N | Cr2O3 | CaF2 | K2O | Al(OH)3 |
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H2Se | SeO2 | SiF 4 | LiOH | Ba(OH)2 | Al 2 S 3 |
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Cl2 | H3AsO4 | Rb2O | ClF 3 | NaOH | |||
Ca(OH) 2 | RbOH | ||||||
NaH2PO4 | HPO 3 | ||||||
Be(OH) 2 =H 2 BeO 2 | Al(OH) 3 =H 3 AlO 3 | ||||||
CH 4 | Li2SO3 | ||||||
Ca(HSO 4 ) 2 | |||||||
Exceptions. | Except nia. | Exceptions. | Exceptions. | Exceptions. | Exceptions. | Exceptions. | Exceptions. |
Metal hydrides: | OF 2- oxygen fluoride | ||||||
1 -1 MeH (KH) | H 2 O 2 - hydrogen peroxide | ||||||
2 -1 MeH2(BaH2) | 1 -1 Me 2 O 2 (Na 2 O 2 ) - alkali metal peroxides | ||||||
3 -1 MeH3 (AlH3) | 1 -1 MeO 2 (CaO 2, BaO 2 ) - alkaline earth metal peroxides | ||||||
conclusions : The highest positive oxidation state of most elements is numerically equal to the group number of the table of elements in which it is found. The lowest negative oxidation state of a non-metal element is determined by the number of electrons that are missing to fill the valence layer
We find which of the two elements in the compound is more electronegative. | We determine the numerical value of the oxidation state for the more electronegative element. (See rules) | Determine the total number of negative charges in the compound. | Find the oxidation number of the less electronegative element. |
We place a minus sign (-) above the symbol of the more electronegative element. | To do this, divide the total number of positive charges by the index of a given element. |
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Place a plus sign (+) above the symbol of the less electronegative element. | To do this, we multiply the oxidation state of the more electronegative element by its index. | We remember that the algebraic sum of the oxidation states of the chemical elements in the compound must be equal to =0. |
Consolidation: determine the oxidation states of elements in the given formulas of binary compounds. SiF 4, P 2 O 5, As 2 O 5, CaH 2, Li 3 N, OsF 8, SiCl 4, H 3 P, SCl 4, PCL 3, H 4 C, H 3 As, SF 6, AlN, CuO , Fe
To place correctly oxidation states, you need to keep four rules in mind.
1) In a simple substance, the oxidation state of any element is 0. Examples: Na 0, H 0 2, P 0 4.
2) You should remember the elements that are characteristic constant oxidation states. All of them are listed in the table.
3) The highest oxidation state of an element, as a rule, coincides with the number of the group in which the element is located (for example, phosphorus is in group V, the highest s.d. of phosphorus is +5). Important exceptions: F, O.
4) The search for oxidation states of other elements is based on a simple rule:
In a neutral molecule, the sum of the oxidation states of all elements is zero, and in an ion - the charge of the ion.
A few simple examples for determining oxidation states
Example 1. It is necessary to find the oxidation states of elements in ammonia (NH 3).
Solution. We already know (see 2) that Art. OK. hydrogen is +1. It remains to find this characteristic for nitrogen. Let x be the desired oxidation state. We create the simplest equation: x + 3 (+1) = 0. The solution is obvious: x = -3. Answer: N -3 H 3 +1.
Example 2. Indicate the oxidation states of all atoms in the H 2 SO 4 molecule.
Solution. The oxidation states of hydrogen and oxygen are already known: H(+1) and O(-2). We create an equation to determine the oxidation state of sulfur: 2 (+1) + x + 4 (-2) = 0. Solving this equation, we find: x = +6. Answer: H +1 2 S +6 O -2 4.
Example 3. Calculate the oxidation states of all elements in the Al(NO 3) 3 molecule.
Solution. The algorithm remains unchanged. The composition of the “molecule” of aluminum nitrate includes one Al atom (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. The corresponding equation is: 1 (+3) + 3x + 9 (-2) = 0. Answer: Al +3 (N +5 O -2 3) 3.
Example 4. Determine the oxidation states of all atoms in the (AsO 4) 3- ion.
Solution. In this case, the sum of oxidation states will no longer be equal to zero, but to the charge of the ion, i.e., -3. Equation: x + 4 (-2) = -3. Answer: As(+5), O(-2).
What to do if the oxidation states of two elements are unknown
Is it possible to determine the oxidation states of several elements at once using a similar equation? If we consider this problem from a mathematical point of view, the answer will be negative. A linear equation with two variables cannot have a unique solution. But we are solving more than just an equation!
Example 5. Determine the oxidation states of all elements in (NH 4) 2 SO 4.
Solution. The oxidation states of hydrogen and oxygen are known, but sulfur and nitrogen are not. A classic example of a problem with two unknowns! We will consider ammonium sulfate not as a single “molecule”, but as a combination of two ions: NH 4 + and SO 4 2-. The charges of ions are known to us; each of them contains only one atom with an unknown oxidation state. Using the experience gained in solving previous problems, we can easily find the oxidation states of nitrogen and sulfur. Answer: (N -3 H 4 +1) 2 S +6 O 4 -2.
Conclusion: if a molecule contains several atoms with unknown oxidation states, try to “split” the molecule into several parts.
How to arrange oxidation states in organic compounds
Example 6. Indicate the oxidation states of all elements in CH 3 CH 2 OH.
Solution. Finding oxidation states in organic compounds has its own specifics. In particular, it is necessary to separately find the oxidation states for each carbon atom. You can reason as follows. Consider, for example, the carbon atom in the methyl group. This C atom is connected to 3 hydrogen atoms and a neighboring carbon atom. Along the C-H bond, the electron density shifts towards the carbon atom (since the electronegativity of C exceeds the EO of hydrogen). If this displacement were complete, the carbon atom would acquire a charge of -3.
The C atom in the -CH 2 OH group is bonded to two hydrogen atoms (a shift in electron density towards C), one oxygen atom (a shift in electron density towards O) and one carbon atom (it can be assumed that the shift in electron density in this case not happening). The oxidation state of carbon is -2 +1 +0 = -1.
Answer: C -3 H +1 3 C -1 H +1 2 O -2 H +1.
Do not confuse the concepts of “valency” and “oxidation state”!
Oxidation number is often confused with valence. Don't make this mistake. I will list the main differences:
- the oxidation state has a sign (+ or -), the valence does not;
- the oxidation state can be zero even in a complex substance; valence equal to zero means, as a rule, that an atom of a given element is not connected to other atoms (we will not discuss any kind of inclusion compounds and other “exotics” here);
- oxidation state is a formal concept that acquires real meaning only in compounds with ionic bonds; the concept of “valence,” on the contrary, is most conveniently applied in relation to covalent compounds.
The oxidation state (more precisely, its modulus) is often numerically equal to the valency, but even more often these values do NOT coincide. For example, the oxidation state of carbon in CO 2 is +4; the valence of C is also equal to IV. But in methanol (CH 3 OH), the valency of carbon remains the same, and the oxidation state of C is equal to -1.
A short test on the topic "Oxidation state"
Take a few minutes to check your understanding of this topic. You need to answer five simple questions. Good luck!
Modern formulation of the Periodic Law, discovered by D. I. Mendeleev in 1869:
The properties of elements are periodically dependent on the ordinal number.
The periodically repeating nature of changes in the composition of the electronic shell of atoms of elements explains the periodic change in the properties of elements when moving through the periods and groups of the Periodic System.
Let us trace, for example, the change in higher and lower oxidation states of elements of groups IA – VIIA in the second – fourth periods according to Table. 3.
Positive All elements exhibit oxidation states except fluorine. Their values increase with increasing nuclear charge and coincide with the number of electrons at the last energy level (with the exception of oxygen). These oxidation states are called highest oxidation states. For example, the highest oxidation state of phosphorus P is +V.
Negative oxidation states are exhibited by elements starting with carbon C, silicon Si and germanium Ge. Their values are equal to the number of electrons missing up to eight. These oxidation states are called inferior oxidation states. For example, the phosphorus atom P at the last energy level is missing three electrons to eight, which means that the lowest oxidation state of phosphorus P is – III.
The values of higher and lower oxidation states are repeated periodically, coinciding in groups; for example, in the IVA group, carbon C, silicon Si and germanium Ge have the highest oxidation state +IV, and the lowest oxidation state – IV.
This periodicity of changes in oxidation states is reflected in the periodic changes in the composition and properties of chemical compounds of elements.
A periodic change in the electronegativity of elements in the 1st-6th periods of groups IA–VIA can be similarly traced (Table 4).
In each period of the Periodic Table, the electronegativity of elements increases with increasing atomic number (from left to right).
In each group In the periodic table, electronegativity decreases as the atomic number increases (from top to bottom). Fluorine F has the highest, and cesium Cs has the lowest electronegativity among the elements of the 1st-6th periods.
Typical nonmetals have high electronegativity, while typical metals have low electronegativity.
Examples of tasks for parts A, B1. In the 4th period the number of elements is equal to
2. Metallic properties of elements of the 3rd period from Na to Cl
1) get stronger
2) weaken
3) do not change
4) I don’t know
3. Nonmetallic properties of halogens with increasing atomic number
1) increase
2) decrease
3) remain unchanged
4) I don’t know
4. In the series of elements Zn – Hg – Co – Cd, one element not included in the group is
5. The metallic properties of elements increase in a number of ways
1) In – Ga – Al
2) K – Rb – Sr
3) Ge – Ga – Tl
4) Li – Be – Mg
6. Non-metallic properties in the series of elements Al – Si – C – N
1) increase
2) decrease
3) do not change
4) I don’t know
7. In the series of elements O – S – Se – Those sizes (radii) of an atom
1) decrease
2) increase
3) do not change
4) I don’t know
8. In the series of elements P – Si – Al – Mg, the dimensions (radii) of an atom are
1) decrease
2) increase
3) do not change
4) I don’t know
9. For phosphorus the element with less electronegativity is
10. A molecule in which the electron density is shifted towards the phosphorus atom is
11. Higher The oxidation state of elements is manifested in a set of oxides and fluorides
1) ClO 2, PCl 5, SeCl 4, SO 3
2) PCl, Al 2 O 3, KCl, CO
3) SeO 3, BCl 3, N 2 O 5, CaCl 2
4) AsCl 5, SeO 2, SCl 2, Cl 2 O 7
12. Lowest oxidation state of elements - in their hydrogen compounds and set fluorides
1) ClF 3, NH 3, NaH, OF 2
2) H 3 S + , NH +, SiH 4 , H 2 Se
3) CH 4, BF 4, H 3 O +, PF 3
4) PH 3, NF+, HF 2, CF 4
13. Valency for a multivalent atom is the same in a series of compounds
1) SiH 4 – AsH 3 – CF 4
2) PH 3 – BF 3 – ClF 3
3) AsF 3 – SiCl 4 – IF 7
4) H 2 O – BClg – NF 3
14. Indicate the correspondence between the formula of a substance or ion and the oxidation state of carbon in it
The oxidation state +2 in all compounds exhibits
Answer:4
Explanation:
Of all the proposed options, only zinc exhibits the +2 oxidation state in complex compounds, being an element of the secondary subgroup of the second group, where the maximum oxidation state is equal to the group number.
Tin is an element of the main subgroup of group IV, a metal, exhibiting oxidation states 0 (in a simple substance), +2, +4 (group number).
Phosphorus is an element of the main subgroup of the main group, being a non-metal, exhibiting oxidation states from -3 (group number - 8) to +5 (group number).
Iron is a metal, the element is located in a secondary subgroup of the main group. Iron is characterized by oxidation states: 0, +2, +3, +6.
The compound of KEO 4 composition forms each of two elements:
1) phosphorus and chlorine
2) fluorine and manganese
3) chlorine and manganese
Answer: 3
Explanation:
The salt of the composition KEO 4 contains an acid residue EO 4 -, where oxygen has an oxidation state of -2, therefore, the oxidation state of the element E in this acid residue is +7. Of the proposed options, chlorine and manganese are suitable - elements of the main and secondary subgroups of group VII, respectively.
Fluorine is also an element of the main subgroup of group VII, however, being the most electronegative element, it does not exhibit positive oxidation states (0 and -1).
Boron, silicon and phosphorus are elements of the main subgroups of groups 3, 4 and 5, respectively, therefore in salts they exhibit the corresponding maximum oxidation states of +3, +4, +5.
Answer: 4
Explanation:
The same highest oxidation state in the compounds, equal to the group number (+5), is exhibited by P and As. These elements are located in the main subgroup of group V.
Zn and Cr are elements of secondary subgroups of groups II and VI, respectively. In compounds, zinc exhibits the highest oxidation state of +2, chromium - +6.
Fe and Mn are elements of the secondary subgroups of groups VIII and VII, respectively. The highest oxidation state for iron is +6, for manganese - +7.
The compounds exhibit the same highest oxidation state
Answer: 4
Explanation:
P and N exhibit the same highest oxidation state in compounds, equal to the group number (+5). These elements are located in the main subgroup of group V.
Hg and Cr are elements of secondary subgroups of groups II and VI, respectively. In compounds, mercury exhibits the highest oxidation state of +2, chromium - +6.
Si and Al are elements of the main subgroups of groups IV and III, respectively. Consequently, for silicon the maximum oxidation state in complex compounds is +4 (the number of the group where silicon is located), for aluminum - +3 (the number of the group where aluminum is located).
F and Mn are elements of the main and secondary subgroups of group VII, respectively. However, fluorine, being the most electronegative element of the Periodic Table of Chemical Elements, does not exhibit positive oxidation states: in complex compounds its oxidation state is −1 (group number −8). The highest oxidation state of manganese is +7.
Nitrogen exhibits oxidation state +3 in each of two substances:
1) HNO 2 and NH 3
2) NH 4 Cl and N 2 O 3
Answer: 3
Explanation:
In nitrous acid HNO 2, the oxidation state of oxygen in the acid residue is -2, that of hydrogen is +1, therefore, in order for the molecule to remain electrically neutral, the oxidation state of nitrogen is +3. In ammonia NH 3, nitrogen is a more electronegative element, so it attracts an electron pair of a covalent polar bond and has a negative oxidation state of -3, the oxidation state of hydrogen in ammonia is +1.
Ammonium chloride NH 4 Cl is an ammonium salt, therefore the oxidation state of nitrogen is the same as in ammonia, i.e. is equal to -3. In oxides, the oxidation state of oxygen is always -2, so for nitrogen it is +3.
In sodium nitrite NaNO 2 (a salt of nitrous acid), the oxidation degree of nitrogen is the same as in nitrogen in nitrous acid, because is +3. In nitrogen fluoride, the oxidation state of nitrogen is +3, since fluorine is the most electronegative element of the Periodic Table and in complex compounds exhibits a negative oxidation state of -1. This answer option satisfies the conditions of the task.
In nitric acid, nitrogen has the highest oxidation state equal to the group number (+5). Nitrogen as a simple compound (since it consists of atoms of one chemical element) has an oxidation state of 0.
The highest oxide of a group VI element corresponds to the formula
Answer: 4
Explanation:
The highest oxide of an element is the oxide of the element with its highest oxidation state. In a group, the highest oxidation state of an element is equal to the group number, therefore, in group VI, the maximum oxidation state of an element is +6. In oxides, oxygen exhibits an oxidation state of -2. The numbers below the element symbol are called indices and indicate the number of atoms of that element in the molecule.
The first option is incorrect, because. the element has an oxidation state of 0-(-2)⋅6/4 = +3.
In the second version, the element has an oxidation state of 0-(-2) ⋅ 4 = +8.
In the third option, the oxidation state of the element E: 0-(-2) ⋅ 2 = +4.
In the fourth option, the oxidation state of the element E: 0-(-2) ⋅ 3 = +6, i.e. this is the answer you are looking for.
The oxidation state of chromium in ammonium dichromate (NH 4) 2 Cr 2 O 7 is equal to
Answer: 1
Explanation:
In ammonium dichromate (NH 4) 2 Cr 2 O 7 in the ammonium cation NH 4 +, nitrogen, as a more electronegative element, has a lower oxidation state of -3, hydrogen is positively charged +1. Therefore, the entire cation has a charge of +1, but since there are 2 of these cations, the total charge is +2.
In order for the molecule to remain electrically neutral, the acidic residue Cr 2 O 7 2− must have a charge of -2. Oxygen in acidic residues of acids and salts always has a charge of -2, so the 7 oxygen atoms that make up the ammonium bichromate molecule are charged -14. There are 2 chromium atoms in molecules, therefore, if the charge of chromium is designated as x, then we have:
2x + 7 ⋅ (-2) = -2, where x = +6. The charge of chromium in the ammonium dichromate molecule is +6.
The +5 oxidation state is possible for each of two elements:
1) oxygen and phosphorus
2) carbon and bromine
3) chlorine and phosphorus
Answer: 3
Explanation:
In the first proposed answer, only phosphorus, as an element of the main subgroup of group V, can exhibit an oxidation state of +5, which is its maximum. Oxygen (an element of the main subgroup of group VI), being an element with high electronegativity, exhibits an oxidation state of -2 in oxides, as a simple substance - 0 and in combination with fluorine OF 2 - +1. The oxidation state +5 is not typical for it.
Carbon and bromine are elements of the main subgroups of groups IV and VII, respectively. Carbon has a maximum oxidation state of +4 (equal to the group number), and bromine exhibits oxidation states of -1, 0 (in the simple compound Br 2), +1, +3, +5 and +7.
Chlorine and phosphorus are elements of the main subgroups of groups VII and V, respectively. Phosphorus exhibits a maximum oxidation state of +5 (equal to the group number); chlorine, similar to bromine, has oxidation states of -1, 0 (in a simple compound Cl 2), +1, +3, +5, +7.
Sulfur and silicon are elements of the main subgroups of groups VI and IV, respectively. Sulfur exhibits a wide range of oxidation states from -2 (group number − 8) to +6 (group number). For silicon, the maximum oxidation state is +4 (group number).
Answer: 1
Explanation:
In sodium nitrate NaNO 3, sodium has an oxidation state of +1 (group I element), there are 3 oxygen atoms in the acidic residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of: 0 − (+ 1) − (−2)·3 = +5.
In sodium nitrite NaNO 2, the sodium atom also has an oxidation state of +1 (an element of group I), there are 2 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of: 0 − (+1) − (−2) 2 = +3.
NH 4 Cl – ammonium chloride. In chlorides, chlorine atoms have an oxidation state of −1, hydrogen atoms, of which there are 4 in the molecule, are positively charged, therefore, in order for the molecule to remain electrically neutral, the oxidation state of nitrogen is: 0 − (−1) − 4 · (+1) = −3. In ammonia and ammonium salt cations, nitrogen has a minimum oxidation state of −3 (the number of the group in which the element is located is 8).
In the molecule of nitric oxide NO, oxygen exhibits a minimum oxidation state of −2, as in all oxides, therefore, the oxidation state of nitrogen is +2.
0EB205
Nitrogen exhibits its highest oxidation state in a compound whose formula is
Answer: 1
Explanation:
Nitrogen is an element of the main subgroup of group V, therefore, it can exhibit a maximum oxidation state equal to the group number, i.e. +5.
One structural unit of iron nitrate Fe(NO 3) 3 consists of one Fe 3+ ion and three nitrate ions. In nitrate ions, nitrogen atoms, regardless of the type of counterion, have an oxidation state of +5.
In sodium nitrite NaNO2, sodium has an oxidation state of +1 (an element of the main subgroup of group I), there are 2 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of 0 − ( +1) − (−2)⋅2 = +3.
(NH 4) 2 SO 4 – ammonium sulfate. In sulfuric acid salts, the SO 4 2− anion has a charge of 2−, therefore, each ammonium cation has a charge of 1+. Hydrogen has a charge of +1, so nitrogen has a charge of −3 (nitrogen is more electronegative, so it attracts the common electron pair of the N–H bond). In ammonia and ammonium salt cations, nitrogen has a minimum oxidation state of −3 (the number of the group in which the element is located is 8).
In the molecule of nitrogen oxide NO2, oxygen exhibits a minimum oxidation state of −2, as in all oxides, therefore, the oxidation state of nitrogen is +4.
28910E
In compounds of the composition Fe(NO 3) 3 and CF 4, the oxidation states of nitrogen and carbon are equal, respectively
Answer: 4
Explanation:
One structural unit of iron (III) nitrate Fe(NO 3) 3 consists of one iron ion Fe 3+ and three nitrate ions NO 3 −. In nitrate ions, nitrogen always has an oxidation state of +5.
In carbon fluoride CF 4, fluorine is a more electronegative element and attracts the common electron pair of the C-F bond, exhibiting an oxidation state of -1. Therefore, carbon C has an oxidation state of +4.
A32B0B
Chlorine exhibits an oxidation state of +7 in each of two compounds:
1) Ca(OCl) 2 and Cl 2 O 7
2) KClO 3 and ClO 2
3) BaCl 2 and HClO 4
Answer: 4
Explanation:
In the first variant, chlorine atoms have oxidation states +1 and +7, respectively. One structural unit of calcium hypochlorite Ca(OCl) 2 consists of one calcium ion Ca 2+ (Ca is an element of the main subgroup of group II) and two hypochlorite ions OCl −, each of which has a charge of 1−. In complex compounds, except OF 2 and various peroxides, oxygen always has an oxidation state of −2, so it is obvious that chlorine has a charge of +1. In chlorine oxide Cl 2 O 7, as in all oxides, oxygen has an oxidation state of −2, therefore, the chlorine in this compound has an oxidation state of +7.
In potassium chlorate KClO 3, the potassium atom has an oxidation state of +1, and oxygen - -2. In order for the molecule to remain electrically neutral, chlorine must exhibit an oxidation state of +5. In chlorine oxide ClO 2, oxygen, as in any other oxide, has an oxidation state of −2; therefore, for chlorine its oxidation state is +4.
In the third option, the barium cation in the complex compound is charged +2, therefore, a negative charge of −1 is concentrated on each chlorine anion in the BaCl 2 salt. In perchloric acid HClO 4 the total charge of 4 oxygen atoms is −2⋅4 = −8, the charge on the hydrogen cation is +1. For the molecule to remain electrically neutral, the charge of chlorine must be +7.
In the fourth variant, in the magnesium perchlorate molecule Mg(ClO 4) 2 the charge of magnesium is +2 (in all complex compounds, magnesium exhibits an oxidation state of +2), therefore, for each ClO 4 − anion there is a charge of 1−. In total, 4 oxygen ions, each exhibiting an oxidation state of −2, are charged −8. Therefore, for the total charge of the anion to be 1−, the chlorine must have a charge of +7. In chlorine oxide Cl 2 O 7, as explained above, the charge of chlorine is +7.