The structure of chemical substances, the composition of chemical substances

There is no definite radius for the movement of electrons around the nucleus, and there is a chance that electrons will appear anywhere in space. The probability of going beyond a certain range is almost zero. So the nearest is considered zero.

So it has a specific radius, and a specific radius should be a certain range. The above explanation is my own opinion, and there is no textbook basis for it. If it works.

General knowledge of science and technology - [common test chemical substances] 1Oxygen: a relatively reactive gas, with oxidizing and flammability, it is a commonly used oxidizing agent.

It burns in oxygen to produce water. 3.Diamond (c) is found in nature.

1.Every carbon atom in benzene is sp2 hybridized, and there are 4 electrons in the outermost shell of benzene, and now only 3 are used, so there is still 1 left in the p orbital, and the 6 carbon atoms are not left with 6 electrons, forming a delocalized bond. The p orbitals are perpendicular to each other.

2。Bond length: The equilibrium distance between the nuclei of two bonding atoms A and B The larger the atomic radius, the longer the bond is, of course.

3.The van der Waals force includes dispersion force, orientation force, induction force, generally speaking, between polar molecules and polar molecules, orientation force, inducing force, and dispersion force all exist; Between polar molecules and non-polar molecules, there are induction and dispersion forces; Between non-polar and non-polar molecules, only dispersion forces exist. The proportional magnitude of these three types of forces is determined by the polarity and deformability of the interacting molecules.

The greater the polarity, the more important the role of the orientation force; The greater the deformability, the more important the dispersion force is; Induction is related to both factors. But for most molecules, the dispersion force is the main one. The magnitude of the intermolecular forces can be reflected in the interaction energy.

The greater the relative mass, the more work you have to do to overcome it, that is, the greater the relative mass, the stronger the van der Waals force.

4.The essence of hydrogen bonding is the electrostatic attraction between the hydrogen nucleus on a strong polar bond (a-h) and the electronegative atom b with lone electron pairs and partial negative charge. So the greater the electronegativity, the smaller the radius and the stronger the hydrogen bond.

Because the smaller the radius, the easier it is to approach the atom.

Categories: Education, Science, >> Learning Aid.

Problem description: When organic matter containing only carbon, hydrogen, and oxygen is completely burned, the volume ratio of oxygen consumed to carbon dioxide produced is 1:2. then in this type of organic matter ( ).

A The compound with the smallest relative molecular mass has the molecular formula CH4O

The molecular formula of the compound with the smallest relative molecular mass is CH2O

c The difference in the relative molecular mass of each compound containing the same number of carbon atoms is a multiple of 18.

Analysis: Looking at the beginning and end of combustion, it can be found that in organic matter.

C is eventually converted to CO2 and H is eventually converted to H2O. So 1mol c consumes 1mol O2 for complete combustion, 1mol h consumes O2 for complete combustion, and 1mol o consumes O2 for complete combustion.

Let the carbon, hydrogen and oxygen contents of the organic matter be x, y, zmol. Then the oxygen consumed is x+ mol, and the CO2 produced is xmol

The ratio of the two is 1:2, and the final result can be 2x+y-2z=0 =>y=2z-2x (so a, b are wrong).

The formula of this organic matter is 12x+y+16z=12+(2z-2x)+16z=10x+18z

So C is correct.

For chemistry, the structure and properties of substances should be simpler than organic, so I send some answer templates and fixed routines.

Take the question as an example. (1) Silicon is homogeneous with carbon, and there are also series of hydrides, but silanes are far less than alkanes in terms of types and quantities, because the carbon-carbon single bond and carbon-hydrogen bond are strong, and the alkane formed is stable, while the bond energy of silicon-silicon single bond and silicon-hydrogen bond in silane is low and easy to break, resulting in the formation of long-chain silane.

Note: When encountering such a and considering the strength of the valence bond, it is related to the bond length and bond angle (if you are afraid of answering wrong, you can only write the key) The bond energy is large and stable; The bond energy is small and unstable.

2) NF is not easy to form a coordination bond with Cu2+, but it can form [Cu(NH3)4]2+ (tetraammonia copper ion), because the nitrogen atom of electronegative F>N>H, NH is -3 valence, while in NF nitrogen is +3 valence, which is highly electron-deficient and is not easy to provide electrons to copper ions to form a coordination bond.

Note: Compared with the size of electronegativity like this, it is still relatively small.

3) Manganese and calcium belong to the same period, but the melting and boiling point of manganese metal is higher than that of calcium, because the atomic radius of manganese is small and there are many valence electrons, and the metal bond is strong.

Note: If you test metal, you generally test metal bonds, metal bonds and atomic radius are related to the number of valence electrons, the atomic radius is small, the number of valence electrons is more, and the metal bond is stronger.

4) Flame color reaction: The electrons outside the nucleus of the nucleus are arranged in a certain orbital order, and the farther the orbital is from the nucleus, the higher the energy. When burning, the electron gains energy and jumps from the inner orbital to another orbital, and the electron that jumps to the new orbital is in an unstable state, and it randomly jumps back to the original orbital and releases energy to the outside world.

Light energy)5) The angle between the h-n-h bonds in [Cu(nH3)4]2+ is greater than that between the h-n-h in the ammonia molecule, because during the formation of [Cu(nH3)4]2+, the lone electrons of the nitrogen atom in ammonia form coordination bonds with copper ions, which are converted into bond-forming electron pairs, and the repulsion to other bond-forming electron pairs is reduced, so the bond angle increases.

6) The thermal decomposition temperature of CaC is less than that of SRCo3 (strontium carbonate), because the atomic radius of calcium ions is smaller than that of strontium ions, and the lattice energy of Cao is greater than that of SRO lattice energy, so calcium carbonate is easier to decompose.

7) The reason why NICL has a higher melting point than NaCl is that the more charge and the smaller the radius of the ion, the greater the lattice energy and the higher the melting point.

Note: For ionic crystals, answer the question from the perspective of lattice energy.

To sum up, ask the question about the melting and boiling point of a molecular crystal from the perspective of van der Waals force and hydrogen bonding, and ask about the bond energy of other molecular crystals. Metal crystals are answered from metal bonds, and ionic crystals are answered from ionic bonds, and lattice can be answered.

In the SO2 molecule, S

Sp2 hybridization is adopted, and the molecule is V-shaped, or angular shaped.

Using the theory of valence shell electron pair repulsion, it is easy to judge, s

is the central atom, there are 6 electrons, o is a coordination atom, no electrons are don't be donated, then s is 6 2 3 pairs of electrons around s, 3 pairs of electrons in space, to have the least repulsive force, they are arranged in a triangle, and the corresponding hybridization mode is sp2

A molecule bound by a non-polar bond is not necessarily a non-polar molecule This statement is true.

To consider the extreme case, that is.

O3, the middle o takes sp2

Hybridization, with the surrounding 2 O atoms, has a slight difference in the ability to attract electrons, so although the covalent bonds in the molecule are both non-polar bonds, the whole molecule has less polarity.

The way of intramolecular bonding is more complex and is generally not required in middle school.

1. About hybridization... The focus is on the valence electron pairs.

It is calculated based on whether valence electrons are provided or not.

When doing intermediate atoms, all the valence electrons are generally donated.

However, when you make a coordination atom, you have to look at the minimum number of electrons.

s is the middle atom, with 6 pips involved. However, oxygen is not involved, so if the valence electron pair is 3, it is a plane triangle, but the number of atoms of SO2 is not enough to form a plane triangle, which is a V-shape.

In other words. SO3 is a planar triangle, and SO2 has lone pairs, so it is a polar molecule.

SO3 has no lone pairs. He is a non-polar molecule.

2. If all of them are built by Sigma, there should be no problem.

But the number of atoms increases, and bonds appear.

The problem is not simple... For example, O3. I won't explain the rest.

Ag+, Cu+, Au+, and the coordination number is usually 2

Cu2+, Zn2+, Hg2+, CD2+, Al3+, Ni (elemental), etc., the coordination number is usually 4

Fe3+, Fe2+, CO2+, CO3+, Ni2+, Pt4+, etc., the coordination number is usually 6

As can be seen from the above, there is no strict relationship between the number of coordination numbers and the valence state of the ions.

In addition, there is a difference between the number of coordinates and the number of ligands, if it is a single-tooth ligand (that is, a ligand has only one coordination atom and only provides a pair of electron pairs), the number of coordinates and the number of ligands are equal, but if it is a multi-tooth ligand, the number of coordinates and the number of ligands are not equal. For example, ethylenediamine is a two-tooth ligand, and when it coordinates with Cu2+, it is a copper ion that binds to two ethylenediamines, and the number of ligands is 2, but the coordination number of Cu2+ is still 4.

There is no direct relationship between the number of ligands and the valence state of the central atom.

The number of ligands in a central atomic energy band is mainly related to the number of empty orbitals it can provide. If the number of empty orbitals of the central atom is large, the number of ligands carried by the formed complex is also more.

For example, [co(H2O)6]2+ and [coCl4]2- are the same central atom, both are +2 valence, and the number of ligands is different.

As for the common complexes, you can go and make a separate note, there are not many in high school, but there is no pattern at all.

Common complexes: [Cu(NH3)4 ]2+ ,Cu(H2O)4]2+ ,AG(NH3)2]+ Fe(CN)6]3-(FeF6)3- That's all I have in mind for the time being, let's write it here first.

Hope it helps

If you have any doubts, please feel free to ask.

If you remember that the culamina ion is four ligands, you remember that the divalent ion is four ligands.

Let me answer your question first, there is a kind of b20 in the elemental substance of boron, which has an icosahedral structure, which is an atomic crystal, and its characteristic is that the crystal structure is a spatial network structure, which is the basic concept of atomic crystals in the book, diamond, crystalline silicon are also similar. I think we should choose a compound with electron deficiency in both A and Al formation, which is related to the type of chemical bond formation in B20. This kind of topic is mainly the knowledge points in the book, option A is the similarity of the properties of the elements of the same family, and option C is the general structural characteristics of atomic crystals.

It is more difficult to talk about organic, there is actually no difficulty in the college entrance examination organic, the most important thing is to keep in mind the relevant reaction conditions in the book, and if you have the ability, it is best to clarify the reaction mechanism. There is generally a breakthrough point in organic inference problems, such as a process like [o] [o], which can determine that there is a structure of -CH2OH and primary alcohol in the original structure, because only such a structure can be oxidized in two steps.

The organic framework is relatively large, and it is difficult to explain clearly at once, you can go and buy a reference book like the new edition of "Dragon Gate", which has a detailed combing.

As for the last two questions, the selection and calculation still need to be practiced, in fact, there are many questions that are repeated. The main ones are the conservation method, the limit method, etc., and it is difficult to involve the problem of equivalent equilibrium. In the case of multiple-choice questions, I think it is also a Shanghai candidate, it is best to stabilize at about 55 points, and if you have the ability, it is best to rush 60, because there are many concept questions that can be scored, and many questions can also be done by elimination, but there will be controversial topics in the college entrance examination, and even some questions will be out of the program.

I can't go into detail here, but if you don't understand, you can just ask me and I'll give you a detailed explanation......

c The crystal structure of boron element is a spatial network structure.

The crystal structure of 1,c boron element is a spatial network structure.

This kind of question should consider what kind of crystal it is!

2. Organic chemistry in high school is very easy, as long as you understand what reaction occurs in what tube-energy group, the bond breaking position of the substance, the reaction conditions, etc., you can solve it by associating it with the simple organic matter in the textbook and analogy with the past!

A is a good sentence, but the similarity of the properties of the elements leads to the similarity of chemical properties, while the melting and boiling point is a physical property, and the donkey's lips are not against the horse's mouth.

b This sentence is to remind you of the knowledge of the high melting and boiling point of carbon, the neighbors are similar to this is not bad, d to see the question clearly, the question is the reason, there are several isotopes and what is the relationship between the melting and boiling point, and the later questions will be a lot, you may not listen to it. I'll just tell you a few points: for the question just now, first of all, don't be afraid because you have given some new information, and see clearly the essence of the problem, which is what to ask.

This calculation problem is also applicable, until you master the logical and chemical calculations after the problem, of course, remember to summarize the various calculation methods, because this will make you faster than others. It also takes a lot of practice. The so-called skills of the experimental questions are all small paths, and the key is to do more, remember more, and think more.

It's easy to memorize, but everyone has a little trick, and there is not much actual high school content.

bc can hi me if there is any problem.