How does chemistry determine a chemical bond? 》How to determine the type of chemical bond

Chemical bonds are divided into ionic and covalent bonds.

Ionic bonds: Metals or NH4+ form ionic bonds with non-metals.

Covalent bonds: ionic bonds formed between non-metals, if they are non-polar covalent bonds formed by the same atoms. Polar covalent bonds are formed between different atoms.

Chemical bonds: Strong interactions between adjacent atoms.

Classification: ionic, covalent, and metallic bonds.

Ionic bonding: the electrostatic action that combines anions and cations into compounds, such as NaCl, Na2O, Na2O2, NaOH, Na2SO4, etc.

Covalent bonds: interactions between atoms formed by sharing electrons, in molecules that are not metallic elements (except for noble gases): such as O2 F2 H2 C60

Non-metal-formed compounds such as SO2 CO2 CH4 H2O2 CS2

In some ionic compounds, such as SO42- in Na2SO4, there is a covalent bond in OH- in NaOH, and there is a covalent bond in NH4CL in NH4C.

Covalent bonds are classified into non-polar covalent bonds and polar covalent bonds according to the nature of bonding, that is, polar and non-polar.

If you still don't understand, I have a copy of this information and can send it to you.

Symmetrical is a non-polar bond, asymmetrical is a polar bond, and what can be ionized in water is an ionic bond.

Question 1: How to determine what chemical bonds are contained in a compound If there are active metal elements such as NA, K, CA, etc. or ammonium ions in the compound, there must be ionic bonds.

If there is an atomic cluster in the compound, since the atomic cluster is mainly formed by non-metallic elements, there must be a covalent bond inside the atomic cluster, so there must be a covalent bond in the compound.

If all the compounds are non-metallic elements (except for ammonium salts), then there must be covalent bonds in the compound, and there are no ionic bonds.

Special case: Alcl3 is formed by covalent bonds.

All chemical bonds are formed as a result of the simultaneous attraction of electrons to two or more nuclei of atoms. There are 4 limit types of chemical bonds: ionic bonds, covalent bonds, metallic bonds, and coordination bonds.

Ionic bonds] The interaction between oppositely charged ions is called ionic bonding, and the essence of bonding is that the electrostatic interaction between anion and cation ions is the electrostatic interaction between anion and ion.

Covalent bonds] are the attraction of two or more atoms by sharing electron pairs, and a typical covalent bond is formed by two atoms by attracting a pair of bonding electrons.

Metallic bonds] are interactions that bind metal atoms together and can be seen as highly delocalized covalent bonds.

Coordination bond] Coordination bond, is a type of chemical bond, two or more atoms use their outer electrons together, ideally to reach a state of electron saturation, thus forming a relatively stable chemical structure.

Question 2: How to distinguish the types of chemical bonds Types of chemical bonds: ionic bonds, covalent bonds (including coordination bonds), and metallic bonds.

Chemical bonds: When atoms combine into molecules, there is a strong interaction between adjacent atoms. (Chemical bonding begins with emphasizing the forces between adjacent atoms within the molecule.)

Van der Waals force or hydrogen bonds generally do not belong to the category of chemical bonds, and according to the type of bonding, chemical bonds can be divided into ionic bonds, covalent bonds (including coordination bonds), and metallic bonds. In ionic compounds, covalent compounds, or elemental compounds, there are chemical bonds between atoms and ions

The combined bond residue containing active metal elements must contain ionic bonds (AlCl3 is a covalent bond), and there is an ionic bond between the ammonium ion and the acid ion in the ammonium salt. Are the chemical bonds formed by non-metals and non-metals covalent bonds, or are the above two special cases; In a covalent bond, a non-polar bond is formed between the same elements, and a polar bond is formed between different elements. ）

Problem 3: How to accurately determine the type of chemical bond There must be ionic bonds in ionic compounds, and there may be covalent bonds.

You need to be familiar with the number of outermost electrons of the element.

1. For example, H O, H has one electron in the outermost shell, and O has 6 electrons in the outermost shell, and after the two are combined to form the H O molecule, the only electron in H is bonded with an electron in the outermost shell of O. So there are no electrons in the outer shell of H, and there are still 4 electrons in the outer shell of O, which are two pairs of lone pairs of electrons.

2. BeCl, the outermost two electrons of Be, the outermost 7 electrons of Cl, after bonding, the electrons of the outermost wax layer of Be are bonded, and each Cl has 6 electrons (3 pairs of electrons), that is, the whole molecule has 6 pairs of lone pairs of electrons.

No lone electron pairs are paired valence electrons that are bound or shared with other atoms. It is present in the outermost electronic shell of an atom. The presence and distribution of lone pairs of electrons in the molecule affect the shape of the molecule, especially on the molecule composed of light atoms.

Refers to unbonded pairs of valence electrons in a molecule.

1.A single key.

Bonds can be formed between p orbitals and are called p-p bonds.

Also available in. The orbitals are formed between them, called p-d bonds

This can be judged from the type of orbital in which electrons outside the nucleus are involved in bonding.

2.The delocalized key (i.e., the major key).

Expressed as (n,m

====This is only required by the competition, and you can not watch it if you do not participate in the competition training.

This needs to judge that the molecule is planar, the number of atoms participating in the bond == that is, that n, and then calculate the number of all the outermost electrons in the molecule, count the number of bonding electrons and the number of unshared electrons, and the rest is the number of electrons involved in the formation of the large bond ===, that is, that m(m<2n).

As in SO2, the molecule is planar and the S atom takes SP2

Hybridization. The molecule is.

V-shaped, there are 3 atoms involved in the formation of large bonds.

The outermost number of electrons is the total number of valence electrons.

6*3=18.

The common electron pairs formed (i.e., sigma bonds.

There are two pairs, and there are unshared electron pairs around を.

4 pairs. There is a pair of unshared electron pairs around s, so that there are still electrons left.

Number = 18-7 * 2 = 4.

So there is one in SO2.

3, 4).

1.A single key.

Bonds can be formed between p orbitals and are called p-p bonds.

Also available in. The orbitals are formed between them, called p-d bonds

This can be judged from the type of orbital in which electrons outside the nucleus are involved in bonding.

2.The delocalized key (i.e., the major key).

Expressed as (n,m

====This is only required by the competition, and you can not watch it if you do not participate in the competition training.

This needs to judge that the molecule is planar, the number of atoms participating in the bond == that is, that n, and then calculate the number of all the outermost electrons in the molecule, count the number of bonding electrons and the number of unshared electrons, and the rest is the number of electrons involved in the formation of the large bond ===, that is, that m(m<2n).

As in SO2, the molecule is planar and the S atom takes SP2

Hybridization. The molecule is.

V-shaped, there are 3 atoms involved in the formation of large bonds.

The outermost number of electrons is the total number of valence electrons.

6*3=18.

The common electron pairs formed (i.e., sigma bonds.

There are two pairs, and there are unshared electron pairs around を.

4 pairs. There is a pair of unshared electron pairs around s, so that there are still electrons left.

Number = 18-7 * 2 = 4.

So there is one in SO2.

3, 4).

In addition to the bonds of diatomic compounds, there may be large bonds in polyatomic and matter, such as one bond and two bonds in Co, and one bond and two large P-P bonds in CO2.

It's easy to judge this in high school.

You only need to look at the types of elements that make up the chemical bonds to make up the chemical bonds. Of course, this is not the essence here, so there are many anti-shirt state examples, which will be mentioned later.

Non-metallic + non-metallic.

Covalent bond. For example.

Both of HCl's bonding elements are non-metals, so they are covalent bonds.

Metal + non-metal.

Ionic bond. For example, NACL

A piece of metal. A non-metal, so it's an ionic bond.

You also mentioned that when the ammonium ion is taken as a whole, it is considered as a metal cation.

For example. nh3

NH4+ is a combination of NOT + NEL, so both are covalent bonds and NH4Cl is regretted

, there is an ionic bond between the ammonium ion and the chloride ion. In one is that aluminum chloride is a covalent bond.

As for the accurate judgment, you need to take the book Structure of Matter (high school) to learn, using the difference in electronegativity to judge, the covalent bond that is less than.

greater than the ionic bonds. If you don't learn this textbook, don't master this Sun Zheng.

Your last question is about the writing of the electronic formula, if the anion and cation in the electronic formula are more complex, and the outer layer needs to be enclosed in middle brackets if there are electrons in the outer shell. The more complicated here is actually a standard, as long as there are electrons in the outer shell, you have to add, na+, and there are no electrons in the outermost shell, so you don't need chloride ions.

Bromine ions. The outermost layer is 8, so parentheses should be added, and the ammonium root should naturally be added.

H O, the central atom is O, O belongs to the VIA group, the outermost number of electrons is 6, minus the number of electrons required for two H (since the ground state H atom has only 1s energy level, one more electron can be stable) 6-2 1=4, the number of lone pairs of electrons of O is 4 2=2.

Coupled to form two bonds with two Hs, the number of valence shell electron pairs of O is 2+2=4. The central atom O is sp3 hybridized, the VSEPR configuration is tetrahedral, and the collision row configuration ignores the lone pairs and is V-shaped.

The number of valence shell electron pairs of the central atom (note that the number of valence electrons is not a smile) = the number of lone electrons of the central atom + the number of electron pairs of the central atom and the surrounding atoms.

1. Find the central atom first, generally the valency value is larger, the bound atom is more, and the electronegativity is less (except for H).

2. Look at which main group the central atom is in, and the group number is the number of outermost electrons.

3. Then look at the number of electrons required by the atom that binds to the central atom (8-its group number).

4. The number of outermost electrons of the central atom minus the total number of electrons required by the surrounding atoms, and then divided by 2 is the number of lone electron pairs of the central atom.

5. Count how many atoms are combined with the central atom around the number, and the number of electron pairs between the central atom and the surrounding atoms is the number.

The method of judging the key and the bond is: spatial configuration, energy difference.

1. Spatial configuration.

The spatial configuration of the keys and keys is different, so their existence can be judged by the spatial configuration. Since the bond is axial, the atoms on both sides of the chemical bond can rotate around the bond axis without changing the presence of the bond. Since the bond is on a flat surface, the atoms on both sides of the chemical bond cannot rotate around the bond axis, otherwise the bond will break.

For example, a carbon-carbon single bond in a benzene molecule is a bond, while a carbon-carbon double bond is a bond. The planar structure of the benzene molecule is such that the carbon atom on the double bond cannot rotate around the double bond axis, so the carbon-carbon double bond in the benzene molecule is a bond.

2. Energy differences.

The energy difference between the bonds and the bonds can also be used to judge their existence. The bonding energy of the bond is higher than that of the bond because the electron cloud density of the bond is higher and the mutual attraction is stronger. Therefore, when a chemical bond is broken, the bond requires a higher amount of energy than the bond.

For example, when the chemical bond of a hydrogen molecule is broken, the energy required is 945kj mol when the bond is broken in the hydrogen molecule.

The key and the key are the same

Both bonds and bonds have the characteristic of sharing electron pairs. In bonding, the electron orbitals of two atoms overlap into a region to form a shared electron pair. This shared electron pair is formed by a closed positive direct overlap between two atoms, so the bond is also known as a direct bond.

In the bond mill, the electron orbitals of the two atoms are parallel to each other to form a shared electron pair.

Both bonds and keys have the characteristics of bond energy. Bond energy refers to the amount of energy that needs to be expended in forming a chemical bond. In bonds, the electron clouds between atoms overlap to form a stable shared electron pair, so the bond energy of the bond is higher.

In bonds, the overlap of the electron orbitals is weaker than the bonds, so the bond energy of the bonds is lower. However, even in the bond, the overlapping of the electron orbitals leads to the loss of bond energy.

Delocalized key 2009-01-23 09:08:08|Classification:

Structure |Tags: |Font size: Large, Medium and Small subscription1. Definition of Delocalized Bonds (B-level Emphasis on Mastery) Delocalized bonds are defined by p-orbitals of multiple atoms (three or more)."Shoulder to shoulder"The way of overlapping to form a covalent bond.

Three-dimensional pattern of delocalized bonds in a benzene molecule Enlarged Figure 2: Formation conditions (B-level key mastery) To form a delocalized bond, the following conditions must be met: (1) the atoms involved in the formation of the major bond must be in the same plane (i.e., only the hybridization of sp, sp2 and dsp2 is possible to form a large bond); (2) each atom has a parallel p-orbital (they must all be p-orbitals that are not hybridized); (3) The total number of p electrons is less than twice the number of p orbitals.

Usually the central atom is a second-period element, such as b, c, n, o(s is also acceptable), and the coordination atom is also a second-period element, such as n, o, (f), and large bonds are easily formed between them. Food for thought: Why is it impossible for a single nucleated molecule with sp3, sp3d, sp3d2 hybrid types to exist in a large bond?

B-level mastery) 3: Representation method (B-level key mastery) It is usually represented by the AB symbol, where A represents the number of atoms involved in the formation of the large bond, and B represents the number of electrons in the large bond. Such as 34, 46, etc.

4. Methods for judging the existence of major bonds in molecules (1) First, the valence electron repulsion theory is used to determine the hybrid type of molecules. (2) If the hybrid type is either SP, SP2 or DSP2, there may be a major bond (3) Subtract the total number of valence electrons of the central atom or ion (subtract the charge number if it is a cation, add the charge number for anions) by the number of electrons that have formed a bond (including bond formation and lone pairs), if this value is 0, there is no major bond; If it is not 0, it means that there is a large bond, and this value is the number of electrons provided by the central atom to form the large bond.

4) The number of electrons provided by the central atom + the number of electrons provided by each coordination atom The number of coordination atoms = the total number of electrons in the major bond.