Large key electronic number algorithm

The algorithm of big password electronic number is as follows:

Calculation method of electron number of large π bond: the electron number of large π bond is equal to the sum of all bonded P orbital electrons. Aniline, the six p orbitals of C, each contains one electron, n 1 p orbitals, and two electrons (N atom has five electrons, one bonds with C, two bonds with two H respectively, and the remaining pair is in P specification), so one of π bonds of * * * yoke has 6 * 1+60.

1. Calculate the number of electrons of atoms with lone pair in the reactant.

2. Calculate the number of electrons of atoms with empty orbits in reactants.

3. Determine the distance between the lone pair and the empty orbit in the reactant.

4. Calculate the bond energy and bond length of the pie key.

The formation of large π bonds (called * * * yoke π bonds, or delocalized π bonds) is characterized by that many molecules are SP2-hybridized (SP2 degenerate orbitals form the σ skeleton of molecules, and the remaining P orbitals form * * * yoke π bonds).

To judge the number of large π bonds, we must first judge the hybridization type of each atom. If all atoms in the same region are SP2-hybridized (such as benzene and naphthalene, the atoms in the molecule are basically SP2-hybridized, so it is the same * * * yoke system), then a large π bond will be formed, and if there is an interruption in the middle, it will be counted separately (such as C6H5-CH2-C6H5, and the methylene in the middle is SP3-hybridized, which is a * * yoke system).

In addition, we also need to understand the formation and characteristics of large π bonds. For example, a large π bond is usually shared by two electrons in the P orbit between two atoms, and its forming ability is stronger than that of a single σ bond, which can affect the chemical properties and reactivity of some substances.

Formation conditions of pie bonds

1, the atoms are all on the same chemical plane, which is the prerequisite for the formation of pie bond;

2. Atoms have mutually parallel atomic orbits, which is a necessary condition for forming pie bonds;

3. The total number of electrons remaining in atomic orbitals is less than twice the number of atomic orbitals, which is the final condition for forming a big pie bond.