Question

Какие орбитали участвуют в образовании молекул Cl2, H2, O2, и N2. Напишите структурные формулы этих молекул. Сколько электронных пар участвует в образовании связей между атомами хлорв,водорода,кмслорода,азота? Укажите типы связей между атомами в этих молекулах.

Answer 1

Orbitals involved in the formation of Cl2, H2, O2, and N2 molecules

The molecules Cl2, H2, O2, and N2 are formed through the combination of atoms and the sharing of electrons. The orbitals involved in the formation of these molecules are as follows:

1. Cl2 (Chlorine): - Chlorine is a halogen with atomic number 17. It has 7 valence electrons in its outermost shell. - In the formation of Cl2, each chlorine atom contributes one unpaired electron from its 3p orbital. - These two unpaired electrons from each chlorine atom overlap to form a sigma bond (σ bond) between the two atoms. - The structural formula of Cl2 is: Cl-Cl.

2. H2 (Hydrogen): - Hydrogen is the simplest element with atomic number 1. It has 1 valence electron in its outermost shell. - In the formation of H2, each hydrogen atom contributes one unpaired electron from its 1s orbital. - These two unpaired electrons from each hydrogen atom overlap to form a sigma bond (σ bond) between the two atoms. - The structural formula of H2 is: H-H.

3. O2 (Oxygen): - Oxygen is a non-metal with atomic number 8. It has 6 valence electrons in its outermost shell. - In the formation of O2, each oxygen atom contributes two unpaired electrons from its 2p orbitals. - These four unpaired electrons from each oxygen atom overlap to form two sigma bonds (σ bonds) between the two atoms. - The structural formula of O2 is: O=O.

4. N2 (Nitrogen): - Nitrogen is a non-metal with atomic number 7. It has 5 valence electrons in its outermost shell. - In the formation of N2, each nitrogen atom contributes three unpaired electrons from its 2p orbitals. - These six unpaired electrons from each nitrogen atom overlap to form three sigma bonds (σ bonds) between the two atoms. - The structural formula of N2 is: N≡N.

Number of electron pairs involved in bond formation and types of bonds

The number of electron pairs involved in the formation of bonds between the atoms of chlorine, hydrogen, oxygen, and nitrogen can be determined based on their valence electrons and the octet rule:

1. Cl2 (Chlorine): - Each chlorine atom contributes one electron, resulting in a total of two electrons involved in bond formation. - The type of bond formed between the chlorine atoms in Cl2 is a single covalent bond.

2. H2 (Hydrogen): - Each hydrogen atom contributes one electron, resulting in a total of two electrons involved in bond formation. - The type of bond formed between the hydrogen atoms in H2 is a single covalent bond.

3. O2 (Oxygen): - Each oxygen atom contributes two electrons, resulting in a total of four electrons involved in bond formation. - The type of bond formed between the oxygen atoms in O2 is a double covalent bond.

4. N2 (Nitrogen): - Each nitrogen atom contributes three electrons, resulting in a total of six electrons involved in bond formation. - The type of bond formed between the nitrogen atoms in N2 is a triple covalent bond.

In summary, the orbitals involved in the formation of Cl2, H2, O2, and N2 molecules are the 3p orbital for chlorine, the 1s orbital for hydrogen, and the 2p orbital for oxygen and nitrogen. The number of electron pairs involved in bond formation varies for each molecule, resulting in different types of bonds (single, double, or triple covalent bonds).