I am having some chemistry problems.
3. The graph compares the 1s orbital energies for the F atom (Z = 9), the Ne+ ion (Z = 10), and the Na++ ion (Z = 11).
(the graph shows F having an Orbital Energy of 692.45, Ne+ with 47.74, and Na++ with 33.52)
a. How many electrons does each species have?
b. According to the graph, which species is most stable?
c. Use Coulomb's law to explain the energy measurements shown in the graph and how this affects the distribution of electrons.
To answer these questions, we need to understand the concept of orbital energies and the relationship between electron count and stability. Additionally, we will use Coulomb's law to explain the energy measurements shown in the graph and the distribution of electrons.
a. To determine the number of electrons each species has, we need to refer to their respective atomic numbers (Z).
- The F atom (Z = 9) has 9 electrons.
- The Ne+ ion (Z = 10) has one less electron than neon (which has 10), giving it a total of 9 electrons.
- The Na++ ion (Z = 11) has two fewer electrons than sodium (which has 11), resulting in 9 electrons.
b. To identify which species is most stable based on the graph, we need to look at the orbital energy values. The lower the orbital energy, the more stable the species.
- From the graph, we observe that the Na++ ion (with an orbital energy of 33.52) has the lowest energy, indicating it is the most stable species compared to both F and Ne+.
c. Coulomb's law explains the energy measurements shown in the graph and how they affect electron distribution. According to Coulomb's law, the energy of interaction between two charged particles (in this case, electrons and the nucleus) is directly proportional to the product of their charges and inversely proportional to the distance between them.
- In this scenario, the measured orbital energies represent the attraction between the electrons and the nucleus of each species.
- As the atomic number (Z) increases, the positive charge in the nucleus also increases. This greater positive charge exerts a stronger attractive force on the electrons, resulting in lower orbital energies.
- From F to Ne+ to Na++, the atomic number increases gradually. Therefore, the Na++ ion, with the highest atomic number, experiences the strongest attractive force from its nucleus, resulting in the lowest orbital energy amongst the three species.
- Lower orbital energy suggests that the electrons in the Na++ ion are more closely held by the nucleus compared to F and Ne+ ions, implying a more stable electron distribution.
In summary, the F atom, Ne+ ion, and Na++ ion all have 9 electrons. Among these species, Na++ is the most stable based on its lower orbital energy. The energy measurements in the graph are explained by Coulomb's law, which states that higher positive charge in the nucleus leads to stronger electron attraction, resulting in lower orbital energies and a more stable electron distribution.
a. To determine the number of electrons each species has, we can use their respective atomic numbers.
- F atom (Z = 9): The atomic number of fluorine is 9, which means it has 9 electrons.
- Ne+ ion (Z = 10): The atomic number of neon is 10. When it loses one electron to become a Ne+ ion, it will have 9 electrons.
- Na++ ion (Z = 11): The atomic number of sodium is 11. When it loses two electrons to become a Na++ ion, it will have 9 electrons.
Therefore, each species has 9 electrons.
b. According to the graph, the Ne+ ion (Z = 10) is the most stable species. This is because it has the lowest orbital energy value (47.74) among the three species shown on the graph.
c. Coulomb's law states that the attractive or repulsive force between two charged particles is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them.
In the case of the graph, the energy measurements shown represent the orbital energies of the electrons in each species. These values depend on the attractive or repulsive forces experienced by the electrons due to the positively charged nucleus.
As the atomic number (Z) increases from F (Z = 9) to Ne+ (Z = 10) to Na++ (Z = 11), the number of protons in the nucleus increases. This results in an increase in the positive charge of the nucleus. According to Coulomb's law, as the charge of the nucleus increases, the attractive force experienced by the electrons also increases.
The distribution of electrons is affected by the balance between the attractive force of the nucleus and the repulsive forces between the electrons themselves. When the attractive force of the nucleus is stronger than the repulsive forces among the electrons, the electrons are more tightly bound and occupy lower energy orbitals. This leads to a more stable electron configuration and lower orbital energy measurements on the graph.
In the graph, the Ne+ ion (Z = 10) has the most stable electron configuration because it has the lowest orbital energy. This is because, despite having one less electron compared to the F atom (Z = 9), the attractive force from the increased positive charge of the nucleus in the Ne+ ion outweighs the repulsive forces among the electrons.