Hello guys, any help is appreciated. Thank you so much
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).
a. How many electrons does each species have? (1 point)
b. According to the graph, which species is most stable? (1 point)
c. Use Coulomb's law to explain the energy measurements shown in the graph and how this affects the distribution of electrons. (2 points)
a. How many electrons does each species have? (1 point)
LOL
9 = 9
10 - 1 = 9
11 - 2 = 9
If it has a charge of +1, then it lost one electron and has one more positive protons than it has negative electrons
I can not see your pictures so that is all I can help with.
See my response to Bill just one or two posts below.
Next, explain in some detail what you don't understand about your problem. Remember we can't see the graph you are seeing.
Thanks! I was just confused about Na++, I didn't know that it could exist. That is very helpful!! The graph shows F having the highest orbital energy, Ne+ having the second-highest, and Na++ having the lowest
a. To determine the number of electrons for each species, we need to look at their atomic number (Z). The atomic number represents the number of protons in an atom's nucleus, which is also equal to the number of electrons in a neutral atom.
For the F atom (Z = 9): The F atom has 9 electrons.
For the Ne+ ion (Z = 10): The Ne+ ion has a charge of +1, which means it lost one electron. So, the Ne+ ion has 10 - 1 = 9 electrons.
For the Na++ ion (Z = 11): The Na++ ion has a charge of +2, which means it lost two electrons. So, the Na++ ion has 11 - 2 = 9 electrons.
Therefore, all three species (F atom, Ne+ ion, and Na++ ion) have 9 electrons each.
b. To determine which species is most stable, we need to compare their orbital energies on the graph. The lower the energy level, the more stable the species.
Looking at the graph, if the F atom has the lowest orbital energy among the three species, it would be the most stable. Similarly, if the Ne+ ion or the Na++ ion has the lowest orbital energy, it would be the most stable.
c. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In the context of atoms and ions, this law helps explain their energy measurements and electron distribution.
In the graph, as the atomic number (Z) increases from F (Z = 9) to Ne+ (Z = 10) to Na++ (Z = 11), the positive charge in the nucleus increases. According to Coulomb's law, the force between the positive nucleus and the negatively charged electrons becomes stronger as the charge in the nucleus increases.
As a result, the energy required to remove an electron from the atom or ion increases. This is because the electron experiences a stronger attraction to the positively charged nucleus, making it more difficult to remove.
The energy measurements shown in the graph reflect this. When the electron is closer to the nucleus, it experiences a stronger attraction, leading to higher energy levels. On the other hand, when the electron moves further away from the nucleus, the attraction is weaker, resulting in lower energy levels.
This energy difference affects the distribution of electrons. Electrons will fill the lower energy levels first before occupying higher energy levels. The strength of the attraction between the nucleus and electrons determines the arrangement of electrons in different orbitals, shells, and subshells within an atom or ion.
Therefore, the energy measurements on the graph and Coulomb's law explain the relationship between atomic number, orbital energies, and the distribution of electrons in the F atom, Ne+ ion, and Na++ ion.