which of the following sets of quantum numbers (n,l,ml,ms) describes the electron which is most easily removed when a sodium aton is converted to a sodium ion?
a)2,1,0,-1/2
b)2,0,0,-1/2
c)3,1,1,+1/2
d)3,2,1,-1/2
e)3,0,0,+1/2
can anyone explain this so I understand it?
I can.
Na is atomic number 11 meaning it has 11 electrons. How are they arranged.
1s2 2s2 2p6 3s1. So the outside electron is a 3s electron.
N = 3 is the third shell.
l can have values of 0, 1, 2, 3, etc up to a maximum of n-1 (in this case it can have 0 since there is only 1 electron with which to contend.)
ml can have values of -l to +l in increments of 1 unit and zero is included. When l is 0, ml can be ONLY 0 (-0 to +0 doesn't mans much sense).
ms is EITHER +1/2 or -1/2. With only one electron we usually give it the +1/2 for the first one and -1/2 for the second one.
So the values are
n=3
l=0
ml=0
ms=+1/2
The answer can be ONLY e.
To determine which set of quantum numbers describes the electron that is most easily removed when a sodium atom is converted to a sodium ion, we need to understand the concept of ionization energy.
Ionization energy is the amount of energy required to remove an electron from an atom or ion. The electron that is most easily removed has the highest ionization energy.
In general, the ionization energy increases as we move from left to right across a period on the periodic table, and decreases as we move from top to bottom within a group.
Now let's analyze the given sets of quantum numbers:
a) 2, 1, 0, -1/2
b) 2, 0, 0, - 1/2
c) 3, 1, 1, +1/2
d) 3, 2, 1, -1/2
e) 3, 0, 0, +1/2
The most important quantum number for determining ionization energy is the principal quantum number (n). As the value of n increases, the ionization energy generally decreases. This is because electrons that are further from the nucleus experience weaker attraction and are therefore easier to remove.
Therefore, we can eliminate options c) and d) since they have higher n values.
Now, we need to consider the azimuthal quantum number (l). For a given value of n, the ionization energy is lowest for the electron with the highest value of l. This is because electrons with higher values of l are located further from the nucleus and experience weaker attraction.
Considering the remaining options a), b), and e), we can see that option e) has the highest value of l (l=0), compared to options a) and b) (l=1 for both). Therefore, option e) is the answer.
So, the set of quantum numbers that describes the electron that is most easily removed when a sodium atom is converted to a sodium ion is e) 3, 0, 0, +1/2.
To determine which set of quantum numbers describes the electron that is most easily removed when a sodium atom is converted to a sodium ion, we need to understand the concept of electron configuration and the relationship between the quantum numbers.
1. Quantum Numbers:
- Principal quantum number (n): Determines the energy level or shell of an electron. It can take on positive integer values (1, 2, 3, ...).
- Azimuthal quantum number (l): Determines the shape of the electron's orbital. It can take on values ranging from 0 to (n-1).
- Magnetic quantum number (ml): Specifies the orientation of the orbital in space. It can take on values ranging from -l to +l.
- Spin quantum number (ms): Describes the spin state of the electron. It can be either +1/2 (spin-up) or -1/2 (spin-down).
2. Ionization:
When an atom is converted into an ion, one or more electrons are removed from its outermost energy level (valence shell). The electron that is most easily removed is the one with the highest energy, meaning it is located in the outermost shell and has the greatest value for the principal quantum number (n).
3. Analysis of the Given Options:
a) Set: 2, 1, 0, -1/2
b) Set: 2, 0, 0, -1/2
c) Set: 3, 1, 1, +1/2
d) Set: 3, 2, 1, -1/2
e) Set: 3, 0, 0, +1/2
The principal quantum number (n) is the same for options c, d, and e. Therefore, we need to focus on the values of the azimuthal quantum number (l) within these options.
Option c has the highest value for l (l=1), followed by option d (l=2), and then option e (l=0). According to the Aufbau principle, electrons fill orbitals from lowest to highest energy levels. The higher the azimuthal quantum number, the higher the energy level of the electron within the same principal quantum number (n). Hence, option e has the lowest energy level among these three options.
Thus, option e is describing the electron with the highest energy level and is most easily removed when a sodium atom is converted to a sodium ion.
Therefore, the correct answer is e) 3, 0, 0, +1/2.