Answer :
To determine the correct electron configuration for nobelium (No), we should consider its position in the periodic table and its atomic number.
Nobelium has an atomic number of 102. This means it has 102 electrons. Here is the step-by-step method to find the electron configuration for nobelium:
1. Identify the previous noble gas: The noble gas preceding nobelium in the periodic table is radon (Rn). Radon has an atomic number of 86.
2. Determine the remaining electrons: Subtract the atomic number of radon from that of nobelium:
[tex]\[ 102 - 86 = 16 \][/tex]
So, we need to find how these 16 electrons are distributed.
3. Fill the following orbitals according to the Aufbau principle:
- After [Rn], the next orbitals to be filled are in this order: 7s, 5f, and 6d if necessary.
4. Placement of electrons in the orbitals:
- Fill the 7s orbital:
[tex]\[ 7s^2 \][/tex]
This orbital will take 2 of the 16 electrons.
- Fill the 5f orbital:
[tex]\[ 5f^{14} \][/tex]
This orbital will take the remaining 14 electrons.
Therefore, the complete electron configuration for nobelium (No) is:
[tex]\[ [Rn]7s^2 5f^{14} \][/tex]
This matches the first option in the provided multiple-choice list. Thus, the correct answer is:
[tex]\[ [Rn]7s^2 5f^{14} \][/tex]
Nobelium has an atomic number of 102. This means it has 102 electrons. Here is the step-by-step method to find the electron configuration for nobelium:
1. Identify the previous noble gas: The noble gas preceding nobelium in the periodic table is radon (Rn). Radon has an atomic number of 86.
2. Determine the remaining electrons: Subtract the atomic number of radon from that of nobelium:
[tex]\[ 102 - 86 = 16 \][/tex]
So, we need to find how these 16 electrons are distributed.
3. Fill the following orbitals according to the Aufbau principle:
- After [Rn], the next orbitals to be filled are in this order: 7s, 5f, and 6d if necessary.
4. Placement of electrons in the orbitals:
- Fill the 7s orbital:
[tex]\[ 7s^2 \][/tex]
This orbital will take 2 of the 16 electrons.
- Fill the 5f orbital:
[tex]\[ 5f^{14} \][/tex]
This orbital will take the remaining 14 electrons.
Therefore, the complete electron configuration for nobelium (No) is:
[tex]\[ [Rn]7s^2 5f^{14} \][/tex]
This matches the first option in the provided multiple-choice list. Thus, the correct answer is:
[tex]\[ [Rn]7s^2 5f^{14} \][/tex]
