which of this has principal quantum number of 3 and an angular momentum quantum number of a) 3s b)3d c) 4f d) 3f
This question is not complete.
3s
To determine which option has a principal quantum number (n) of 3 and an angular momentum quantum number (l), we need to understand the rules for assigning these values.
The principal quantum number (n) represents the energy level or shell of an electron. It can have any positive whole number value starting from 1. The angular momentum quantum number (l) represents the orbital shape and can have values ranging from 0 to (n-1).
Given the options:
a) 3s
b) 3d
c) 4f
d) 3f
Let's go through each option:
a) 3s:
In the case of the 3s orbital, the principal quantum number (n) is 3, which matches the requirement. However, the angular momentum quantum number (l) for the s orbital is always 0. Therefore, the correct answer for this option would be "no."
b) 3d:
For the 3d orbital, the principal quantum number (n) is 3, which matches the requirement. The angular momentum quantum number (l) for the d orbital can have values ranging from 0 to 2, corresponding to the d orbitals being labeled as dxy, dxz, and dyz. Therefore, the correct answer for this option would be "yes."
c) 4f:
In the case of the 4f orbital, the principal quantum number (n) is 4, which does not match the requirement of being equal to 3. Therefore, the correct answer for this option would be "no."
d) 3f:
Similar to the previous option, the principal quantum number (n) for the 3f orbital is 3, which matches the requirement. However, the angular momentum quantum number (l) for the f orbital can have values ranging from 0 to 3, corresponding to the f orbitals being labeled as fx(x^2-y^2), fz(3z^2-r^2), fxz, and fyz. Therefore, the correct answer for this option would be "yes."
In summary, the correct options that have a principal quantum number (n) of 3 and an angular momentum quantum number (l) are b) 3d and d) 3f.