To determine the quantum numbers of the valence electron of vanadium (V), we need to consider its electron configuration.
The electron configuration of vanadium is [Ar] 3d³ 4s². This configuration indicates that vanadium has two electrons in the 4s orbital and three electrons in the 3d orbital.
The valence electron of an atom refers to the outermost electron involved in chemical bonding. In the case of vanadium, the valence electron would be in the 4s orbital since it has a higher principal quantum number (n) than the 3d orbital.
The quantum numbers of an electron include the principal quantum number (n), azimuthal quantum number (ℓ), magnetic quantum number (mℓ), and spin quantum number (ms).
For the valence electron of vanadium, the quantum numbers are:
- Principal quantum number (n): 4
- Azimuthal quantum number (ℓ): 0 (since the 4s orbital has ℓ = 0)
- Magnetic quantum number (mℓ): 0 (since there is only one 4s orbital with mℓ = 0)
- Spin quantum number (ms): The spin of an electron can be either +1/2 or -1/2.
Therefore, the quantum numbers of the valence electron of vanadium are: n = 4, ℓ = 0, mℓ = 0, and ms = ±1/2.