what is the angular momentum quantum number for the orbital from which an Mg atom loses two electrons to form Mg2+ ion.
12Mg = 1s2 2s2 2p6 3s2
It loses the two 3s electrons. The angular momentum quantum number is l = 0 for the s orbital and that is in units of h/2*pi
To determine the angular momentum quantum number for the orbital from which an Mg atom loses two electrons to form the Mg2+ ion, we need to understand the electron configuration of an Mg atom and the process of ionization.
The electron configuration of an Mg atom is 1s² 2s² 2p⁶ 3s².
When two electrons are removed from the Mg atom to form the Mg2+ ion, the last two electrons that are lost are from the 3s orbital.
The angular momentum quantum number (denoted by the letter ℓ) describes the shape of an orbital and can have values ranging from 0 to n - 1, where n is the principal quantum number.
For the 3s orbital, the principal quantum number (n) is 3. Thus, the possible values for the angular momentum quantum number (ℓ) in the 3s orbital are 0, indicating an s-orbital.
Therefore, the angular momentum quantum number for the orbital from which an Mg atom loses two electrons to form the Mg2+ ion is 0.