What is the element whose 4+ cation has the following set of quantum numbers, n=3, l=2, ml=0, and m.s=+1/2 for its highest energy electron?
I know, based on the quantum numbers, that its referring to a 3d sublevel.
How can I find the solution?
Thanks
Is it Ge?
To find the answer, we will need to understand the electronic configuration of elements and how it relates to quantum numbers. The set of quantum numbers provided describes the highest energy electron of the 4+ cation of the element.
Firstly, let's identify the electronic configuration of the 4+ cation. Since it is a 4+ cation, it means that the atom has lost four electrons. Therefore, we need to determine the electronic configuration of the neutral atom.
The quantum numbers given are n=3, l=2, ml=0, and m.s=+1/2. We know that n represents the principle quantum number, l represents the azimuthal quantum number, ml represents the magnetic quantum number, and m.s represents the spin quantum number.
For n=3, it means the electron is in the third energy level (n=3).
For l=2, it means the electron is in the d-sublevel. The d-sublevel has a total of five orbitals, labeled as dxy, dyz, dxz, dx^2-y^2, and dz^2.
For ml=0, it means the electron is in the dx^2-y^2 orbital. The dx^2-y^2 orbital has its lobes aligned along the x and y axes.
For m.s=+1/2, it means the electron has a positive spin (+1/2). This represents the clockwise direction of electron spin.
Now that we know the electron is in the dx^2-y^2 orbital, we can determine the element in question. The d-sublevel is a part of the transition metals, which are located in the middle of the periodic table.
Among the transition metals, the d-sublevel starts filling from scandium (Sc) at element number 21. Going from scandium (Sc) to the element we are looking for, there are a total of 17 elements with the 4+ cation in the d-sublevel.
Therefore, the element with the given set of quantum numbers for its highest energy electron in the 4+ cation is the 17th element after scandium (Sc), which is
Vanadium (V).