Solution of the schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behavior of the electron. each function is characterized by 3 quantum numbers: n, l, and ml. if the value of n = 1 the quantum number l can have values from to . the total number of orbitals possible at the n = 1 energy level is . if the value of l = 1 the quantum number ml can have values from to . the total number of orbitals possible at the l = 1 sublevel is .

If n=1, l can only have the value l=0 and the total number of orbitals is 0. If l=1, ml can be -1,0 or 1 and the total number of orbitals at l=1 is 3.

Explanation:

When solving the radial part of Schrödinger equation, one needs to expand in power series that lead to Laguerre polynomials of the form , being n and l the quantum numbers. As is known, in the Laguerre polynomials the subindex must be greater than or equal to 0, which implies . So if n=1, the only possible value of l is l=0.

Likewise, when solving the angular part, one gets the spherical harmonics can be . Therefore the sublevel l=| has 3 possible orbitals.

the answer is c. nacl