Answer Happy

Question 7 (a) The electron configuration of an atom with the following four quantum numbers; n = 3:1 = 2; m₁ = 0; m,= -1/ for the last electron is (b) And the atom with this configuration is. O(a) 1s² 2s² (b) Be (a) 1s² 2s2 2p 3s² 3p6 4s¹ 3d10 (b) Cu o(a) 1s² 2s22p² (b) C o(a) 1s² 2s2 2p 3s 3p 4s² 3d8 (b) Ni O You selected the incorrect answer. The principal quantum number, n, equals 3. The azimuthal or secondary quantum number, I, is 2; therefore, the electrons are filling the 3d atomic o The magnetic quantum number, m,, for the d atomic orbitals take on values of -2, -1, 0, +1, and +2 There are five degenerate atomic orbitals which can be represented by the following with eight electrom Degenerate "d" orbitals-1.jpg| The first five electrons that enter the atomic orbital have spin quantum numbers of +1/2.