Question

#1

1. 
   In the Sr2 + cation, in its ground state, how many electrons have ml = 0 and s = -1/2?

2. 
   In the anion As3-, in its ground state, how many electrons have l = 1 and ml = 0?

3. 
   In the Co4 + cation, in its ground state, how many electrons have ml = 0?

4. 
   In the Zn2 + cation, in its ground state, how many electrons have ml = 0 and s = +1/2?

Answer 1

1) The given cation is ₃₈Sr²⁺ so, there is a total (38-2) = 36 electrons. Now, the electronic configuration of Sr in its ground state is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p^{6​​​​​} 5s² so the ground state electronic configuration of Sr²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p^{6​​​​​} so, now for a l value ml can have -l to 0 to +l values. So, for 1s, l = 0, ml = 0, and 1 electron with -½ then for 2s, l = 0, ml = 0 and 1 electron with -½ then for 2p, l = 1, and for ml = 0, 1 electron would be -½ then 3s, l = 0, ml = 0 and 1 electron with -½ then for 3p, l = 0, ml = 0 and 1 electronic would be -½ and for 3d, l = 2, for ml = 1 electron would be there with -½ then for 4s, l = 0, ml = 0 and 1 electronic would be -½ and for 4p, l = 1, ml = 0 and 1 electronic would be -½ so total from each orbital one electronic is possible for which ml = 0 and s = -½ so total number of that kind of electrons is (1+1+1+1+1+1+1+1) = 8 such kind of electrons are there.

2) The given anion is ₃₃As^3- so here total (33+3) = 36 electrons are present in the anion. The electronic configuration of As in its ground state is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³ so the ground state electronic configuration of As^3- is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ so for 1s, l ≠ 1, for 2s, l ≠ 1, for 2p, l = 1 and ml = 0 possible for 2 electrons, for 3s, l ≠ 1, for 3p, l = 1 and ml = 0 possible for 2 electrons, for 4s, l ≠ 1, for 3d, l ≠ 1, and for 4p, l = 1 and ml = 0 possible for 2 electrons. So, total a number of (2+2+2) = 6 electrons would possible here for which l = 1 and ml = 0.

3) The given cation is ₂₇Co⁴⁺ so here total (27-4) = 23 electrons,are present in the given cation. Now, the ground state electronic configuration of Co is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷ and the ground state electronic configuration of Co⁴⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵so now for each orbital ml = 0 possible like for 1s, ml can be 0 for 2 electrons, for 2s, ml can be 0 for 2 electrons, for 2p, ml can be 0 for 2 electrons, for 3s, ml can be 0 for 2 electrons, for 3p ml can be 0 for 2 electrons, and for 3d ml can be 0 for 1 electron. So, total a number of (2+2+2+2+2+1) = 11 electrons are there for which ml = 0.

4) The given cation is ₃₀Zn²⁺ so here total (30-2) = 28 electrons are present. The ground state electronic configuration of Zn is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ and the ground state electronic configuration of Zn²⁺ ion is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ so, here from each orbital at least one electron possible for which ml = 0 and s = +½ is possible. So, here a total (1+1+1+1+1+1) = 6 electrons have the condition that ml = 0, s = +½.