A divalent ion M ²⁺ belongs to the 3 d transition series and has an atomic number Z < 25 . The spin-only magnetic moment of this ion is 3.87 B.M. What is the total number of electrons with azimuthal quantum number l = 0 in the ground state of the neutral atom M ?

Step-by-step solution

The spin-only magnetic moment is given by = n(n+2) B.M. Given = 3.87 B.M. , the number of unpaired electrons n = 3 . For a divalent ion M ²⁺ in the 3 d series with Z The ground state electronic configuration of the neutral Vanadium atom ( V ) is 1 s ^2 2 s ^2 2 p ^6 3 s ^2 3 p ^6 3 d ^3 4 s ^2 . The azimuthal quantum number l = 0 corresponds to s -orbitals. The electrons present in s -orbitals are in the 1 s , 2 s , 3 s , and 4 s subshells. Total number of electrons with l = 0 is 2 + 2 + 2 + 2 = 8 . If the 4 s elec