What element has the following set of four quantum numbers for its last auf bau electron?
please explain and answer
- 9 years agoFavorite Answer
4, 2, -1, 1/2 means it's outermost electron is 4d2
the electron configuration is [36Kr] 5s2 4d2
the atomic number is 52 (Z = 52)
therefore, the element is Zr (zirconium)
hope this helps!
- mcarthurLv 44 years ago
chlorine's atomic variety is 17. so the final electron is in n=3 (in n=a million we've 2 electrons and in n=2 we've 8 electrons. so there is 10 electron in this 2 layer) in n=3 we've s,p and d and 7 electron that's accessible. 2 of them pass in orbital s. we are in a position to have 5 others. for p, l=1and the third orbital has ml=+a million