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Sagot :
If the question is asking us to find the element based on the quantum numbers of the last electron, we have these informations:
n = 4
l = 1
m = +1
s = -1/2
n represents the shell or level value, this value will generally go from 1 to 7, and since our value is 4, we are talking about an element whose valence shell is 4, so this element is in the 4th period
l represents the value of the orbital, this value can range from 0 to 3,
0 is the s orbital
1 is the p orbital
2 the d orbital
3 = f orbital
In out question we have l = 1, therefore the last electron of this element is in the p orbital and in the 4th period, now we are down to 6 elements, Gallium, Germanium, Arsenic, Selenium, Bromine and Krypton, all 6 elements are in the 4th period and have the p orbital as its last one
m represents the location of the electron within the orbitals, each orbital can hold 2 electrons only, p orbitals can hold 6 electrons in 3 pairs of 2, and the m value represents in which orbital the electron is, for l = 1, m can be -1, 0, +1
and in our question we have m = +1, therefore the electron is located in the last orbital, it can only be the 3rd electron or the 6th electron now, and now we are down to Arsenic and Krypton
s is the spin value of the electron, this value will tell us if the electron is pointing upwards or downwards in the orbital, this value can only be +1/2 (upwards) and -1/2 (downwards), and in our question, we have -1/2, which is the last electron of the orbital
Based on all that information, we conclude that this element is Krypton, with 36 of atomic number and electron configuration of [Ar]3d10 4s2 4p6
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