Which of the following has the largest effective nuclear charge experienced by the valence electrons? *
A.O
B.C
C.Sn
D.Rb
Is this B?
Are you supposed to do this as an estimate or are you to use Slater's Rule.
Zeff = Z - S where Z is the nuclear charge (the number of protons) and S is the shielding of the valence electrons from the nucleus. For the estimated number S is simply the number of inner electrons not in the valence shell and that's pretty easy to do. For a more accurate numbef for Zeff you can look up Slater's rule. There is a set protocol for doing this AND there is a formula for calculating based upon numbers supplied by Slater. For example an s electron use 0.5 but for a p electrons it's a different value. At any rate follow what you've been taught in your class. If you have trouble find info on Slater's Rule let me know and I'll look too.
I'm thinking that probably is the simpler approach so I'll show those below.
Zeffective = Z - S where Z is the nuclear charge;i.e., the number of protons and S is the shielding; i.e., all of the electrons EXCEPT the outside one(s). So for 8O16, that is 1s2 2s2 2p4
Zeff = 8 - 2 = 6
6C is 1s2 2s2 2p2 so Zeff = 6 - 2 = 4
50Sn is [Kr] 4d10 5s2 5p2 Zeff = 50-(36+10) = 4
37Rb is [Kr] 5s1 Zeff = 37-36 = 1
To determine which of the given elements has the largest effective nuclear charge experienced by the valence electrons, we need to consider the atomic number and electron configuration of each element.
The effective nuclear charge is the net positive charge experienced by an electron in an atom. It is affected by the number of protons in the nucleus and the shielding effect of inner electrons.
Let's analyze each element:
A. O (oxygen) has an atomic number of 8. Its electron configuration is 1s² 2s² 2p⁴. Oxygen has six valence electrons.
B. C (carbon) has an atomic number of 6. Its electron configuration is 1s² 2s² 2p². Carbon has four valence electrons.
C. Sn (tin) has an atomic number of 50. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p². Tin has two valence electrons.
D. Rb (rubidium) has an atomic number of 37. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s¹. Rubidium has one valence electron.
The larger the atomic number, the larger the effective nuclear charge experienced by the valence electrons. This is because the number of protons in the nucleus increases, increasing the positive charge experienced by the valence electrons.
Among the given elements, Sn (tin) has the largest atomic number (50), and therefore it experiences the largest effective nuclear charge among the valence electrons.
So, the correct answer is C. Sn.