In: Chemistry
1)Answer the following questions.
What is the effective nuclear charge for strontium. Calculation: Z* = =
What is the effective nuclear charge for chlorine. Calculation: Z* = =
What is the effective nuclear charge for iodine. Calculation: Z* = =
Does strontium or iodine have the larger atomic radius?
Does chlorine or iodine have the smaller ionization energy?
2)Why do the elements in groups 6 and 11 not have two electrons in their outermost s sublevel as we would expect?
3)Why did the pickle glow yellow when we sent electricity through it?
The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. The term "effective" is used because the shielding effect of negatively charged electrons prevents higher orbital electrons from experiencing the full nuclear charge.
The effective nuclear charge on an electron is given by the following equation:
Zeff = Z - S
where Z is the number of protons in the nucleus (atomic number), and S is the number of electrons between the nucleus and the electron in question (the number of nonvalence electrons).
(a) For Strontium
Strontium atoms have 38 electrons and the shell structure is 2.8.18.8.2.
For outermost shell, n=5
Zeff = Z - S
= 38 - 36
= +2
For n=4
Zeff = Z - S
= 38 - 28
= +10
For n=3
Zeff = Z - S
= 38 - 10
= +28
For n=2
Zeff = Z - S
= 38 - 2
= +36
For n=1
Zeff = Z - S
= 38 - 0
= +38
(b) For Chlorine
Chlorine atoms have 17 electrons and the shell structure is 2.8.7
For outermost shell, n=3
Zeff = Z - S
= 17 - 10
= +7
For n=2
Zeff = Z - S
= 17 - 2
= +15
For n=1
Zeff = Z - S
= 17 - 0
= +17
(c) For Iodine
Iodine atoms have 53 electrons and the shell structure is 2.8.18.18.7.
For outermost shell, n=5
Zeff = Z - S
= 53 - 46
= +7
For n=4
Zeff = Z - S
= 53 - 28
= +25
For n=3
Zeff = Z - S
= 53 - 10
= +43
For n=2
Zeff = Z - S
= 53 - 2
= +51
For n=1
Zeff = Z - S
= 53 - 0
= +53
(d) Strontium and Iodine are in same period.
so as we move left to right, Atomic size decreases.
So strontium is bigger