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CHEM 1A Lecture Notes - Lecture 8: Diamagnetism, Paramagnetism, Ion

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18 Oct 2016

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CHEM 1A Full Course Notes

CHEM 1A Full Course Notes

Document Summary

S number of electrons in lower energy levels trend is s > p > d > f. S = 2- (3 electrons, one in the valence shell) S = 10- n=1 : 2- n=2 : 8- n=3 doesn"t count. Filling orbitals d orbitals half lled and lled is favorable sometimes this will cause the orbitals to ll other than how you would expect them too. Cations positive charged decreases in size compared to its atom as you go down a group - the cation begins to get less small than its atom compared to higher atoms in a group. Anions: negative charge increases in size compared to its atom as you go down a group it gets less big than atoms/anions of a atom higher in the group. First ionization energy energy to remove electron from neutral atom all atoms have the rst ionization energy.

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Related Questions

Slater’s rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the “shielding constant,” S, is calculated. The effective nuclear charge is then the difference between S and the atomic number, Z.

The shielding constant, S, is calculated using the following rules:

(1) The electrons of an atom are grouped as follows: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d), and so on.

(2) Electrons in higher groups (to the right) do not shield those in the lower groups.

(3) For ns and np valence electrons

a) Electrons in the same ns, np group contribute 0.35 (for 1s 0.30 works better).

b) Electrons in n − 1 groups contribute 0.85.

c) Electrons in n − 2 groups (and lower) contribute 1.00.

(4) For nd and nf electrons, electrons in the same nd or nf group contribute 0.35, and those in groups to the left contribute 1.00.

As an example, let us calculate Z* for the outermost electron of oxygen:

Here is the calculation for a d electron in Ni:

and for an s electron Ni:

(Here 3s, 3p, and 3d electrons are in the (n − 1) groups.)

a) Calculate Z* for F and Ne. Relate the Z* values for O, F, and Ne to their relative atomic radii and ionization energies.

b) Calculate Z* for one of the 3d electrons of Mn, and compare this with Z* for one of the 4s electrons of the element. Do the Z* values give us some insight into the ionization of Mn to give the cation?

Slater’s rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the “shielding constant,” S , is calculated. The effective nuclear charge is then the difference between S and the atomic number, Z . (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) Z * = Z ‒ S The shielding constant, S , is calculated using the following rules: (1) The electrons of an atom are grouped as follows: (1 s ) (2 s , 2 p ) (3 s , 3 p ) (3 d ) (4 s , 4 p ) (4 d ), and so on. (2) Electrons in higher groups (to the right) do not shield those in the lower groups. (3) For ns and np valence electrons a) Electrons in the same ns, np group contribute 0.35 (for 1 s 0.30 works better). b) Electrons in n ‒ 1 groups contribute 0.85. c) Electrons in n ‒ 2 groups (and lower) contribute 1.00. (4) For nd and nf electrons, electrons in the same nd or nf group contribute 0.35, and those in groups to the left contribute 1.00. As an example, let us calculate Z* for the outermost electron of oxygen: S = (2 × 0.85) + (5 × 0.35) = 3.45 Z * = 8 ‒ 3.45 = 4.55 Here is a calculation for a d electron in Ni: Z * = 28 ‒ [18 × 1.00] ‒ [7 × 0.35] = 7.55 and for an s electron in Ni: Z * = 28 ‒ [10 × 1.00] ‒ [16 × 0.85] ‒ [1 × 0.35] = 4.05 (Here 3 s , 3 p , and 3 d electrons are in the ( n ‒ 1) groups.) a) Calculate Z * for F and Ne. Relate the Z * values for O, F, and Ne to their relative atomic radii and ionization energies. b) Calculate Z * for one of the 3 d electrons of Mn, and compare this with Z * for one of the 4 s electrons of the element. Do the Z * values give us some insight into the ionization of Mn to give the cation?

amethystcockroach626

1)The NaCl2 ionic salt does not exist. Which of the followingstatements best explains why?

A. Because the formation of Na2+ is an unfavorable process due tothe great amount of energy required to remove the 2nd electron fromthe Na atom (the 2nd electron would have to be removed from thecore electrons of the Na atom).
B. Because the Cl- anion is too large for there to be two of themattached to one Na+ ion.
C. Because NaCl2 is actually a polyatomic ion with a -1 charge, andnot an ionic salt. In the presence of another Na+ ion, the Na2Cl2salt would form.
D. Because Cl2 has no charge and therefore would not be attractedelectrostatically to the positively charged Na2+ cation.

2)Select the false statement below.

A. The first ionization energy of an atom is smaller than thesecond ionization energy of the same atom.
B. Phosphorus has a greater first ionization energy than magnesium,and the reason for this is because P has a greater effectivenuclear charge (Zeff) than Mg.
C. The following process is an endothermic process and representsthe first ionization energy of an atom A:A(g) ? A1+(g) + e1-
D. Fluorine has a greater first ionization energy than iodine, andthe reason for this is because fluorine has a greater amount ofshielding of outer electrons by inner electrons than doesiodine.

3)The Group 5A elements N,P,As have lower (less negative) electronaffinities than most of the Group 4A or Group 6A elements becauseN,P,As have half-filled subshells.

True
False

4)A fluorine atom is smaller than a bromine atom, and a potassiumatom is larger than a lithium atom.

True
False

5)A certain main-group element exhibits the following successiveionization energies in kJ/mol:

IE1 = 100; IE2 = 1200; IE3 = 2300; IE4 = 3100; IE5 = 4200

Select the false statement about this element below.

A. This is likely a Group 1A element.
B. This element is typically found uncombined in nature (that is,it is not usually found in compounds, but rather as a freeelement).
C. If this element is given a symbol of "J", it would form an ioniccompound with sulfur as J2S.
D. If this element was in period 2 of the periodic table, it wouldbe lithium.

6)Select the false statement below.

A. Generally speaking, non-metallic elements have greater effectivenuclear charge (Zeff) than metallic elements in the sameperiod.
B. Generally speaking, as you go down a group in the periodictable, there is successively greater shielding of the outer-shellelectrons by the inner-shell electrons.
C. Generally speaking, the heavier alkali metals can be predictedto exhibit greater shielding of the outer-shell electrons by theinner-shell electrons than lighter alkali metals.
D. Generally speaking, as you go across a period in the periodictable, the effective nuclear charge (Zeff) decreases because eachsuccessive element contains more protons in the nucleus than theprevious element and the outer electrons are all in the same shellas you go across a period.

7)Place the following in order of increasing size:

O2-, F1-, Li1+

A. F1- < Li1+ < O2-
B. Li1+ < O2- < F1-
C. Li1+ < F1- < O2-
D. F1- < O2- < Li1+

8)Select the false statement below.

A. The following set of quantum numbers is allowed:
n = 4, l = 2, ml = -2, ms = +1/2
B. The n = 1 shell of any given atom can accomodate up to 2electrons.
C. The n = 4 shell of any given atom can accomodate up to 16electrons.
D. In any given atom, the l = 2 subshell can accomodate up to 5electrons that have negative spins (ms =