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Periodic Trends.notebook 1 January 26, 2016 Periodic Trends The trends we will study all have to do with the valence electrons in one way or another. Two key ideas: Nuclear Charge = the number of protons in the nucleus. This is the positive charge acting on valence (outermost energy level) electrons. This can be thought of as the positive “pull” on the negative electron(s) (opposites attract). The greater the nuclear charge (i.e. the greater the number of protons), the greater the “pull” or attraction on the valence electron(s). Shielding of inner level (or core) electrons. Inner level electrons block or “shield” the effects of the positive “pull” of the nuclear charge on valence electrons. The greater the number of levels of electrons between the nucleus and the valence electrons the greater the shielding and the weaker the nuclear “pull” on the valence electrons.

Periodic Trends - Scarsdale Public Schools · Periodic Trends.notebook 1 January 26, 2016 Periodic Trends The trends we will study all have to do with the valence electrons in one

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Page 1: Periodic Trends - Scarsdale Public Schools · Periodic Trends.notebook 1 January 26, 2016 Periodic Trends The trends we will study all have to do with the valence electrons in one

Periodic Trends.notebook

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January 26, 2016

Periodic Trends

The trends we will study all have to do with the valence electrons in one way or another.

Two key ideas:

Nuclear Charge = the number of protons in the nucleus. This is the positive charge acting on valence (outermost energy level) electrons. This can be thought of as the positive “pull” on the negative electron(s) (opposites attract). The greater the nuclear charge (i.e. the greater the number of protons), the greater the “pull” or attraction on the valence electron(s).

Shielding of inner level (or core) electrons. Inner level electrons block or “shield” the effects of the positive “pull” of the nuclear charge on valence electrons. The greater the number of levels of electrons between the nucleus and the valence electrons the greater the shielding and the weaker the nuclear “pull” on the valence electrons.

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Let’s consider what happens down a Group: (the number below the element is the nuclear charge)

Li(3+)

Na(11+)

K(19+)

Rb(37+)

Notice that the valence electron gets further from the nucleus and the shielding of inner levels (underlined) increases; the shielding more than offsets the increasing nuclear charge so that the dominant factor going down a group is shielding.

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Let’s consider what happens across a Period:

Li(3+)

Be(4+)

B(5+)

C(6+)

N(7+)

O(8+)

F(9+)

Ne(10+)

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Notice that even though we get an extra sublevel when going from Be to B, the valence electrons remain in the second level and there is no increased shielding because extra energy levels; the dominant factor here is the increasing nuclear charge.

Properties & Trends Explained

Ionization Energy (IE): X(g) + IE à X+(g) + e­

the energy needed to remove the outermost (valence) electron of an element in its gaseous phase

IE decreases down a group because as you go down a group you add energy levels and thus increase the shielding; the valence electron is farther away from the nuclear “pull” and is increasingly shielded from it.

IE generally increases across a period because across a period you stay in the same level and the shielding thus stays the same but you increase in nuclear charge. This increase in nuclear charge without increase in shielding has an increasing “pull” on valence electrons.

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Electronegativity (EN) – a measure of how an atom attracts electrons when forming chemical bonds. This is directly related to IE and is affected by the same factors. The greater the EN, the greater the IE.

Down a group EN decreases because of increasing shielding of inner level electrons.

Across a period EN increases because across a period the shielding stays the same but the nuclear charge increases.

Atomic Size (Radius)

Down a group atomic radius increases simply because you are adding energy levels.

Across a period atomic size decreases because you stay in the same level but nuclear charge increases, pulling in the outermost electrons closer to the nucleus.

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Ionic Radii

Metals lose electrons and the resulting cations are smaller than the atoms. This is because in losing electron(s) the metal loses a sublevel. Consider Na vs Na+

• If the electron­proton ration decreases, the radius decreases

Nonmetals gain electrons and the resulting anions increase in size. This is the result of increased electron­electron repulsions in the ion.

• If the electron­proton ration increases, the radius increaes

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Isoelectronic Species

Ions having the same number of total electrons are said to be isoelectronic. Consider sulfide ion, S2­. When neutral sulfur (16 electrons) gains two electrons to become sulfide ion, it now has a total of 18 electrons – the same as argon. Hence sulfide ion and argon are isoelectronic. Isolectronic species will not have the same radius (size) despite having the same number of electrons. This is because they have differing numbers of protons (nuclear charge). For species that are isoelectronic, what is the relation between radius and nuclear charge?

Arrange the following by increasing radius (i.e. smallest to largest):

P3­, Ar, Cl­, Ca2+, S2­, K+

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Origin of the Elements

Interstellar gas (hydrogen) and dust (frozen CO2, CH4, etc) begin to contract under the force of gravity: This huge, whirling, contracting gaseous "cloud" is a protostar. As the gravitational collapse continues, the temperature and pressure of the gas increases. When the temperature reaches 10 million K, the electrical barrier repelling protons is overcome and nuclear fusion occurs. The net fusion reaction is: 4 1H à 2He + ENERGY (radiation) 1 4 (protons) (helium)

In the process 0.7% of the mass of hydrogen is lost. It is converted to energy. (E = mc2).

When

OUTWARD INWARD RADIATION = GRAVITATIONAL PRESSURE PRESSURE

equilibrium is reached and A STAR IS BORN.This equilibrium continues for many millions of years. After the hydrogen fuel is used up, the He core contracts. The temperature rises to 100 million K and then the fusion of helium to carbon occurs. After the He fuel is used up, what happens next depends on the original mass of the star.

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In small mass stars (up to 1.4 solar masses), the carbon core contracts and begins to heat up. The temperature required to overcome the electrical barrier of 6 positive charges and the fusion of carbon nuclei is not reached. Thus, the carbon core contracts to an extremely dense white dwarf which radiates its last energy and becomes a black dwarf (which is really a giant diamond in space).

In very massive stars (greater than 4 solar masses), the heat generated from the contraction of the carbon core reaches 600 million K. At this temperature carbon fuses to form magnesium and heavier elements up to Fe. Fusion continues until the core is mostly Fe. The core contracts for the last time and explodes violently in a SUPERNOVA. The temperatures reached in a supernova are high enough for the elements heavier than Fe to form. These heavier elements are shot out into space to become part of the interstellar gas and dust and the cycle continues.

You, chemistry students, are part of that cycle. YOU ARE STAR STUFF.

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Characteristics of elements

1. a. How many protons does an atom of bromine have? __________

b. In which group is bromine found? __________

c. What is the name of its chemical family? __________

d How many valence electrons does it have? __________

e. Is it a metal, nonmetal or semimetal? __________

f. How does its atomic radius compare to chlorine’s? __________

g. How does its atomic radius compare to arsenic’s? __________

h. What ion is it most likely to form in compounds? __________

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2. a. How many protons does an atom of magnesium have? __________

b. In which group is magnesium found? __________

c. What is the name of its chemical family? __________

d How many valence electrons does it have? __________

e. Is it a metal, nonmetal or semimetal? __________

f. How does its atomic radius compare to calcium’s? __________

g. How does its atomic radius compare to aluminum’s?__________

h. What ion is it most likely to form in compounds? __________

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3. a. How many protons does an atom of potassium have? __________

b. In which group is potassium found? __________

c. What is the name of its chemical family? __________

d How many valence electrons does it have? __________

e. Is it a metal, nonmetal or semimetal? __________

f. How does its atomic radius compare to sodium’s? __________

g. How does its atomic radius compare to selenium? __________

h. What ion is it most likely to form in compounds? __________

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Review Questions

Which atom in each pair has the larger atomic radius?

1. Li or K __________

2. Ca or Ni __________

3. Ga or B __________

4. O or C __________

5. Cl or Br __________

6. Be or Ba __________

7. Si or S __________

Which ion in each pair has the smaller atomic radius?

8. K+ or O2­ __________

9. Ba2+ or I­ __________

10. Al3+ or P3­__________

11. K+ or Cs+ __________

12. F­ or S2­ __________

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Which atom or ion in each pair has the larger ionization energy?

13. Na or O __________

14. Be or Ba __________

15. Ar or F __________

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Practice Problems

1. Chlorine, selenium, and bromine are located near each other on the periodic table. Which of these elements is

a) the smallest atom?

b) the atom with the highest ionization energy?

2. Phosphorous, sulfur, and selenium are located near each other on the periodic table. Which of these elements is

a) the largest atom?

b) the atom with the smallest ionization energy?

3. Scandium,. Yttrium, and lanthanum are located near each other on the periodic table. Which of these elements is

a) the largest atom?

b) the atom with the smallest ionization energy?

4. Which of the elements vanadium, chromium, or tungsten

a) is the smallest

b) has the highest ionization energy?

5. Which of the elements nitrogen, phosphorous, or arsenic

a) is the smallest?

b) has the smallest ionization energy?

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