7. how does atomic size/radius change within groups and across periods?
8. arrange these elements in order of decreasing (largest to smallest) atomic size/radius:
a. sulfur, chlorine, aluminum ad sodium
b. be, mg, sr
c. bi, cs, ba
d. f, cl, br
9. how do ionization energies vary within the groups and across periods?
10. arrange these elements in order of increasing (smallest to largest) ionization energy:
a. be, mg, sr
b. bi, cs, ba
c. na, al, s
d. f, cl, br
11. how does ionic radius of a typical metal compare with its atomic radius?
12. in each pair, circle the largest ion?
a. cu⁺ and cu²⁺
c. al and al³⁺
b. s and s²⁻
d. ca²⁺ and ca²⁻
13. determine the type of bond (ionic or covalent) that will form between the following atoms.
a. ca and cl
b. c and s
c. mg and f
d. h and o
e. s and o
f. br and cl
g. p and s
h. h and cl
i. c and h
14. name the following compounds:
a. caf₂

Question

7. how does atomic size/radius change within groups and across periods?
8. arrange these elements in order of decreasing (largest to smallest) atomic size/radius:
   a. sulfur, chlorine, aluminum ad sodium
   b. be, mg, sr
   c. bi, cs, ba
   d. f, cl, br
9. how do ionization energies vary within the groups and across periods?
10. arrange these elements in order of increasing (smallest to largest) ionization energy:
    a. be, mg, sr
    b. bi, cs, ba
    c. na, al, s
    d. f, cl, br
11. how does ionic radius of a typical metal compare with its atomic radius?
12. in each pair, circle the largest ion?
    a. cu⁺ and cu²⁺
    c. al and al³⁺
    b. s and s²⁻
    d. ca²⁺ and ca²⁻
13. determine the type of bond (ionic or covalent) that will form between the following atoms.
    a. ca and cl
    b. c and s
    c. mg and f
    d. h and o
    e. s and o
    f. br and cl
    g. p and s
    h. h and cl
    i. c and h
14. name the following compounds:
    a. caf₂

Accepted Answer

### Question 7
# Brief Explanations:
Within a group (vertical column) in the periodic table, atomic size/radius increases from top to bottom. This is because as you go down a group, the number of electron shells increases. Each new shell is further from the nucleus, so the atomic radius gets larger. Across a period (horizontal row) from left to right, atomic size/radius decreases. This is due to the increasing effective nuclear charge (the positive charge felt by the outermost electrons). As you move from left to right across a period, the number of protons in the nucleus increases, pulling the electrons in the same valence shell closer to the nucleus, thus decreasing the atomic radius.
# Answer:
Within groups: atomic size/radius increases from top to bottom (more electron shells). Across periods: atomic size/radius decreases from left to right (increasing effective nuclear charge).

### Question 8
#### Part a
# Brief Explanations:
Sodium (Na), aluminum (Al), sulfur (S), and chlorine (Cl) are in the same period (period 3). Across a period, atomic size decreases from left to right. Sodium is the leftmost, then aluminum, then sulfur, then chlorine. So the order from largest to smallest is based on their positions in period 3.
# Answer:
sodium, aluminum, sulfur, chlorine

#### Part b
# Brief Explanations:
Beryllium (Be), magnesium (Mg), and strontium (Sr) are in the same group (group 2). Within a group, atomic size increases from top to bottom. Be is at the top, then Mg, then Sr. So the order from largest to smallest is Sr, Mg, Be.
# Answer:
Sr, Mg, Be

#### Part c
# Brief Explanations:
Cesium (Cs), barium (Ba), and bismuth (Bi) are in the same period (period 6). Cs is on the left, then Ba, then Bi. Across a period, atomic size decreases from left to right. Also, Cs is in group 1, Ba in group 2, Bi in group 15. So the order from largest to smallest is Cs, Ba, Bi.
# Answer:
Cs, Ba, Bi

#### Part d
# Brief Explanations:
Fluorine (F), chlorine (Cl), and bromine (Br) are in the same group (group 17). Within a group, atomic size increases from top to bottom. F is at the top, then Cl, then Br. So the order from largest to smallest is Br, Cl, F.
# Answer:
Br, Cl, F

### Question 9
# Brief Explanations:
Within a group (top to bottom), ionization energy decreases. This is because the outermost electrons are in higher energy levels (more electron shells) and are farther from the nucleus, so they are easier to remove (lower energy required). Across a period (left to right), ionization energy generally increases. This is due to the increasing effective nuclear charge, which makes the outermost electrons more tightly held by the nucleus, so more energy is required to remove them. Exceptions exist (e.g., between groups 2 and 13, 15 and 16) due to electron configuration stability, but the general trend holds.
# Answer:
Within groups: ionization energy decreases from top to bottom (electrons farther from nucleus). Across periods: ionization energy generally increases from left to right (increasing effective nuclear charge).

### Question 10
#### Part a
# Brief Explanations:
Be, Mg, Sr are in group 2. Within a group, ionization energy decreases from top to bottom (electrons are farther from nucleus, easier to remove). So the order from smallest (easiest to remove, lowest ionization energy) to largest (hardest to remove, highest ionization energy) is Sr, Mg, Be.
# Answer:
Sr, Mg, Be

#### Part b
# Brief Explanations:
Cs, Ba, Bi are in period 6. Cs (group 1) has the lowest ionization energy (easiest to remove electrons), then Ba (group 2), then Bi (group 15). Across a period, ionization energy generally increases. Also, within the group trend (Cs is in group 1, very low ionization energy). So the order from smallest to largest ionization energy is Cs, Ba, Bi.
# Answer:
Cs, Ba, Bi

#### Part c
# Brief Explanations:
Na, Al, S are in period 3. Na (group 1) has lower ionization energy than Al (group 13), and Al has lower than S (group 16). Across a period, ionization energy increases from left to right (due to increasing effective nuclear charge). So the order from smallest to largest is Na, Al, S.
# Answer:
Na, Al, S

#### Part d
# Brief Explanations:
F, Cl, Br are in group 17. Within a group, ionization energy decreases from top to bottom (electrons farther from nucleus, easier to remove). So the order from smallest (lowest ionization energy) to largest (highest ionization energy) is Br, Cl, F.
# Answer:
Br, Cl, F

### Question 11
# Brief Explanations:
A typical metal atom loses electrons to form a cation (positive ion). When a metal loses electrons, it loses one or more valence electron shells. For example, a sodium atom (Na) has 3 electron shells, and the sodium ion (Na⁺) has 2 electron shells. Removing the outermost electrons (and sometimes a shell) makes the ionic radius smaller than the atomic radius. The remaining electrons are pulled closer to the nucleus by the same nuclear charge (since the number of protons doesn't change), resulting in a smaller ionic radius.
# Answer:
The ionic radius of a typical metal (which forms a cation) is smaller than its atomic radius (loses electrons, often a valence shell, and remaining electrons are pulled closer by the nucleus).

### Question 12
#### Part a
# Brief Explanations:
Cu⁺ has lost 1 electron, and Cu²⁺ has lost 2 electrons. Both have the same number of protons (29), but Cu²⁺ has fewer electrons. The fewer electrons mean a greater effective nuclear charge on the remaining electrons, pulling them closer. So Cu⁺ is larger than Cu²⁺.
# Answer:
Cu⁺

#### Part b
# Brief Explanations:
S (atom) has 16 electrons, and S²⁻ (ion) has gained 2 electrons (18 electrons). Gaining electrons increases electron - electron repulsion in the same valence shell, causing the electron cloud to expand. So S²⁻ is larger than S.
# Answer:
S²⁻

#### Part c
# Brief Explanations:
Al (atom) has 13 electrons, and Al³⁺ (ion) has lost 3 electrons (10 electrons). Losing electrons (and a valence shell) makes the remaining electrons be pulled closer by the nucleus. So Al is larger than Al³⁺.
# Answer:
Al

#### Part d
# Brief Explanations:
Ca²⁺ has lost 2 electrons (has 18 electrons), and Ca²⁻ would be a very unstable ion (gaining 2 electrons, has 20 electrons). But if we consider the electron configurations, Ca²⁻ has more electrons. However, in reality, Ca²⁻ is not a common ion. But based on electron count, Ca²⁻ has more electrons, so its electron cloud is more expanded. So Ca²⁻ is larger than Ca²⁺.
# Answer:
Ca²⁻

### Question 13
#### Part a
# Brief Explanations:
Calcium (Ca) is a metal (group 2), chlorine (Cl) is a non - metal (group 17). Ionic bonds form between metals and non - metals (electron transfer). So Ca and Cl form an ionic bond.
# Answer:
ionic

#### Part b
# Brief Explanations:
Carbon (C) and sulfur (S) are both non - metals. Covalent bonds form between non - metals (electron sharing). So C and S form a covalent bond.
# Answer:
covalent

#### Part c
# Brief Explanations:
Magnesium (Mg) is a metal (group 2), fluorine (F) is a non - metal (group 17). Metals and non - metals form ionic bonds (electron transfer). So Mg and F form an ionic bond.
# Answer:
ionic

#### Part d
# Brief Explanations:
Hydrogen (H) and oxygen (O) are both non - metals. Non - metals form covalent bonds (electron sharing). So H and O form a covalent bond.
# Answer:
covalent

#### Part e
# Brief Explanations:
Sulfur (S) and oxygen (O) are both non - metals. Non - metals share electrons to form covalent bonds. So S and O form a covalent bond.
# Answer:
covalent

#### Part f
# Brief Explanations:
Bromine (Br) and chlorine (Cl) are both non - metals. Non - metals form covalent bonds through electron sharing. So Br and Cl form a covalent bond.
# Answer:
covalent

#### Part g
# Brief Explanations:
Phosphorus (P) and sulfur (S) are both non - metals. Non - metals form covalent bonds by sharing electrons. So P and S form a covalent bond.
# Answer:
covalent

#### Part h
# Brief Explanations:
Hydrogen (H) and chlorine (Cl) are both non - metals. Non - metals share electrons to form covalent bonds. So H and Cl form a covalent bond.
# Answer:
covalent

#### Part i
# Brief Explanations:
Carbon (C) and hydrogen (H) are both non - metals. Non - metals form covalent bonds by sharing electrons. So C and H form a covalent bond.
# Answer:
covalent

### Question 14 (Part a)
# Brief Explanations:
For ionic compounds, we name the cation first (using the element name) and then the anion. The cation here is calcium (Ca²⁺), and the anion is fluoride (F⁻, since F is in group 17, the anion is named by changing the ending of the element name to - ide). The formula is CaF₂, so the name is calcium fluoride.
# Answer:
calcium fluoride

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