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

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 8a # Explanation: ## Step1: Recall periodic trend for atomic radius Atomic radius decreases across a period (left to right) and increases down a group (top to bottom). Sodium (Na), Aluminum (Al), Sulfur (S), Chlorine (Cl) are in period 3. Order from left to right: Na, Al, S, Cl. So atomic radius: Na > Al > S > Cl. ## Step2: Arrange in decreasing order Based on the trend, the order from largest to smallest is sodium, aluminum, sulfur, chlorine. # Answer: sodium, aluminum, sulfur, chlorine (Note: There was a typo "ad" in the question, should be "and") ### Question 8b # Explanation: ## Step1: Recall group trend for atomic radius Be, Mg, Sr are in group 2. Atomic radius increases down a group. So Sr (bottom) > Mg > Be (top). ## Step2: Arrange in decreasing order Order: Sr, Mg, Be. # Answer: Sr, Mg, Be ### Question 8c # Explanation: ## Step1: Recall period and group trends Cs, Ba are in period 6, Bi is in period 6 too. Cs is in group 1, Ba group 2, Bi group 15. Across period, atomic radius decreases from left to right. So Cs > Ba > Bi. ## Step2: Arrange in decreasing order Order: Cs, Ba, Bi. # Answer: Cs, Ba, Bi ### Question 8d # Explanation: ## Step1: Recall group trend for atomic radius F, Cl, Br are in group 17. Atomic radius increases down a group. So Br > Cl > F. ## Step2: Arrange in decreasing order Order: Br, Cl, F. # Answer: Br, Cl, F ### Question 9 # Brief Explanations: Ionization energy (IE) is the energy to remove an electron. Across a period (left to right), IE generally increases (due to increasing effective nuclear charge, electrons are harder to remove). Within a group (top to bottom), IE generally decreases (electrons are in higher energy levels, farther from nucleus, easier to remove). Exceptions exist (e.g., half - filled, fully - filled subshells). # Answer: Across a period, ionization energy generally increases; within a group, ionization energy generally decreases. ### Question 10a # Explanation: ## Step1: Recall group trend for ionization energy Be, Mg, Sr are in group 2. IE decreases down a group (electrons are farther from nucleus). So Sr (lowest IE) < Mg < Be (highest IE). ## Step2: Arrange in increasing order Order: Sr, Mg, Be. # Answer: Sr, Mg, Be ### Question 10b # Explanation: ## Step1: Recall period and group trends Cs (group 1), Ba (group 2), Bi (group 15) are in period 6. IE increases across a period. Cs has lowest IE, then Ba, then Bi. ## Step2: Arrange in increasing order Order: Cs, Ba, Bi. # Answer: Cs, Ba, Bi ### Question 10c # Explanation: ## Step1: Recall period trend for ionization energy Na, Al, S are in period 3. IE increases across a period (with exceptions, but here general trend holds). Na (group 1) < Al (group 13) < S (group 16). ## Step2: Arrange in increasing order Order: Na, Al, S. # Answer: Na, Al, S ### Question 10d # Explanation: ## Step1: Recall group trend for ionization energy F, Cl, Br are in group 17. IE decreases down a group. So Br (lowest IE) < Cl < F (highest IE). ## Step2: Arrange in increasing order Order: Br, Cl, F. # Answer: Br, Cl, F ### Question 11 # Brief Explanations: A typical metal (e.g., Na) loses electrons to form a cation (e.g., Na⁺). When a metal loses electrons, the remaining electrons are pulled closer to the nucleus (less electron - electron repulsion, same nuclear charge). So the ionic radius (of the cation) of a typical metal is smaller than its atomic radius. # Answer: The ionic radius of a typical metal (cation) is smaller than its atomic radius. ### Question 12a # Explanation: ## Step1: Recall ion size trend for same element Cu⁺ has one less electron than Cu (neutral), but Cu²⁺ has two less electrons. For the same element, more positive charge (higher oxidation state) means smaller ion (electrons pulled closer). So Cu⁺ > Cu²⁺. ## Step2: Identify the larger ion Larger ion is Cu⁺. # Answer: Cu⁺ ### Question 12b # Explanation: ## Step1: Recall ion size trend for non - metal S (neutral) gains 2 electrons to form S²⁻. Gaining electrons increases electron - electron repulsion, so the ion is larger than the neutral atom. So S²⁻ > S. ## Step2: Identify the larger ion Larger ion is S²⁻. # Answer: S²⁻ ### Question 12c # Explanation: ## Step1: Recall ion size trend for metal Al (neutral) loses 3 electrons to form Al³⁺. Losing electrons makes the ion smaller (less electron - electron repulsion, same nuclear charge). So Al > Al³⁺. ## Step2: Identify the larger ion Larger ion is Al. # Answer: Al ### Question 12d (Note: There is a typo "Ca²⁻", should be maybe "Ca" or another ion, assuming it's a typo and comparing Ca²⁺ with Ca (neutral) or another. If we assume it's a typo and compare Ca²⁺ with Ca (neutral), Ca (neutral) has more electrons, so Ca > Ca²⁺. But since the question has a typo, if we consider a valid pair, for example, if it was Ca and Ca²⁺, larger is Ca. If it was a different pair, but based on the given, assuming a typo, but if we take as is (invalid Ca²⁻), but likely a typo. If we assume it's a different anion, but with the given, we can note the typo. However, if we consider a possible intended pair, say Ca²⁺ and another, but given the error, we can state the typo. But if we proceed with the given (assuming a mistake), but for the purpose of answering, if we consider Ca (neutral) and Ca²⁺, larger is Ca. But the question has an error in "Ca²⁻".) # Explanation: ## Step1: Note the typo There is a typo in "Ca²⁻" (calcium can't form a - 2 ion easily). Assuming it's a typo, and comparing Ca (neutral) with Ca²⁺, Ca (neutral) has more electrons, so atomic radius of Ca is larger than ionic radius of Ca²⁺. ## Step2: Identify the larger ion (assuming typo) If comparing Ca and Ca²⁺, larger is Ca. But due to the typo, the question is flawed. # Answer: (Due to typo in "Ca²⁻", the question is incorrect. If assuming comparison with Ca (neutral), larger is Ca) ### Question 13a # Explanation: ## Step1: Recall bond type rules Ionic bonds form between metals (low electronegativity) and non - metals (high electronegativity). Ca is a metal, Cl is a non - metal. ## Step2: Determine bond type Bond between Ca and Cl is ionic. # Answer: ionic ### Question 13b # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals (or metalloids). C and S are both non - metals. ## Step2: Determine bond type Bond between C and S is covalent. # Answer: covalent ### Question 13c # Explanation: ## Step1: Recall bond type rules Ionic bonds form between metals and non - metals. Mg is a metal, F is a non - metal. ## Step2: Determine bond type Bond between Mg and F is ionic. # Answer: ionic ### Question 13d # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals. H and O are non - metals. ## Step2: Determine bond type Bond between H and O is covalent. # Answer: covalent ### Question 13e # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals. S and O are non - metals. ## Step2: Determine bond type Bond between S and O is covalent. # Answer: covalent ### Question 13f # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals. Br and Cl are non - metals. ## Step2: Determine bond type Bond between Br and Cl is covalent. # Answer: covalent ### Question 13g # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals. P and S are non - metals. ## Step2: Determine bond type Bond between P and S is covalent. # Answer: covalent ### Question 13h # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals. H and Cl are non - metals. ## Step2: Determine bond type Bond between H and Cl is covalent. # Answer: covalent ### Question 13i # Explanation: ## Step1: Recall bond type rules Covalent bonds form between non - metals. C and H are non - metals (C is a non - metal, H is a non - metal). ## Step2: Determine bond type Bond between C and H is covalent. # Answer: covalent ### Question 14a # Explanation: ## Step1: Recall ionic compound naming rules For ionic compounds, name the cation first (Calcium, Ca²⁺) and then the anion. Fluoride (F⁻), and since there are 2 F⁻, but in naming, we use the - ide suffix for the anion. The formula is CaF₂, so name is Calcium fluoride. ## Step2: Write the name Name: Calcium fluoride. # Answer: Calcium fluoride

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QUESTION IMAGE

Question

Explanation:

Question 8a

Step1: Recall periodic trend for atomic radius

Step2: Arrange in decreasing order

Step1: Recall group trend for atomic radius

Step2: Arrange in decreasing order

Step1: Recall period and group trends

Step2: Arrange in decreasing order

Answer:

Question 8b