Questions: Element#3 1. The radius of my most common ion is smaller than my atomic radius. 2. My valence shell contains only s-orbital electrons. 3. I combine with phosphate ions in a 3:2 ratio. 4. I have a lower first ionization energy than calcium. 5. I have a smaller atomic radius than barium. a. What element am I? b. Write my symbol and noble gas electron configuration. Element#9 1. I am a metal 2. My first ionization energy is significantly lower than my second. 3. I am not the smallest or largest atom in my family. 4. I have a radius smaller than only 2 other elements in my family. a. What element am I ? b. Write my symbol and full electron configuration. c. Justify clue #2. Element#5 1. I am a non-metal. 2. I am diatomic. 3. I am the most reactive element in my family. 4. I have a larger ionization energy than oxygen. 5. I combine with aluminum ions in a 1:1 ratio a. What element am I ? b. Write my symbol and orbital diagram. c. Justify clue #4.

Element#3
1. The radius of my most common ion is smaller than my atomic radius.
2. My valence shell contains only s-orbital electrons.
3. I combine with phosphate ions in a 3:2 ratio.
4. I have a lower first ionization energy than calcium.
5. I have a smaller atomic radius than barium.
a. What element am I?
b. Write my symbol and noble gas electron configuration.

Element#9
1. I am a metal
2. My first ionization energy is significantly lower than my second.
3. I am not the smallest or largest atom in my family.
4. I have a radius smaller than only 2 other elements in my family.
a. What element am I ?
b. Write my symbol and full electron configuration.
c. Justify clue #2.

Element#5
1. I am a non-metal.
2. I am diatomic.
3. I am the most reactive element in my family.
4. I have a larger ionization energy than oxygen.
5. I combine with aluminum ions in a 1:1 ratio
a. What element am I ?
b. Write my symbol and orbital diagram.
c. Justify clue #4.

Transcript text: Element\#3 1. The radius of my most common ion is smaller than my atomic radius. 2. My valence shell contains only s-orbital electrons. 3. I combine with phosphate ions in a 3:2 ratio. 4. I have a lower first ionization energy than calcium. 5. I have a smaller atomic radius than barium. a. What element am I? b. Write my symbol and noble gas electron configuration. Element\#9 1. I am a metal 2. My first ionization energy is significantly lower than my second. 3. I am not the smallest or largest atom in my family. 4. I have a radius smaller than only 2 other elements in my family. a. What element am $I$ ? b. Write my symbol and full electron configuration. c. Justify clue \#2. Element\#5 1. I am a non-metal. 2. I am diatomic. 3. I am the most reactive element in my family. 4. I have a larger ionization energy than oxygen. 5. I combine with aluminum ions in a $1: 1$ ratio a. What element am $I$ ? b. Write my symbol and orbital diagram. c. Justify clue \#4.

Solution

Solution Steps

Step 1: Analyze Clues for Element #3

Clue 1: The radius of the most common ion is smaller than the atomic radius, indicating the element forms a cation.
Clue 2: The valence shell contains only s-orbital electrons, suggesting it is an alkali or alkaline earth metal.
Clue 3: Combines with phosphate ions in a 3:2 ratio, indicating a +3 charge.
Clue 4: Has a lower first ionization energy than calcium, suggesting it is in Group 13.
Clue 5: Has a smaller atomic radius than barium, which is consistent with elements in the third period or higher.

From these clues, the element is likely Aluminum (Al).

Step 2: Write Symbol and Electron Configuration for Element #3

Symbol: Al
Noble Gas Electron Configuration: $[ \text{Ne} ] 3s^2 3p^1$

Step 3: Analyze Clues for Element #A

Clue 1: It is a metal.
Clue 2: The first ionization energy is significantly lower than the second, indicating it forms a +1 cation.
Clue 3: Not the smallest or largest atom in its family, suggesting it is in the middle of its group.
Clue 4: Has a radius smaller than only 2 other elements in its family, indicating it is likely in Group 1.

From these clues, the element is likely Sodium (Na).

Step 4: Write Symbol and Electron Configuration for Element #A

Symbol: Na
Full Electron Configuration: $1s^2 2s^2 2p^6 3s^1$

Step 5: Justify Clue #2 for Element #A

The first ionization energy is significantly lower than the second because removing the first electron results in a stable noble gas configuration, making the second electron much harder to remove.

Step 6: Analyze Clues for Element #5

Clue 1: It is a non-metal.
Clue 2: It is diatomic, suggesting it is a halogen.
Clue 3: It is the most reactive element in its family, indicating it is Fluorine.
Clue 4: Has a larger ionization energy than oxygen, consistent with Fluorine.
Clue 5: Combines with aluminum ions in a 1:1 ratio, indicating a -1 charge.

From these clues, the element is likely Fluorine (F).

Step 7: Write Symbol and Orbital Diagram for Element #5

Symbol: F
Orbital Diagram: \[ \begin{array}{c} \uparrow\downarrow \\ 1s \\ \end{array} \quad \begin{array}{c} \uparrow\downarrow \\ 2s \\ \end{array} \quad \begin{array}{ccc} \uparrow & \uparrow & \uparrow \\ 2p_x & 2p_y & 2p_z \\ \end{array} \]

Step 8: Justify Clue #4 for Element #5

Fluorine has a larger ionization energy than oxygen because it is more electronegative and has a smaller atomic radius, resulting in a stronger attraction between the nucleus and the valence electrons.