13. when a chlorine atom forms an ion its radius increases, but when a sodium atom forms an ion its radius decreases. explain this apparent contradiction.
14. choose which has the:
a) lowest first ionization energy: k, na, ca
b) largest radius: s²⁻, cl⁻, cl
c) highest electronegativity: as, sn, s
d) smallest atom: na, li, be
e) highest first ionization energy: c, n, si
15. lithium, cesium, potassium, sodium, rubidium
a) place the elements above in order from smallest atomic radius to largest atomic radius.
b) place the elements above in order from smallest ionization energy to largest ionization energy.
c) place the elements above in order from smallest electronegativity to largest electronegativity.
16. in each of the following pairs, circle the species with the higher ionization energy:
a) li or cs
b) cl or ar
c) ca or br
d) na⁺ or ne
e) b or be
17. determine which element you would expect to have the lowest first ionization energy: li, cs, h, he, ba
18. identify which atom should have the largest value for the electron affinity: he, f, na, si, mn
19. define ionization energy.
20. is it easier to form a positive ion with an element that has high ionization energy or an element that has low ionization energy? explain.
21. what trend in ionization energy do you see as you go down a group/family on the periodic table? what causes this trend?
22. circle the atom in each pair that has the greater electronegativity:
a) ca ga
b) li o
c) cl s
d) br as
e) ba sr
f) o s
24. ca sr ba mg
a) most metallic?
b) least metallic?
c) lowest electronegativity?
d) highest ionization energy?
e) largest atomic radius?
f) which are non - metals?
g) list them in order from small to large.
h) would they gain or lose electrons to form ions? what charge would the ion have?
i) would the ion be larger or smaller than the atom?
k) would the ions be cations or anions?
l) what is the special name given to the group in which these elements belong?

Question

13. when a chlorine atom forms an ion its radius increases, but when a sodium atom forms an ion its radius decreases. explain this apparent contradiction.
14. choose which has the:
    a) lowest first ionization energy: k, na, ca
    b) largest radius: s²⁻, cl⁻, cl
    c) highest electronegativity: as, sn, s
    d) smallest atom: na, li, be
    e) highest first ionization energy: c, n, si
15. lithium, cesium, potassium, sodium, rubidium
    a) place the elements above in order from smallest atomic radius to largest atomic radius.
    b) place the elements above in order from smallest ionization energy to largest ionization energy.
    c) place the elements above in order from smallest electronegativity to largest electronegativity.
16. in each of the following pairs, circle the species with the higher ionization energy:
    a) li or cs
    b) cl or ar
    c) ca or br
    d) na⁺ or ne
    e) b or be
17. determine which element you would expect to have the lowest first ionization energy: li, cs, h, he, ba
18. identify which atom should have the largest value for the electron affinity: he, f, na, si, mn
19. define ionization energy.
20. is it easier to form a positive ion with an element that has high ionization energy or an element that has low ionization energy? explain.
21. what trend in ionization energy do you see as you go down a group/family on the periodic table? what causes this trend?
22. circle the atom in each pair that has the greater electronegativity:
    a) ca  ga
    b) li  o
    c) cl  s
    d) br  as
    e) ba  sr
    f) o  s
24. ca  sr  ba  mg
    a) most metallic?
    b) least metallic?
    c) lowest electronegativity?
    d) highest ionization energy?
    e) largest atomic radius?
    f) which are non - metals?
    g) list them in order from small to large.
    h) would they gain or lose electrons to form ions? what charge would the ion have?
    i) would the ion be larger or smaller than the atom?
    k) would the ions be cations or anions?
    l) what is the special name given to the group in which these elements belong?

Accepted Answer

### Question 13 (Explanation - Step by Step)
# Explanation:
## Step 1: Analyze Chlorine Atom to Ion
A chlorine atom ($Cl$) has 7 valence electrons. When it forms an ion ($Cl^-$), it gains 1 electron. This addition of an electron increases electron - electron repulsions in the valence shell. Also, the effective nuclear charge experienced by the outermost electrons remains relatively the same (since the nuclear charge doesn't change, but the number of electrons increases). The increased repulsions cause the electron cloud to expand, so the radius of $Cl^-$ is larger than that of $Cl$.
## Step 2: Analyze Sodium Atom to Ion
A sodium atom ($Na$) has 1 valence electron. When it forms an ion ($Na^+$), it loses this valence electron. Now, the ion has one less energy level (the outermost shell is removed). Also, the effective nuclear charge experienced by the remaining electrons increases (since there are fewer electrons to shield the nuclear charge). This causes the remaining electrons to be pulled closer to the nucleus, so the radius of $Na^+$ is smaller than that of $Na$.
# Answer:
Chlorine gains an electron (increasing electron - electron repulsions and keeping effective nuclear charge similar, expanding the electron cloud). Sodium loses a valence electron (losing a shell and increasing effective nuclear charge, pulling electrons closer), leading to the apparent contradiction.

### Question 14a (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy Trends
Ionization energy generally decreases as we move down a group and increases as we move across a period from left to right.
## Step 2: Analyze the Elements
$K$ (potassium) and $Na$ (sodium) are in Group 1, with $K$ below $Na$. $Ca$ (calcium) is in Group 2. Down a group, ionization energy decreases, so $K$ has lower ionization energy than $Na$. Also, Group 1 elements have lower ionization energy than Group 2 elements in the same period. So among $K$, $Na$, $Ca$, $K$ has the lowest first ionization energy.
# Answer:
$K$

### Question 14b (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Radius Trends for Atoms and Ions
For anions, the more negative the charge (or the larger the number of electrons relative to protons), the larger the radius. Also, anions are larger than their parent atoms.
## Step 2: Analyze the Species
$S^{2 - }$ has 18 electrons, $Cl^-$ has 18 electrons, and $Cl$ has 17 electrons. The nuclear charge of $S$ is $ + 16$ and for $Cl$ is $ + 17$. Since $S^{2 - }$ has a smaller nuclear charge than $Cl^-$ but the same number of electrons, the electron cloud in $S^{2 - }$ is less tightly held. Also, $Cl^-$ is larger than $Cl$ (anion vs atom). So $S^{2 - }$ has the largest radius.
# Answer:
$S^{2 - }$

### Question 14c (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Electronegativity Trends
Electronegativity generally increases as we move across a period from left to right and decreases as we move down a group.
## Step 2: Analyze the Elements
$As$ (arsenic) is in Group 15, Period 4; $Sn$ (tin) is in Group 14, Period 5; $S$ (sulfur) is in Group 16, Period 3. $S$ is the most to the right and the highest up among the three. So $S$ has the highest electronegativity.
# Answer:
$S$

### Question 14d (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Atomic Radius Trends
Atomic radius generally decreases as we move across a period from left to right and increases as we move down a group.
## Step 2: Analyze the Elements
$Na$ (sodium) is in Period 3, Group 1; $Li$ (lithium) is in Period 2, Group 1; $Be$ (beryllium) is in Period 2, Group 2. $Li$ and $Be$ are in Period 2, with $Be$ to the right of $Li$ (so $Be$ has a smaller radius than $Li$). $Na$ is below $Li$ (so $Na$ has a larger radius than $Li$). So $Be$ is the smallest atom.
# Answer:
$Be$

### Question 14e (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy Trends
Ionization energy generally increases as we move across a period from left to right and decreases as we move down a group.
## Step 2: Analyze the Elements
$C$ (carbon) is in Period 2, Group 14; $N$ (nitrogen) is in Period 2, Group 15; $Si$ (silicon) is in Period 3, Group 14. $N$ is to the right of $C$ in the same period (so $N$ has higher ionization energy than $C$). $Si$ is below $C$ (so $Si$ has lower ionization energy than $C$). So $N$ has the highest first ionization energy.
# Answer:
$N$

### Question 15a (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Atomic Radius Trend in a Group
In a group (Group 1 for these elements: lithium ($Li$), sodium ($Na$), potassium ($K$), rubidium ($Rb$), cesium ($Cs$)), atomic radius increases as we move down the group. This is because as we go down, the number of energy levels (shells) increases, and the outermost electrons are further from the nucleus.
## Step 2: Order the Elements
The order from smallest to largest atomic radius is based on their position in Group 1. The elements in order from top to bottom (smallest to largest radius) are $Li$, $Na$, $K$, $Rb$, $Cs$.
# Answer:
$Li$, $Na$, $K$, $Rb$, $Cs$

### Question 15b (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy Trend in a Group
In a group, ionization energy decreases as we move down the group. This is because as we go down, the outermost electrons are further from the nucleus, and the effective nuclear charge experienced by them is lower (due to increased shielding from inner electrons), so it is easier to remove an electron.
## Step 2: Order the Elements
The order from smallest (lowest) to largest (highest) ionization energy is the reverse of the atomic radius order in a group. So the order is $Cs$, $Rb$, $K$, $Na$, $Li$.
# Answer:
$Cs$, $Rb$, $K$, $Na$, $Li$

### Question 15c (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Electronegativity Trend in a Group
In a group, electronegativity decreases as we move down the group. This is because the outermost electrons are further from the nucleus, so the atom has less of an ability to attract electrons.
## Step 2: Order the Elements
The order from smallest (lowest) to largest (highest) electronegativity is the same as the order of increasing ionization energy (and reverse of atomic radius order) in a group. So the order is $Cs$, $Rb$, $K$, $Na$, $Li$.
# Answer:
$Cs$, $Rb$, $K$, $Na$, $Li$

### Question 16a (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy Trend in a Group
Ionization energy decreases as we move down a group. $Li$ (lithium) is in Period 2, Group 1 and $Cs$ (cesium) is in Period 6, Group 1. So $Li$ is above $Cs$ in the group.
## Step 2: Compare the Two Elements
Since ionization energy decreases down a group, $Li$ has a higher ionization energy than $Cs$. So we circle $Li$.
# Answer:
Circle $Li$

### Question 16b (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy for Noble Gases and Halogens
Noble gases have a stable electron configuration, so their ionization energy is very high. Halogens ($Cl$) need to gain an electron to achieve a stable configuration, while noble gases ($Ar$) have a stable configuration.
## Step 2: Compare $Cl$ and $Ar$
$Ar$ has a higher ionization energy than $Cl$ because it has a stable octet. So we circle $Ar$.
# Answer:
Circle $Ar$

### Question 16c (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy Trend Across a Period
Ionization energy generally increases as we move across a period from left to right. $Ca$ (calcium) is in Group 2, Period 4 and $Br$ (bromine) is in Group 17, Period 4.
## Step 2: Compare $Ca$ and $Br$
Since $Br$ is to the right of $Ca$ in the same period, $Br$ has a higher ionization energy. So we circle $Br$.
# Answer:
Circle $Br$

### Question 16d (Explanation - Step by Step)
# Explanation:
## Step 1: Analyze Electron Configurations
$Na^+$ has the electron configuration of $Ne$ (since $Na$ loses one electron: $Na$: $1s^22s^22p^63s^1$, $Na^+$: $1s^22s^22p^6$). But $Na^+$ has a nuclear charge of $ + 11$ and $Ne$ has a nuclear charge of $ + 10$.
## Step 2: Compare Ionization Energies
The higher nuclear charge in $Na^+$ means that its electrons are more tightly held. So $Ne$ has a lower ionization energy than $Na^+$? Wait, no. Wait, $Na^+$ is already a cation. The ionization energy of $Na^+$ would be the energy to remove another electron from $Na^+$, and $Ne$ is a neutral atom. But since $Na^+$ has a higher nuclear charge and the same electron configuration as $Ne$, the electrons in $Na^+$ are more tightly held. So $Ne$ has a lower ionization energy? Wait, no, the question is about which has higher ionization energy between $Na^+$ and $Ne$. Since $Na^+$ has a higher nuclear charge ($ + 11$ vs $ + 10$ for $Ne$) and the same electron cloud, it is harder to remove an electron from $Na^+$ than from $Ne$. So $Na^+$ has higher ionization energy? Wait, no, let's re - think. $Ne$ is a neutral atom with electron configuration $1s^22s^22p^6$. $Na^+$ is also $1s^22s^22p^6$ but with a $ + 11$ nucleus. The effective nuclear charge for $Na^+$ is higher. So when we consider the ionization energy (energy to remove an electron), for $Ne$, the electron is in a neutral atom, for $Na^+$, the electron is in a cation. So it is harder to remove an electron from $Na^+$ than from $Ne$. So we circle $Na^+$? Wait, no, maybe I made a mistake. Wait, the question is "circle the species with the higher ionization energy". Let's check the electron configurations. $Ne$: $1s^22s^22p^6$ (stable). $Na^+$: $1s^22s^22p^6$ (also stable, but with a higher nuclear charge). The ionization energy of $Na^+$ (second ionization energy of $Na$) is higher than the ionization energy of $Ne$ (first ionization energy of $Ne$) because the nuclear charge is higher. So we circle $Na^+$.
# Answer:
Circle $Na^+$

### Question 16e (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Electron Configurations and Ionization Energy
$Be$ has an electron configuration of $1s^22s^2$ (stable $s$ - orbital configuration), $B$ has an electron configuration of $1s^22s^22p^1$.
## Step 2: Compare $B$ and $Be$
It is easier to remove the $p$ - electron from $B$ than the $s$ - electrons from $Be$. So $Be$ has a higher ionization energy. So we circle $Be$.
# Answer:
Circle $Be$

### Question 17 (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Ionization Energy Trend in a Group
Ionization energy decreases as we move down a group. Also, noble gases ($He$) have very high ionization energy, and hydrogen ($H$) has relatively high ionization energy compared to Group 1 metals.
## Step 2: Analyze the Elements
$Li$ (lithium), $Cs$ (cesium), $Ba$ (barium) are metals. $Cs$ is in Group 1, lower than $Li$ in the group. $Ba$ is in Group 2. Among these, $Cs$ is the lowest in its group (Group 1) and has the lowest first ionization energy. $He$ has very high ionization energy, $H$ has higher than Group 1 metals. So $Cs$ has the lowest first ionization energy.
# Answer:
$Cs$

### Question 18 (Explanation - Step by Step)
# Explanation:
## Step 1: Recall Electron Affinity Trend
Electron affinity generally increases as we move across a period from left to right (excluding noble gases) and decreases as we move down a group. Noble gases ($He$) have very low (or negative) electron affinity. Metals ($Na$, $Mn$) have low electron affinity. Non - metals in the upper right (excluding noble gases) have high electron affinity.
## Step 2: Analyze the Elements
$F$ (fluorine) is in Group 17, Period 2. It is a highly electronegative non - metal and has a very high electron affinity (it strongly attracts an additional electron to achieve a stable octet). $He$ is a noble gas, $Na$ is a metal, $Si$ is a metalloid, $Mn$ is a metal. So $F$ has the largest electron affinity.
# Answer:
$F$

### Question 19 (Explanation - Brief)
# Brief Explanations:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron (the valence electron) from an isolated, gaseous atom or ion in its ground state.
# Answer:
The minimum energy to remove a valence electron from a gaseous atom/ion.

### Question 20 (Explanation - Brief)
# Brief Explanations:
Ionization energy is the energy required to remove an electron. An element with low ionization energy requires less energy to remove an electron, so it is easier to form a positive ion (cation) with an element that has low ionization energy. Elements with high ionization energy require a large amount of energy to remove an electron, making it harder to form a positive ion.
# Answer:
It is easier with an element of low ionization energy (less energy needed to remove an electron to form a positive ion).

### Question 21 (Explanation - Step by Step)
# Explanation:
## Step 1: Identify the Trend
As we go down a group/family on the periodic table, the first ionization energy generally decreases.
## Step 2: Explain the Cause
As we move down a group, the number of energy levels (shells) increases. The outermost electrons are further from the nucleus. Also, the shielding effect (the repulsion between inner electrons and the outermost electrons) increases. The increased distance and shielding effect reduce the effective nuclear charge experienced by the outermost electrons. So it becomes easier to remove an outermost electron, and thus the ionization energy decreases.
# Answer:
Trend: Ionization energy decreases down a group. Cause: Increased energy levels, more shielding, lower effective nuclear charge on outer electrons, easier to remov…

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 13 (Explanation - Step by Step)

Step 1: Analyze Chlorine Atom to Ion

Step 2: Analyze Sodium Atom to Ion

Step 1: Recall Ionization Energy Trends

Step 2: Analyze the Elements

Step 1: Recall Radius Trends for Atoms and Ions

Step 2: Analyze the Species

Answer:

Question 14a (Explanation - Step by Step)