lesson 3 electron configurations practice
date
period
multiple choice
1. how many valence electrons does a chlorine atom have if its electron configuration is ne3s²3p⁵?
a. 3 b. 21 c. 5 d. 7
2. given boron’s electron configuration of he2s²2p¹, which of the following represents its electron - dot structure?
a. ·be· b. ·b· c. b: d. be
3. given beryllium’s electron configuration of 1s²2s², which of the following represents its electron - dot structure?
a. ·be· b. ·b· c. b: d. be
short answer
4. write electron - dot structures for the following atoms.
a. ne3s²3p³ b. ar4s²3d³ c. potassium
5. write electron configurations for the following.
a. calcium
b. zirconium
c. arsenic
d. rubidium

Question

lesson 3 electron configurations practice
date
period
multiple choice
1. how many valence electrons does a chlorine atom have if its electron configuration is ne3s²3p⁵?
 a. 3  b. 21  c. 5  d. 7
2. given boron’s electron configuration of he2s²2p¹, which of the following represents its electron - dot structure?
 a. ·be·  b. ·b·  c. b:  d. be
3. given beryllium’s electron configuration of 1s²2s², which of the following represents its electron - dot structure?
 a. ·be·  b. ·b·  c. b:  d. be
short answer
4. write electron - dot structures for the following atoms.
 a. ne3s²3p³  b. ar4s²3d³  c. potassium
5. write electron configurations for the following.
 a. calcium
 b. zirconium
 c. arsenic
 d. rubidium

Accepted Answer

### Question 1
# Explanation:
## Step1: Identify valence electrons
Valence electrons are in the outermost shell. The electron configuration is $[	ext{Ne}]3s^2 3p^5$. The outermost shell is $n = 3$, with $3s^2$ and $3p^5$.
## Step2: Sum valence electrons
Add the electrons in the outermost shell: $2 + 5 = 7$.
# Answer: d. 7

### Question 2
# Explanation:
## Step1: Determine valence electrons
Boron's electron configuration is $[	ext{He}]2s^2 2p^1$. Valence electrons are $2 + 1 = 3$.
## Step2: Match electron - dot structure
Electron - dot structure shows valence electrons. Boron (B) should have 3 valence electrons represented as $\cdot \dot{	ext{B}}\cdot$. Option b has the correct representation.
# Answer: b. $\cdot \dot{	ext{B}}\cdot$

### Question 3
# Explanation:
## Step1: Find valence electrons
Beryllium's electron configuration is $1s^2 2s^2$. Valence electrons are in the $2s$ subshell, so there are 2 valence electrons.
## Step2: Identify electron - dot structure
The electron - dot structure for beryllium (Be) with 2 valence electrons is $\cdot 	ext{Be}\cdot$. Option a matches.
# Answer: a. $\cdot 	ext{Be}\cdot$

### Question 4
#### Part a: $[	ext{Ne}]3s^2 3p^3$
# Explanation:
## Step1: Identify the element and valence electrons
The electron configuration is for phosphorus (P). Valence electrons are in $3s^2 3p^3$, so total valence electrons $2 + 3=5$.
## Step2: Draw electron - dot structure
Represent the 5 valence electrons around P: $\dot{	ext{P}}\colon$ (with two electrons paired in $3s$ and three unpaired in $3p$, but in electron - dot structure, we show all valence electrons. The correct electron - dot structure is $\cdot \dot{	ext{P}}\colon$ (more accurately, $\dot{	ext{P}}\colon$ with one pair and three single dots, but the standard is to show the valence electrons as $\cdot \dot{	ext{P}}\colon$ or $\colon \dot{	ext{P}}\cdot$? Wait, no. The electron configuration $3s^2 3p^3$ means 2 electrons in $3s$ (paired) and 3 in $3p$ (unpaired). So the electron - dot structure is $\dot{	ext{P}}\colon$ with two electrons (paired) and three single dots. So it can be written as $\cdot \dot{	ext{P}}\colon$ (but the standard way is to have the paired electrons as a pair and the unpaired as single dots. So the electron - dot structure is $\dot{	ext{P}}\colon$ with one pair and three single dots, so $\cdot \dot{	ext{P}}\colon$ (or more precisely, $\colon \dot{	ext{P}}\cdot$? No, let's recall: for $3s^2 3p^3$, the $3s$ has two paired electrons (so a pair of dots) and $3p$ has three unpaired (three single dots). So the electron - dot structure is $\colon \dot{	ext{P}}\cdot$? Wait, no, the correct electron - dot structure for P (phosphorus) with electron configuration $[	ext{Ne}]3s^2 3p^3$ is $\dot{	ext{P}}\colon$ with two electrons (paired) and three unpaired, so the structure is $\cdot \dot{	ext{P}}\colon$ (showing the two in $3s$ as a pair and three in $3p$ as single dots around the symbol P).
# Answer: $\cdot \dot{	ext{P}}\colon$ (or $\colon \dot{	ext{P}}\cdot$ with the correct arrangement of 5 valence electrons)

#### Part b: $[	ext{Ar}]4s^2 3d^3$
# Explanation:
## Step1: Identify the element and valence electrons
The electron configuration is for vanadium (V). Valence electrons are in $4s^2 3d^3$, but in electron - dot structure, we consider the outermost shell valence electrons. However, for transition metals, sometimes we consider the $4s$ electrons as valence. But the electron - dot structure for vanadium (V) with electron configuration $[	ext{Ar}]4s^2 3d^3$: the valence electrons are $2 + 3 = 5$? Wait, no. The electron - dot structure for transition metals can be a bit tricky, but vanadium's electron - dot structure shows the valence electrons. The $4s^2$ electrons and $3d^3$ electrons. But the standard electron - dot structure for V is $\cdot \dot{	ext{V}}\cdot$? No, actually, the correct way is to show the valence electrons. The electron configuration is $[	ext{Ar}]4s^2 3d^3$, so the valence electrons are 5 (2 from $4s$ and 3 from $3d$). So the electron - dot structure is $\cdot \dot{	ext{V}}\colon$ (with two paired and three unpaired? No, better to represent as $\cdot \dot{	ext{V}}\cdot$ with five valence electrons (two in $4s$ and three in $3d$). But more accurately, the electron - dot structure for V is $\dot{	ext{V}}\colon$ with two electrons in $4s$ (paired) and three in $3d$ (unpaired), so the structure is $\cdot \dot{	ext{V}}\colon$ (or similar with five valence electrons).
# Answer: $\cdot \dot{	ext{V}}\colon$ (or a structure showing 5 valence electrons around V)

#### Part c: Potassium
# Explanation:
## Step1: Determine electron configuration and valence electrons
Potassium (K) has an atomic number of 19. Its electron configuration is $[	ext{Ar}]4s^1$. Valence electrons are 1 (in $4s^1$).
## Step2: Draw electron - dot structure
The electron - dot structure for K is $\cdot 	ext{K}$ (with one valence electron).
# Answer: $\cdot 	ext{K}$

### Question 5
#### Part a: Calcium
# Explanation:
## Step1: Determine atomic number and electron configuration
Calcium (Ca) has an atomic number of 20. The electron configuration follows the Aufbau principle. Filling orbitals: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$. Or in noble - gas notation, $[	ext{Ar}]4s^2$.
# Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ (or $[	ext{Ar}]4s^2$)

#### Part b: Zirconium
# Explanation:
## Step1: Determine atomic number and electron configuration
Zirconium (Zr) has an atomic number of 40. The electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^2$. In noble - gas notation, $[	ext{Kr}]5s^2 4d^2$.
# Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^2$ (or $[	ext{Kr}]5s^2 4d^2$)

#### Part c: Arsenic
# Explanation:
## Step1: Determine atomic number and electron configuration
Arsenic (As) has an atomic number of 33. The electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3$. In noble - gas notation, $[	ext{Ar}]4s^2 3d^{10} 4p^3$.
# Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3$ (or $[	ext{Ar}]4s^2 3d^{10} 4p^3$)

#### Part d: Rubidium
# Explanation:
## Step1: Determine atomic number and electron configuration
Rubidium (Rb) has an atomic number of 37. The electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1$. In noble - gas notation, $[	ext{Kr}]5s^1$.
# Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1$ (or $[	ext{Kr}]5s^1$)

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Identify valence electrons

Step2: Sum valence electrons

Step1: Determine valence electrons

Step2: Match electron - dot structure

Step1: Find valence electrons

Step2: Identify electron - dot structure

Answer:

Question 2