7. to which element does each of the following abbreviated electron configurations refer? a. ne 3s¹ 3p¹ b. ar 4s¹ c. ar 4s² 3d¹⁰ 4p⁵ d. kr 5s² 4d¹⁰ 5p¹
8. how many valence electrons does each of the following atoms have? a. rubidium, z = 37 b. arsenic, z = 33 c. aluminum, z = 13 d. nickel, z = 28
9. how many 3d electrons are found in each of the following elements? a. nickel, z = 28 b. vanadium, z = 23 c. manganese, z = 25 d. iron, z = 26
10. for each of the following elements, indicate which set of orbitals is filled last. a. radium, z = 88 b. iodine, z = 53 c. gold, z = 79 d. lead, z = 82
11. give some similarities that exist among the elements of group viia.
12. which elements on the periodic table lose electrons most easily (metals or nonmetals)? why?
13. in each of the following sets of atoms, which element would be expected to have the highest ionization energy? a. cs, k, li b. ba, sr, ca c. i, br, cl d. mg, si, s

Question

Accepted Answer

### Question 7
#### a.
# Explanation:
## Step1: Analyze electron - configuration
The noble - gas core [Ne] has 10 electrons. Adding the electrons from $3s^{2}3p^{1}$ (2 + 1=3 more electrons), the total number of electrons is 10 + 3=13. The element with Z = 13 is aluminum (Al).
## Step2: Conclusion
The element is Al.
#### b.
# Explanation:
## Step1: Analyze electron - configuration
The noble - gas core [Ar] has 18 electrons. Adding the electron from $4s^{1}$ (1 more electron), the total number of electrons is 18+1 = 19. The element with Z = 19 is potassium (K).
## Step2: Conclusion
The element is K.
#### c.
# Explanation:
## Step1: Analyze electron - configuration
The noble - gas core [Ar] has 18 electrons. Then, $4s^{2}$ has 2 electrons, $3d^{10}$ has 10 electrons, and $4p^{5}$ has 5 electrons. The total number of electrons is 18+2 + 10+5=35. The element with Z = 35 is bromine (Br).
## Step2: Conclusion
The element is Br.
#### d.
# Explanation:
## Step1: Analyze electron - configuration
The noble - gas core [Kr] has 36 electrons. Then, $5s^{2}$ has 2 electrons, $4d^{10}$ has 10 electrons, and $5p^{1}$ has 1 electron. The total number of electrons is 36+2 + 10+1=49. The element with Z = 49 is indium (In).
## Step2: Conclusion
The element is In.

### Question 8
#### a.
# Explanation:
## Step1: Determine electron - configuration
Rubidium (Rb) has Z = 37. Its electron - configuration is $[Kr]5s^{1}$. The valence electron is in the 5s orbital, so it has 1 valence electron.
## Step2: Conclusion
1
#### b.
# Explanation:
## Step1: Determine electron - configuration
Arsenic (As) has Z = 33. Its electron - configuration is $[Ar]4s^{2}3d^{10}4p^{3}$. The valence electrons are in the 4s and 4p orbitals. The number of valence electrons is 2 + 3=5.
## Step2: Conclusion
5
#### c.
# Explanation:
## Step1: Determine electron - configuration
Aluminum (Al) has Z = 13. Its electron - configuration is $[Ne]3s^{2}3p^{1}$. The valence electrons are in the 3s and 3p orbitals. The number of valence electrons is 2+1 = 3.
## Step2: Conclusion
3
#### d.
# Explanation:
## Step1: Determine electron - configuration
Nickel (Ni) has Z = 28. Its electron - configuration is $[Ar]4s^{2}3d^{8}$. The valence electrons are in the 4s and 3d orbitals. The number of valence electrons is 2 + 8=10.
## Step2: Conclusion
10

### Question 9
#### a.
# Explanation:
## Step1: Recall electron - configuration
Nickel (Ni) has Z = 28. Its electron - configuration is $[Ar]4s^{2}3d^{8}$, so it has 8 3d electrons.
## Step2: Conclusion
8
#### b.
# Explanation:
## Step1: Recall electron - configuration
Vanadium (V) has Z = 23. Its electron - configuration is $[Ar]4s^{2}3d^{3}$, so it has 3 3d electrons.
## Step2: Conclusion
3
#### c.
# Explanation:
## Step1: Recall electron - configuration
Manganese (Mn) has Z = 25. Its electron - configuration is $[Ar]4s^{2}3d^{5}$, so it has 5 3d electrons.
## Step2: Conclusion
5
#### d.
# Explanation:
## Step1: Recall electron - configuration
Iron (Fe) has Z = 26. Its electron - configuration is $[Ar]4s^{2}3d^{6}$, so it has 6 3d electrons.
## Step2: Conclusion
6

### Question 10
#### a.
# Explanation:
## Step1: Determine electron - configuration
Radium (Ra) has Z = 88. Its electron - configuration is $[Rn]7s^{2}$. The last - filled orbitals are the 7s orbitals.
## Step2: Conclusion
7s
#### b.
# Explanation:
## Step1: Determine electron - configuration
Iodine (I) has Z = 53. Its electron - configuration is $[Kr]5s^{2}4d^{10}5p^{5}$. The last - filled orbitals are the 5p orbitals.
## Step2: Conclusion
5p
#### c.
# Explanation:
## Step1: Determine electron - configuration
Gold (Au) has Z = 79. Its electron - configuration is $[Xe]6s^{1}4f^{14}5d^{10}$. The last - filled orbitals are the 5d orbitals.
## Step2: Conclusion
5d
#### d.
# Explanation:
## Step1: Determine electron - configuration
Lead (Pb) has Z = 82. Its electron - configuration is $[Xe]6s^{2}4f^{14}5d^{10}6p^{2}$. The last - filled orbitals are the 6p orbitals.
## Step2: Conclusion
6p

### Question 11
# Explanation:
## Step1: Recall properties of Group VIIA elements
Elements in Group VIIA (halogens) have 7 valence electrons. They are non - metals. They have high electronegativities. They tend to gain one electron to form a stable octet. They form ionic compounds with metals and covalent compounds with non - metals.
## Step2: Summarize similarities
They have 7 valence electrons, are non - metals, have high electronegativities, and tend to gain an electron to form anions.
# Answer:
Elements in Group VIIA have 7 valence electrons, are non - metals, have high electronegativities, and tend to gain an electron to form anions.

### Question 12
# Explanation:
## Step1: Recall metallic and non - metallic properties
Metals have low ionization energies and tend to lose electrons to form cations because their valence electrons are relatively loosely held. Non - metals have high ionization energies and tend to gain electrons to form anions.
## Step2: Identify the answer
Metals lose electrons most easily because they have low ionization energies.
# Answer:
Metals; because they have low ionization energies.

### Question 13
#### a.
# Explanation:
## Step1: Recall ionization energy trend
Ionization energy increases up a group. Cs, K, and Li are in Group 1. Li is at the top of the group, so it has the highest ionization energy.
## Step2: Conclusion
Li
#### b.
# Explanation:
## Step1: Recall ionization energy trend
Ba, Sr, and Ca are in Group 2. Ionization energy increases up a group. Ca is at the top of the group, so it has the highest ionization energy.
## Step2: Conclusion
Ca
#### c.
# Explanation:
## Step1: Recall ionization energy trend
I, Br, and Cl are in Group 17. Ionization energy increases up a group. Cl is at the top of the group, so it has the highest ionization energy.
## Step2: Conclusion
Cl
#### d.
# Explanation:
## Step1: Recall ionization energy trend
Ionization energy generally increases across a period from left to right. Mg, Si, and S are in Period 3. S is the furthest to the right, so it has the highest ionization energy.
## Step2: Conclusion
S
# Answer:
a. Li
b. Ca
c. Cl
d. S

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QUESTION IMAGE

Question

Explanation:

Question 7

a.

Step1: Analyze electron - configuration

Step2: Conclusion

b.

Step1: Analyze electron - configuration

Step2: Conclusion

c.

Step1: Analyze electron - configuration

Step2: Conclusion

d.

Answer:

Question 12