38. which scientist predicted element for places that were not yet occupied in the periodic table?
39. elements that share very similar chemical properties are most likely belong to the same because they have the same
40. circle the element that is larger in size for each pair?
a) k, ga b) pb, c c) br, cl d) li, na

41. circle the element that has larger ionization energy in each pair?
a) rb, sr b) ge, c c) br, at d) li, cs

42. state the symbols of all the metalloids:

43. what are the 3 characteristic of metals?

44. state the name of group listed below:
group 1
group 2
group 17
group 18

45. what are the elements in the d block and f block called?

46. identify the element for each:
a) 1s²2s²2p⁵
b) 1s²2s²2p⁶3s²3p³
c) 1s²2s²2p⁶3s²3p⁶4s²3d¹
d) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p²

47. what is the octet rule?

Question

38. which scientist predicted element for places that were not yet occupied in the periodic table?
39. elements that share very similar chemical properties are most likely belong to the same  because they have the same
40. circle the element that is larger in size for each pair?
a) k, ga  b) pb, c  c) br, cl  d) li, na

41. circle the element that has larger ionization energy in each pair?
a) rb, sr  b) ge, c  c) br, at  d) li, cs

42. state the symbols of all the metalloids:

43. what are the 3 characteristic of metals?

44. state the name of group listed below:
    group 1
    group 2
    group 17
    group 18

45. what are the elements in the d block and f block called?

46. identify the element for each:
    a) 1s²2s²2p⁵
    b) 1s²2s²2p⁶3s²3p³
    c) 1s²2s²2p⁶3s²3p⁶4s²3d¹
    d) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p²

47. what is the octet rule?

Accepted Answer

### Question 38
# Brief Explanations:
Dmitri Mendeleev is the scientist who predicted elements for unoccupied places in the periodic table while formulating the periodic law and creating the periodic table. He left gaps for elements yet to be discovered and predicted their properties.
# Answer:
Dmitri Mendeleev

### Question 39
# Brief Explanations:
Elements with similar chemical properties belong to the same group (or family) in the periodic table because they have the same number of valence electrons. The number of valence electrons determines an element's chemical reactivity and bonding behavior, so elements in the same group (vertical column) have the same valence electron configuration, leading to similar properties.
# Answer:
Group (or Family); number of valence electrons

### Question 40
# Explanation:
#### Part a) K, Ga
## Step1: Recall atomic size trends
Atomic size increases down a group and decreases across a period (from left to right). K (potassium) is in Group 1, Period 4; Ga (gallium) is in Group 13, Period 4. Since K is to the left of Ga in the same period, K has a larger atomic size.
## Step2: Circle K
Circle K.

#### Part b) Pb, C
## Step1: Recall atomic size trends
Atomic size increases down a group. Pb (lead) is in Group 14, Period 6; C (carbon) is in Group 14, Period 2. Since Pb is below C in the same group, Pb has a larger atomic size.
## Step2: Circle Pb
Circle Pb.

#### Part c) Br, Cl
## Step1: Recall atomic size trends
Atomic size increases down a group. Br (bromine) is in Group 17, Period 4; Cl (chlorine) is in Group 17, Period 3. Since Br is below Cl in the same group, Br has a larger atomic size.
## Step2: Circle Br
Circle Br.

#### Part d) Li, Na
## Step1: Recall atomic size trends
Atomic size increases down a group. Li (lithium) is in Group 1, Period 2; Na (sodium) is in Group 1, Period 3. Since Na is below Li in the same group, Na has a larger atomic size.
## Step2: Circle Na
Circle Na.

# Answer:
a) Circle K
b) Circle Pb
c) Circle Br
d) Circle Na

### Question 41
# Explanation:
Ionization energy is the energy required to remove an electron from an atom. It generally increases across a period (left to right) and decreases down a group (top to bottom).

#### Part a) Rb, Sr
## Step1: Analyze positions
Rb (rubidium) is in Group 1, Period 5; Sr (strontium) is in Group 2, Period 5. Across a period, ionization energy increases from left to right. So Sr has a larger ionization energy.
## Step2: Circle Sr
Circle Sr.

#### Part b) Ge, C
## Step1: Analyze positions
Ge (germanium) is in Group 14, Period 4; C (carbon) is in Group 14, Period 2. Ionization energy decreases down a group (since electrons are farther from the nucleus, easier to remove). So C has a larger ionization energy.
## Step2: Circle C
Circle C.

#### Part c) Br, At
## Step1: Analyze positions
Br (bromine) is in Group 17, Period 4; At (astatine) is in Group 17, Period 6. Ionization energy decreases down a group. So Br has a larger ionization energy.
## Step2: Circle Br
Circle Br.

#### Part d) Li, Cs
## Step1: Analyze positions
Li (lithium) is in Group 1, Period 2; Cs (cesium) is in Group 1, Period 6. Ionization energy decreases down a group. So Li has a larger ionization energy.
## Step2: Circle Li
Circle Li.

# Answer:
a) Circle Sr
b) Circle C
c) Circle Br
d) Circle Li

### Question 42
# Brief Explanations:
Metalloids (or semi - metals) are elements with properties intermediate between metals and non - metals. The metalloids and their symbols are: Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), and Polonium (Po).
# Answer:
B, Si, Ge, As, Sb, Te, Po

### Question 43
# Brief Explanations:
Metals have several characteristic properties. Three common ones are:
1. **Good Conductors**: Metals are good conductors of heat and electricity. The free - moving electrons in the metallic lattice allow for the easy transfer of heat and electric charge.
2. **Malleability and Ductility**: Metals are malleable, meaning they can be hammered or rolled into thin sheets, and ductile, meaning they can be drawn into wires. This is due to the ability of the metal atoms to slide past one another in the metallic crystal structure.
3. **Luster**: Metals have a shiny or metallic luster when freshly cut or polished. The free electrons in the metal can absorb and re - emit light, giving them this characteristic shine.
# Answer:
1. Good conductors of heat and electricity; 2. Malleable (can be hammered into sheets) and ductile (can be drawn into wires); 3. Have a metallic luster (shiny when polished)

### Question 44
# Brief Explanations:
- **Group 1**: Elements in Group 1 of the periodic table are called Alkali Metals. They have one valence electron and are highly reactive.
- **Group 2**: Elements in Group 2 are called Alkaline Earth Metals. They have two valence electrons and are also reactive, but less so than alkali metals.
- **Group 17**: Elements in Group 17 are called Halogens. They have seven valence electrons and are highly reactive non - metals, often forming salts with metals.
- **Group 18**: Elements in Group 18 are called Noble Gases. They have a full valence electron shell (except for helium, which has 2 electrons), making them very stable and generally unreactive.
# Answer:
Group 1: Alkali Metals
Group 2: Alkaline Earth Metals
Group 17: Halogens
Group 18: Noble Gases

### Question 45
# Brief Explanations:
Elements in the d - block of the periodic table (Groups 3 - 12) are called Transition Metals. They are characterized by the filling of d - orbitals in their electron configuration. Elements in the f - block are called Inner Transition Metals. They are further divided into Lanthanides (elements with atomic numbers 58 - 71, filling 4f orbitals) and Actinides (elements with atomic numbers 90 - 103, filling 5f orbitals).
# Answer:
Elements in the d - block: Transition Metals; Elements in the f - block: Inner Transition Metals (Lanthanides and Actinides)

### Question 46
# Explanation:
To identify an element from its electron configuration, we calculate the total number of electrons (which is equal to the atomic number for a neutral atom) and then find the element with that atomic number.

#### Part a) $1s^{2}2s^{2}2p^{5}$
## Step1: Calculate total electrons
Sum the exponents: $2 + 2+5=9$. The atomic number is 9, so the element is Fluorine (F).
## Step2: Identify the element
Fluorine (F).

#### Part b) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
## Step1: Calculate total electrons
Sum the exponents: $2 + 2+6 + 2+3 = 15$. The atomic number is 15, so the element is Phosphorus (P).
## Step2: Identify the element
Phosphorus (P).

#### Part c) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{1}$
## Step1: Calculate total electrons
Sum the exponents: $2+2 + 6+2+6 + 2+1=21$. The atomic number is 21, so the element is Scandium (Sc).
## Step2: Identify the element
Scandium (Sc).

#### Part d) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{2}$
## Step1: Calculate total electrons
Sum the exponents: $2+2 + 6+2+6 + 2+10+2 = 32$. The atomic number is 32, so the element is Germanium (Ge).
## Step2: Identify the element
Germanium (Ge).

# Answer:
a) Fluorine (F)
b) Phosphorus (P)
c) Scandium (Sc)
d) Germanium (Ge)

### Question 47
# Brief Explanations:
The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a stable electron configuration with eight valence electrons (an octet). This is similar to the electron configuration of noble gases (except for helium, which has a duet of 2 electrons). Atoms form chemical bonds (ionic or covalent) to satisfy this rule, as having eight valence electrons gives them a stable, low - energy state.
# Answer:
The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with 8 valence electrons (like noble gases, except He with 2).

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QUESTION IMAGE

Question

Explanation:

Question 38

Part a) K, Ga

Step1: Recall atomic size trends

Step2: Circle K

Part b) Pb, C

Step1: Recall atomic size trends

Step2: Circle Pb

Part c) Br, Cl

Step1: Recall atomic size trends

Step2: Circle Br

Part d) Li, Na

Step1: Recall atomic size trends

Step2: Circle Na

Answer:

Question 39