
(a)
Interpretation:
The way by which the ionization energy of the main group elements influences their metallic character is to be determined.
Concept introduction:
Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.
(a)

Answer to Problem 9.1P
The metallic character of the main group elements decreases with an increase in ionization energy.
Explanation of Solution
On moving across the period in the periodic table, the size of the atoms decreases. Thus the outermost electrons in the atom are closer to the nucleus and are thus harder to be pulled out of the atom. Metals have the specific property of losing electrons. The easier it is for an element to lose an electron, the more is the metallic character of the element. Since a large value of the ionization energy implies more difficulty in extracting an electron from an atom, it thus also indicates a low metallic character as well. Hence, with an increase in the ionization energy of an element, the metallic character decreases.
The metallic character of the main group elements decreases with an increase in ionization energy.
(b)
Interpretation:
The way by which the atomic radius of the main group elements influences their metallic character is to be determined.
Concept introduction:
The atomic radius of an element is defined as the distance of the outermost electron in the atom from its nucleus.
The types of atomic radii are as follows:
1) Covalent radius – Covalent radius is calculated as one half of the distance of the two atoms of the same element that are covalently bonded to each other.
2) Van der Waals radius – Van der Waals radius is calculated as one half the distance between two nuclei of two atoms of the same element that are not bonded to each other.
3) Metallic radius – Metallic radius is calculated as one half the distance between the nuclei of two metallic atoms or ions in the metallic lattice.
(b)

Answer to Problem 9.1P
The metallic character of the main group elements increases with an increase in the atomic radius.
Explanation of Solution
In the periodic table, on moving across the period, the radius of the elements decreases. As the radius decreases, the distance of the outermost electrons from the nucleus of the atom decreases. At a smaller distance from the nucleus, the outermost electrons experience greater forces of attraction from the nucleus and hence are harder to be knocked out of the atom. The atoms of an element have a greater metallic character if they can lose their outermost electrons easily. Hence with an increase in the atomic radius of an element, the metallic character increases.
The metallic character of the main group elements increases with an increase in the atomic radius.
(c)
Interpretation:
The way by which the number of outer electrons of the main group elements influences their metallic character is to be determined.
Concept introduction:
The atomic radius of an element is defined as the distance of the outermost electron in the atom from its nucleus.
The types of atomic radii are as follows:
1) Covalent radius – Covalent radius is calculated as one half of the distance of the two atoms of the same element that are covalently bonded to each other.
2) Van der Waals radius – Van der Waals radius is calculated as one half the distance between two nuclei of two atoms of the same element that are not bonded to each other.
3) Metallic radius – Metallic radius is calculated as one half the distance between the nuclei of two metallic atoms or ions in the metallic lattice.
(c)

Answer to Problem 9.1P
The metallic character decreases with an increase in the number of outermost electrons on moving across a period in the periodic table.
Explanation of Solution
While moving across a period from left to right in the periodic table, the radius of the elements decreases. This happens because the increase in the number of electrons and the protons is the same, whereas on moving down a group in the periodic table, the outermost electrons due to electron shielding experience much lesser nuclear charge and hence are easily knocked out.
Thus while moving across a period, with the increase in the number of outermost electrons, the metallic character decreases due to a decrease in the atomic radius and hence an increase in the ionization potential.
The metallic character decreases with an increase in the number of outermost electrons on moving across a period in the periodic table.
(d)
Interpretation:
The way by which the effective nuclear charge of the main group elements influences their metallic character is to be determined.
Concept introduction:
The effective nuclear charge is the net nuclear charge an electron in an atom experiences. The electrons at the outermost orbitals experience lesser nuclear charge compared to the electrons in the inner orbitals. Thus the inner electrons shield the outer electrons from the attractive forces of the atomic nucleus.
The effective nuclear charge is calculated as follows:
Here,
(d)

Answer to Problem 9.1P
The metallic character of an element decreases with an increase in the effective nuclear charge.
Explanation of Solution
In an atom, as the effective nuclear charge experienced by the outermost electrons increases, the electrons experience more attraction from the nucleus. The electrons experiencing greater nuclear charge are more firmly held in the atom and are thus harder to be knocked out. Elements, in which the outermost electrons are difficult to be knocked out, have decreased metallic character. Therefore, an increase in the effective nuclear charge decreases the metallic character.
The metallic character of an element decreases with an increase in the effective nuclear charge.
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Chapter 9 Solutions
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- 个 Stuck x ctc xALE X A ALE × A ALE X Lab x (19: x - G www-awu.aleks.com/alekscgi/x/Isl.exe/10_u-lgNslkr7j8P3jH-1Q1g8NUi-mObka ZLx2twjEhK7mVG6PUUIO06 Chapter 12 HW 三 Question 26 of 39 (4 points) 1 Question Attempt: 1 of Unlimited Answer the following questions about the given alkane. Part: 0 / 2 Part 1 of 2 Give the IUPAC name. Skip Part 2 53°F Clear Check × Q Search hp hp 02arrow_forwardCalculate the equilibrium constant at 25.0 oC for the following equation. Cd(s) + Sn+2(aq) ↔Cd+2(aq) + Sn(s) Group of answer choices 3.11x104 1.95x1018 9.66x108 1.40x109arrow_forwardWhat is the pH at the cathode for the following cell written in line notation at 25.0 oC with a Ecell = -0.2749 V? Ni(s)|Ni+2(aq, 1.00 M)||H+1(aq, ?M)|H2(g, 1.00 atm)|Pt(s)arrow_forward
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- < Done 19:22 www-awu.aleks.com Chapter 12 HW Question 4 of 39 (2 points) | Question Attempt: 5 of Unlimited : .. LTE סוי 1 ✓ 2 ✓ 3 = 4 ✓ 5 ✓ 6 ✓ 7 ✓ 8 ✓ 9 = 10 11 ✓ 12 Consider the molecule (CH3)2CHCH2CHCн for the following questions. Part 1 of 2 Which of the following molecules is/are constitutional isomer(s) to (CH3)2CHCH2CH2CH3? Check all that apply. Part 2 of 2 (CH3),C(CH2)2CH3 CH3 H,C-CH-CH-CH, CH 3 None of the above. ☑ Which of the following molecules is/are identical molecules to (CH3)2CHCH2CH2CH₁₂? Check all that apply. CH3 H,C-CH-CH₂-CH2-CH, CH3(CH2)2CH(CH3)2 CH2-CH2-CH3 HỌC-CH=CH, 乂 ☑ а None of the above Check Save For Later Submit Assignment © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibilityarrow_forward18:11 LTE ا... US$50 off hotels is waiting for you Book now, hotels in Nashville are going fast QUTSLIVII 25 61 69 points) | QuestIVIT ALLēm... now Give the IUPAC name for each compound. Part 1 of 3 Part 2 of 3 X ☑ Х Check Save For Later Submit © 2025 McGraw Hill LLC. All Rights Reserved. TOMS CT US ...vacy Center | Accessibilityarrow_forwardDone 19:17 www-awu.aleks.com Chapter 12 HW Question 29 of 39 (6 points) | Question Attempt: 1 of Unlimited .III LTE סוי 27 28 = 29 30 31 32 = 33 34 35 Consider this structure. CH3CH2CH2 Part 1 of 3 3 CH2 CH2CH3 - C-CH2CH 3 H CH₂ Give the IUPAC name of this structure. 3-ethyl-3,4-dimethylheptane Part: 1/3 Part 2 of 3 Draw the skeletal structure. Skip Part < Check Click and drag to start drawing a structure. Save For Later Submit © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility Хarrow_forward
- 18:57 .III LTE www-awu.aleks.com Chapter 12 HW Question 31 of 39 (8 points) | Question Attem... Give the IUPAC name of each compound. Part 1 of 4 Part 2 of 4 Х Х Check Save For Later Submit © 2025 McGraw Hill LLC. All Rights Reserved. TOMS OF US vacy Center | Accessibilityarrow_forwardWhat is the missing reactant in this organic reaction? CH3-C-CH2-NH2 + R - CH3 O: 0 CH3-N-CH2-C-NH-CH2-C-CH3 + H2O Specifically, in the drawing area below draw the condensed structure of R. If there is more than one reasonable answer, you can draw any one of them. If there is no reasonable answer, check the No answer box under the drawing area. Note for advanced students: you may assume no products other than those shown above are formed. Explanation Check Click anywhere to draw the first atom of your structure. C © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accesarrow_forwardDone 18:17 • www-awu.aleks.com Chapter 12 HW Question 24 of 39 (4 points) | Question Attempt: 1 of Unlimited ▼ 20 ✓ 21 × 22 23 24 25 26 raw the structure corresponding to each IUPAC name. Part 1 of 2 .III LTE 22 27 28 סוי 29 29 3 A skeletal structure corresponding to the IUPAC name 3-ethyl-4-methylhexane. Part 2 of 2 Click and drag to start drawing a structure. A condensed structure corresponding to the IUPAC name 2,2,4- trimethylpentane. Click anywhere to draw the first atom of your structure. Check Save For Later Submit < Х ப: G © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility : Garrow_forward
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