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Concept explainers
(a)
Interpretation:
The ground-state electron configuration of the aluminum atom is to be written. Valence electrons and core electrons in aluminum are to be stated.
Concept introduction:
The arrangement of electrons in atomic orbitals is called the atom’s ‘electron configuration’. The most stable electron configuration of an element is called the ‘ground state’ configuration. Valence electrons are the electrons occupying the highest energy shell while core electrons are the electrons occupying the remaining lower energy shells of an atom.
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Answer to Problem 1.40P
For aluminum, the group-state electron configuration is
Explanation of Solution
The
Aluminum has three valence electrons and ten core electrons.
(b)
Interpretation:
The ground-state electron configuration of the sulfur atom is to be written. Valence electrons and core electrons in sulfur are to be stated.
Concept introduction:
The arrangement of electrons in atomic orbitals is called the atom’s ‘electron configuration’. The most stable electron configuration of an element is called the ‘ground state’ configuration. Valence electrons are the electrons occupying the highest energy shell while core electrons are the electrons occupying the remaining lower energy shells of an atom.
![Check Mark](/static/check-mark.png)
Answer to Problem 1.40P
For the sulfur atom, the ground-state electron configuration is
Explanation of Solution
The atomic number of sulfur is
Sulfur has six valence electrons and ten core electrons.
(c)
Interpretation:
The ground-state electron configuration of the oxygen atom is to be written. Valence electrons and core electrons in oxygen are to be stated.
Concept introduction:
The arrangement of electrons in atomic orbitals is called the atom’s ‘electron configuration’. The most stable electron configuration of an element is called the ‘ground state’ configuration. Valence electrons are the electrons occupying the highest energy shell while core electrons are the electrons occupying the remaining lower energy shells of an atom.
![Check Mark](/static/check-mark.png)
Answer to Problem 1.40P
For the oxygen atom, the ground-state electron configuration is
Explanation of Solution
The atomic number of oxygen is
Oxygen has six valence electrons and two core electrons.
(d)
Interpretation:
The ground-state electron configuration of the nitrogen atom is to be written. Valence electrons and core electrons in nitrogen are to be stated.
Concept introduction:
The arrangement of electrons in atomic orbitals is called the atom’s ‘electron configuration’. The most stable electron configuration of an element is called the ‘ground state’ configuration. Valence electrons are the electrons occupying the highest energy shell while core electrons are the electrons occupying the remaining lower energy shells of an atom.
![Check Mark](/static/check-mark.png)
Answer to Problem 1.40P
For the nitrogen atom, the ground-state electron configuration is
Explanation of Solution
The atomic number of Nitrogen is
Nitrogen has five valence electrons and two core electrons.
(e)
Interpretation:
The ground-state electron configuration of the fluorine atom is to be written. Valence electrons and core electrons in fluorine are to be stated.
Concept introduction:
The arrangement of electrons in atomic orbitals is called the atom’s ‘electron configuration’. The most stable electron configuration of an element is called the ‘ground state’ configuration. Valence electrons are the electrons occupying the highest energy shell while core electrons are the electrons occupying the remaining lower energy shells of an atom.
![Check Mark](/static/check-mark.png)
Answer to Problem 1.40P
For the fluorine atom, the ground-state electron configuration is
Explanation of Solution
The atomic number of fluorine is
Fluorine has seven valence electrons and two core electrons.
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Chapter 1 Solutions
Organic Chemistry: Principles And Mechanisms: Study Guide/solutions Manual (second)
- Complete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forwardNonearrow_forward
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
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