
(a)
Interpretation:
Number of electrons must be gained/lost by lithium to achieve a noble gas configuration should be determined.
Concept Introduction:
Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. The electronic configuration of these noble gases are as follows:
the electronic configuration of He =
the electronic configuration of Ne =
the electronic configuration of Ar =
An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.
There are two types of ions, cationic and anionic.
Cations are formed by losing electrons, thus they have fewer electrons than protons and are positively charged.
Anions are formed by gaining electrons, thus they have more electrons than protons and are negatively charged.
For example, sodium (Na) atom has 11 electrons (

Answer to Problem 33P
1 electron must be lost.
Explanation of Solution
The electronic configuration of Li is
The nearest noble gas to Li is helium (He), whose electronic configuration is
Therefore, Li must lose one electron to achieve He atom's electronic configuration.
Because an electron is lost, the result is Li+ cation.
(b)
Interpretation:
Number of electrons must be gained/lost by iodine to achieve a noble gas configuration should be determined.
Concept Introduction:
Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. The electronic configuration of these noble gases are as follows:
the electronic configuration of He =
the electronic configuration of Ne =
the electronic configuration of Ar =
An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.
There are two types of ions, cationic and anionic.
Cations are formed by losing electrons;thus they have fewer electrons than protons and are positively charged.
Anions are formed by gaining electrons;thus they have more electrons than protons and are negatively charged.
For example, Fluorine (F) atom has 9 electrons (

Answer to Problem 33P
1 electron must be gained
Explanation of Solution
The electronic configuration of I is
The nearest noble gas to Li is xenon (Xe), whose electronic configuration is
Therefore, I must gain one electron to achieve Xe atom's electronic configuration.
Because an electron is lost, the result is I- anion.
(c)
Interpretation:
Number of electrons must be gained/lost by sulfur to achieve a noble gas configuration should be determined.
Concept Introduction:
Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. The electronic configuration of these noble gases are as follows:
the electronic configuration of He =
the electronic configuration of Ne =
the electronic configuration of Ar =
An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.
There are two types of ions, cationic and anionic.
Cations are formed by losing electrons, thus they have fewer electrons than protons and are positively charged.
Anions are formed by gaining electrons, thus they have more electrons than protons and are negatively charged.
For example, Fluorine (F) atom has 9 electrons (

Answer to Problem 33P
2 electrons must be gained
Explanation of Solution
The electronic configuration of S is
The nearest noble gas to S is argon (Ar), whose electronic configuration is
Therefore, S must gain two electrons to achieve Ar atom's electronic configuration.
Because two electrons are lost, the result is S2-anion.
(d)
Interpretation:
Number of electrons must be gained/lost by strontium (Sr) to achieve a noble gas configuration should be determined.
Concept Introduction:
Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. The electronic configuration of these noble gases are as follows:
the electronic configuration of He =
the electronic configuration of Ne =
the electronic configuration of Ar =
An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.
There are two types of ions, cationic and anionic.
Cations are formed by losing electrons, thus they have fewer electrons than protons and are positively charged.
Anions are formed by gaining electrons, thus they have more electrons than protons and are negatively charged.
For example, Sodium (Na) atom has 11 electrons (

Answer to Problem 33P
2 electrons must be lost
Explanation of Solution
The electronic configuration of Sr is
The nearest noble gas to Sr is krypton (Kr), whose electronic configuration is
Therefore, Sr must lose two electrons to achieve Kr atom's electronic configuration.
Because two electrons are lost, the result is Sr2+cation.
Want to see more full solutions like this?
Chapter 3 Solutions
General, Organic, and Biological Chemistry - 4th edition
- Experiment 27 hates & Mechanisms of Reations Method I visual Clock Reaction A. Concentration effects on reaction Rates Iodine Run [I] mol/L [S₂082] | Time mo/L (SCC) 0.04 54.7 Log 1/ Time Temp Log [ ] 13,20] (time) / [I] 199 20.06 23.0 30.04 0.04 0.04 80.0 22.8 45 40.02 0.04 79.0 21.6 50.08 0.03 51.0 22.4 60-080-02 95.0 23.4 7 0.08 0-01 1970 23.4 8 0.08 0.04 16.1 22.6arrow_forward(15 pts) Consider the molecule B2H6. Generate a molecular orbital diagram but this time using a different approach that draws on your knowledge and ability to put concepts together. First use VSEPR or some other method to make sure you know the ground state structure of the molecule. Next, generate an MO diagram for BH2. Sketch the highest occupied and lowest unoccupied MOs of the BH2 fragment. These are called frontier orbitals. Now use these frontier orbitals as your basis set for producing LGO's for B2H6. Since the BH2 frontier orbitals become the LGOS, you will have to think about what is in the middle of the molecule and treat its basis as well. Do you arrive at the same qualitative MO diagram as is discussed in the book? Sketch the new highest occupied and lowest unoccupied MOs for the molecule (B2H6).arrow_forwardQ8: Propose an efficient synthesis of cyclopentene from cyclopentane.arrow_forward
- Q7: Use compound A-D, design two different ways to synthesize E. Which way is preferred? Please explain. CH3I ONa NaOCH 3 A B C D E OCH3arrow_forwardPredict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forward(10 pts) The density of metallic copper is 8.92 g cm³. The structure of this metal is cubic close-packed. What is the atomic radius of copper in copper metal?arrow_forward
- Predict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forwardPredict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forwardQ3: Rank the following compounds in increasing reactivity of E1 and E2 eliminations, respectively. Br ca. go do A CI CI B C CI Darrow_forward
- Q5: Predict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2). H₂O דיי "Br KN3 CH3CH2OH NaNH2 NH3 Page 3 of 6 Chem 0310 Organic Chemistry 1 HW Problem Sets CI Br excess NaOCH 3 CH3OH Br KOC(CH3)3 DuckDuckGarrow_forwardQ4: Circle the substrate that gives a single alkene product in a E2 elimination. CI CI Br Brarrow_forwardPlease calculate the chemical shift of each protonsarrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningLiving By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHERChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning



