
(a)
Interpretation:
The element that is less metallic between
Concept introduction:
The metallic character is characterized by the tendency of metals to lose their outermost valence shell electrons. Greater the ease of electron removal, higher will be the electropositivity of the corresponding elements and vice versa.
The metallic character increases from top to bottom in a group because the electrons are less tightly held by the nucleus of the atom and therefore removed easily.
The metallic character decreases from left to right in a period because the electrons are more tightly held with the nucleus of the atom and therefore the removal becomes difficult.
(b)
Interpretation:
The element that is less metallic between
Concept introduction:
The metallic character is characterized by the tendency of metals to lose their outermost valence shell electrons. Greater the ease of electron removal, higher will be the electropositivity of the corresponding elements and vice versa.
The metallic character increases from top to bottom in a group because the electrons are less tightly held by the nucleus of the atom and therefore removed easily.
The metallic character decreases from left to right in a period because the electrons are more tightly held with the nucleus of the atom and therefore the removal becomes difficult.
(c)
Interpretation:
The element that is less metallic between
Concept introduction:
The metallic character is characterized by the tendency of metals to lose their outermost valence shell electrons. Greater the ease of electron removal, higher will be the electropositivity of the corresponding elements and vice versa.
The metallic character increases from top to bottom in a group because the electrons are less tightly held by the nucleus of the atom and therefore removed easily.
The metallic character decreases from left to right in a period because the electrons are more tightly held with the nucleus of the atom and therefore the removal becomes difficult.

Want to see the full answer?
Check out a sample textbook solution
Chapter 8 Solutions
Chemistry: The Molecular Nature of Matter and Change
- :0: :0: Select to Add Arrows :0: (CH3)2NH :0: ■ Select to Add Arrows :0: :0: (CH3)2NH ■ Select to Add Arrowsarrow_forwardDraw the product of the following H action sequence. Ignore any inorganic byproducts formed. 1. (CH3CH2)2CuLi, THF 2. CH3Br Q Atoms, Bonds and Rings H Charges ㅁarrow_forwardPlease help me with this the problem is so confusingarrow_forward
- 14 Question (1 point) Disiamylborane adds to a triple bond to give an alkenylborane. Upon oxidation with OH, H2O2, the alkenylborane will form an enol that tautomerizes to an aldehyde. In the first box below, draw the mechanism arrows for the reaction of disiamylborane with the alkyne, and in the last box draw the structure of the aldehyde. 4th attempt Feedback i > 3rd attempt OH, H2O2 i See Periodic Table See Hintarrow_forwardanswer with mechanisms and steps. handwritten please!arrow_forwardHello I need some help with Smartwork. For drawing structure B, I know the correct answer is CH₃B₂, but when I try to type it in, it keeps giving me CH₄BH₃ instead. Do you know how I should write it properly? Should I use a bond or something else?arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY





