The elements which have the tendency to form an ionic bond.
Answer to Problem 6RQ
The elements that have high difference in the value of electronegativity usually show a tendency to form an ionic bond.
Explanation of Solution
An ionic bond is a bond present between ions. The atoms involved in ionic bonds must have a high difference in electronegativity. The atoms with low electronegativity have a tendency to give out electrons and generate a positive ion and the atoms with high electronegativity have a tendency to take up electrons and generate a negative ion. The positive and the negative ions tend to attract together by the special type of attractive electrostatic force and that generates an ionic bond between them.
The elements that have high difference in the value of electronegativity usually show a tendency to form an ionic bond. As the metals being electropositive are present on the extreme left part of the periodic table and can form a positive ion by losing an electron whereas the nonmetals are present at the right side of the periodic table and can generate a negative ion by gaining electrons.
Thus, metals and nonmetals tend to form ionic bonds.
Conclusion:
The elements that have high difference in the value of electronegativity usually show a tendency to form an ionic bond.
Chapter 18 Solutions
Conceptual Physical Science Explorations
Additional Science Textbook Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Microbiology: An Introduction
Human Physiology: An Integrated Approach (8th Edition)
Human Biology: Concepts and Current Issues (8th Edition)
Campbell Biology in Focus (2nd Edition)
Anatomy & Physiology (6th Edition)
- I need correct answer not chatgptarrow_forwardWhat is the resistance (in (2) of a 27.5 m long piece of 17 gauge copper wire having a 1.150 mm diameter? 0.445 ΧΩarrow_forwardFind the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring). d. Ag dFe = 2.47 ×arrow_forward
- Find the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring). d Ag = 2.51 dFe ×arrow_forwardShow that the units 1 v2/Q = 1 W, as implied by the equation P = V²/R. Starting with the equation P = V²/R, we can get an expression for a watt in terms of voltage and resistance. The units for voltage, V, are equivalent to [? v2 v2 A, are equivalent to J/C ✓ X . Therefore, 1 = 1 = 1 A V1 J/s Ω V-A X = 1 W. . The units for resistance, Q, are equivalent to ? The units for current,arrow_forwardPlease solve and answer the question correctly please. Thank you!!arrow_forward
College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
University Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSON
Introduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Lecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON