Concept explainers
(a)
Interpretation:
The percent ionic character for LiBr and NaCl bonds separately needs to be determined.
Concept introduction:
Several empirical equations have been proposed to calculate the percentage ionic character in the bond. It is important to note that no bond (even ionic) in chemistry is 100% ionic.
In a bond, the dipole moment can be calculated as follows:
Here,
µ = Dipole mment
δ = Partial charge
d = Distance
(b)
Interpretation:
The calculated ionic characters of LiBr and NaCl needs to be compared with given theoretical values of figure based on electronegativity values.
Concept introduction:
Several empirical equations have been proposed to calculate the percentage ionic character in the bond. It is important to note that no bond (even ionic) in chemistry is 100% ionic.
In a bond, the dipole moment can be calculated as follows:
Here,
µ = Dipole mment
δ = Partial charge
d = Distance
(c)
Interpretation:
The reason for equality of calculated percentage ionic values and given data based on electronegativity needs to be explained.
Concept introduction:
Several empirical equations have been proposed to calculate the percentage ionic character in the bond. It is important to note that no bond (even ionic) in chemistry is 100% ionic.
In a bond, the dipole moment can be calculated as follows:
Here,
µ = Dipole mment
δ = Partial charge
d = Distance
Want to see the full answer?
Check out a sample textbook solutionChapter 10 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)
- Which of these statements about resonance is true?(a) When you draw resonance structures, it is permissibleto alter the way atoms are connected.(b) The nitrate ion has one long N¬O bond and two shortN¬O bonds.(c) “Resonance” refers to the idea that molecules areresonating rapidly between different bonding patterns.(d) The cyanide ion has only one dominant resonancestructure.(e) All of the above are true.arrow_forward18.) Choose all the statements that are correct. (1) Like atomic size, electronegativity decreases going across a period and increases going down a group. (2) The second most electronegative element is chlorine. (3) Electronegativity is directly proportional to atomic number. (4) Like ionization energy, electronegativity increases going across a period and decreases going down a group. (5) Electronegativity is a measure of the ability of an atom to attract electrons and form a negative ion. (6) Electronegativity is a measure of the ability of an atom in a molecule to attract electrons go itself. (7) Electronegativity was first proposed by Linus Pauling. Group of answer choices (2) (1) (4) (3) (5) (7) (6)arrow_forwardUse principles of atomic structure to answer each of the following: (1] (a) The radius of the Ca atom is 197 pm; the radius of the Ca2* ion is 99 pm. Account for the difference. (b) The lattice energy of CaO(s) is –3460 kJ/mol; the lattice energy of K20 is –2240 kJ/mol. Account for the difference. (c) Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies. Element First lonization Energy (kJ/mol) Second lonization Energy (kJ/mol) K 419 3050 Ca 590 1140 (d) The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference.arrow_forward
- Rank the bonds in each set in order of decreasing bond length and decreasing bond strength: (a) Si-F, Si-C, Si-O; (b) N=N, N-N, N=N.arrow_forwardCyanogen (CN)2 is known as pseodohalogen because it has some properties like halogens. It is composed of two CN’s joined together.(i) Draw the Lewis structure for all the possible combination for (CN)2.(ii) Calculate the formal charge and determine which one of the structures that you have drawn is most stable.(iii) For the stable structure, determine the geometry around the two central atoms.(iv) For the stable structure, draw the dipole arrows for the bonds.(v) Base on the stable structure, determine the polarity of molecule and state your reason.arrow_forwardIn addition to ammonia, nitrogen forms three other hydrides: hydrazine (N2H4), diazene (N2H2), and tetrazene (N4H4).(a) Use Lewis structures to compare the strength, length, and order of the nitrogen-nitrogen bonds in hydrazine, diazene, and N2.(b) Tetrazene (atom sequence H2NNNNH2) decomposes above 08C to hydrazine and nitrogen gas. Draw a Lewis structure for tetrazene, and calculate ΔH°rxn for this decomposition.arrow_forward
- A resident expert on electronegativity comes up to visit with you. He makes two claims (seen below) about electronegativity with relation to covalent bonding. Is the expert correct or can you refute him with your knowledge of electronegativity? (a) If a diatomic molecule is made up of atoms X and Y, which have different electronegativities, the molecule must be polar. (b) The farther two atoms are apart in a bond, the larger the dipole moment will be.arrow_forwardWrite the Lewis dot symbols of the reactants and products in the followingreactions. (First balance the equations.)(a) Sr + Se → SrSe(b) Ca + H2 → CaH2(c) Li + N2 → Li3N(d) Al + S → Al2S3arrow_forwardWrite Lewis structures for the following molecules or ions. (Assign lone pairs, radical electrons, and atomic charges where appropriate.) (a) SbO43− (b) ICl6− (c) SO32- (d) HOBrOarrow_forward
- 19. :O: || :0-N- O: Which of the following statements, if true, would support the claim that the NO3 ion, represented above, has three resonance structures? (A) The NO3 ion is not a polar species. (B) The oxygen-to-nitrogen-to-oxygen bond angles are 90°. (C) One of the bonds in NO3 is longer than the other two. (D) One of the bonds in NO3¯ is shorter than the other two.arrow_forwardDraw Lewis diagrams for the following ions. In the formula the symbol of the central atom is given first. (Hint:The valence octet may be expanded for the central atom.)(a) BrO4 - (b) PCl6 - (c) XeF6+arrow_forward(c) Draw the orbital diagrams and Lewis symbols to depict the formation of Na* and CI ions from the atoms. Give the formula of the compound formed. (d) The predicted bond length for HF is 109 pm (the sum of the covalent radii of H, 37 pm and F. 72 pm), however the actual bond length for HF is shorter (92 pm). It was observed that the difference between predicted and actual bond lengths becomes smalleor going down the halogen group from HF to HI Describe these observationsarrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning