Skip to main content
Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory

Chapter 6, Problem 11

Which molecular scale image best represents the ionic com-pound that forms between cesium and chlorine? (Cesium is represented by red circles, and chlorine is represented by blue circles.) (LO 6.12) (a)

(b)

(c)

(d)

Verified Solution
Video duration:
1m
This video solution was recommended by our tutors as helpful for the problem above.
272
views
Was this helpful?

Video transcript

Hey everyone, we're asked to identify the molecular diagram that depicts the ionic compound that sodium and roaming form and we're told that sodium will equal. Are gray circles and roaming will equal are red circles first. Let's go ahead and write out our reaction. We were told that we had sodium react with bromine. When these two react, we end up with our ionic compound of sodium bromide. As we can see right here through our equation, we have a 1-1 ratio between sodium and roaming. Looking at our diagrams, it looks like answer choice B. And answer choice D. Cannot be our answer based solely on that 1 to 1 ratio. We see a 2 to 1 ratio in B and D. Now comparing A. N. C. We know that an ionic compound tends to be solid. So this means it will be arranged in an array. Now using this knowledge and applying it to our diagrams, it looks like answer choice A fits our description. Well now I hope that made sense. And let us know if you have any questions
Related Practice
Textbook Question
Which of the following processes will release the most energy? (LO 6.9) (a) (b) (c) (d)
336
views
Textbook Question
Elements that have large negative electron affinities generally have (LO 6.10) (a) high values for Zeff and a vacancy in a valence orbital. (b) low values for Zeff and a vacancy in a valence orbital. (c) high values for Zeff and filled valence orbitals. (d) low values for Zeff and filled valence orbitals.
442
views
Textbook Question

Predict the formula of the ionic compound that forms between potassium and sulfur. (LO 6.11) (a) KS (b) KS2 (c) K2S2 (d) K2S

828
views
Textbook Question
For a multielectron atom, a 3s orbital lies lower in energy than a 3p orbital because (LO 5.16) (a) a 3p orbital has more nodal surfaces than a 3s orbital. (b) an electron in a 3p orbital has a higher probability of being closer to the nucleus than an electron in a 3s orbital. (c) inner electrons shield electrons in a 3p orbital more effec-tively than electrons in a 3s orbital. (d) the energy of the electron can be spread between three 3p orbitals instead of only one 3s orbital.
1780
views
Textbook Question

Given the following information, construct a Born–Haber cycle to calculate the lattice energy of CaCl2(s). (LO 6.13)

Net energy change for the formation of CaCl2(s) form Ca(s) and Cl2(g) = -795.4 kJ/mol

Heat of sublimation for Ca(s) = +178 kJ/mol

Ei1 for Ca(s) = +590 kJ/mol

Ei2 for Ca(g) = +1145 kJ/mol

Bond dissociation energy for Cl2(g) = +243 kJ/mol

Eea1 for Cl(g) = -348.6 kJ/mol

(a) 2603 kJ/mol (b) 2254 kJ/mol (c) 2481 kJ/mo (d) 1663 kJ/mol

707
views
Textbook Question
Which element has the largest atomic radius? (LO 5.20) (a) Rb (b) Co (c) Mg d) As
916
views