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Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory

Chapter 6, Problem 10

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

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Hey everyone R. Question here wants us to provide the name and formula of the compound form by magnesium and iodine. Now we know that magnesium has a plus two charge since it's in our group to A and iodine has a minus one charge since it's in our group seven A. And when we combine the two we can simply use our criss cross method and we end up with a formula of M. G. I. To. Now when we name this, per our naming rules are medal is going to be named as is which is magnesium and for our non metal which is I dine we're going to replace its ending of I N. E to I D. E. So our name for our compound is going to be magnesium. I died. So I hope this made sense. And let us know if you have any questions.
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The successive ionization energies for a second-period element are given. What is the identity of the element? (LO 6.8) Ea1 = 1402 kJ/mol Ea2 = 2856 kJ/mol Ea3 = 4578 kJ/mol Ea4 = 7475 kJ/mol Ea5 = 9445 kJ/mol Ea6 = 53,266 kJ/mol Ea7 = 64,630 kJ/mol (a) Be (b) C (c) N (d) F
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Which of the following processes will release the most energy? (LO 6.9) (a) (b) (c) (d)
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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.
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Textbook Question

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)

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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.
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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

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