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Ch.8 - Basic Concepts of Chemical Bonding

Chapter 8, Problem 90b

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (b) Which element in NaAlH4 is the most electronegative? Which one is the least electronegative? Which element in NaAlH4 is the least electronegative?

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Hey everyone in this example, we need to identify the least electro negative atom from the set of elements below. We have selenium, lead, antimony and chlorine as our given atoms. So what we should do is first recall our trend for electro negativity on our periodic tables. And we should recall that electro negativity is going to increase as we go towards the top right of our periodic tables. And we want to recall that flooring is our most electro negative element And it's located in group seven a. Which we can say is around here Across period two of our periodic tables. We also want to recall that francie um is going to be the least electro negative element on our periodic tables. And we would recall that francie um is located across period seven And it's located in Group one a. of our periodic tables. Now what we want to do is consider the location for each of our given atoms. So beginning with selenium, we would recognize that selenium is located in Group six A. Of our periodic tables and across period four We would also take note of its atomic number, which in the periodic table is given as 34. Moving on to our next atom, we have led, we should recall that lead Is located in Group four A. Across Period six. And it's given with the atomic number 82. Moving on to our next atom were given antimony, We would recognize that antimony is located in group five a. Across period five. And it's given with the Atomic # 51. And lastly we have the atom chlorine which we would recognize is located in group seven a. Across period three. And it has the atomic # 17. So, thankfully all of our given atoms are in different period numbers. And that means to tell which atom is the most electro negative. We want to find the atom that has the lowest period number because it means it would be higher up on our periodic table corresponding to a higher electro negativity value. So looking at our period numbers, we would see that chlorine has the lowest period number being at period three, meaning it's higher up on our periodic tables in group seven a. And so it corresponds to a higher election negativity value. So we would say that chlorine is the most electro negative, but we need to identify the least electro negative atom. So we want to find the atom that has the highest period number, which on the other hand would be Period six here listed for our Adam lead. So we would see that because it's in period six, it means that it's lower on our periodic table corresponding to a lower value for election negativity. and so because it's in period six, we would say that lead is the least electro negative of our given atoms. And this would be our final answer to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. Otherwise I'll see everyone in the next practice video.
Related Practice
Textbook Question

The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of 1.24 D, and the Cl — O bond length is 160 pm. (b) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule?

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Textbook Question

(b) Using these partial charges and the atomic radii given in Figure 7.8, estimate the dipole moment of the molecule.

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Textbook Question

(c) The measured dipole moment of BrCl is 0.57 D. If you assume the bond length in BrCl is the sum of the atomic radii, what are the partial charges on the atoms in BrCl using the experimental dipole moment?

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Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion.

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Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (d) What is the formal charge on hydrogen in the polyatomic ion?

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Textbook Question

Structures A, B, and C show the connectivity of the atoms in three different molecules that are isomers of C3H4O. By completing the Lewis structures of these molecules, complete the information in the following table:

Isomer A Isomer B Isomer C

Number of single bonds

Number of double bonds

Number of triple bonds

Number of nonbonding pairs