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Chapter 8: Problem 13

Write the Lewis symbol for atoms of each of the following elements: (a) \(\mathrm{Al}\), (b) \(\mathrm{Br}\), (c) \(\mathrm{Ar}\), (d) \(\mathrm{Sr}\).

Short Answer

Expert verified
The Lewis symbols for the given elements are: (a) Al: \[\chemfig{Al(-[:0,.5]-[,0.5]H)(-[:90,.5]-[,0.5]H)(-[:180,.5]-[,0.5]H)}\] (b) Br: \[\chemfig{Br(-[:0,.5]-[,0.5]H)(-[:90,.5]-[,0.5]H)(-[:180,.5]-[,0.5]H)(-[:225,.5]-[,0.5]H)(-[:270,.5]-[,0.5]H)(-[:315,.5]-[,0.5]H)}\] (c) Ar: \[\chemfig{Ar(-[:0,.5]-[,0.5]H)(-[:45,.5]-[,0.5]H)(-[:90,.5]-[,0.5]H)(-[:135,.5]-[,0.5]H)(-[:180,.5]-[,0.5]H)(-[:225,.5]-[,0.5]H)(-[:270,.5]-[,0.5]H)(-[:315,.5]-[,0.5]H)}\] (d) Sr: \[\chemfig{Sr(-[:0,.5]-[,0.5]H)(-[:45,.5]-[,0.5]H)}\]

Step by step solution

01

Determine the number of valence electrons

Aluminium is in group 13 of the periodic table, which means it has 3 valence electrons.
02

Write the Lewis symbol

To write the Lewis symbol, place the chemical symbol in the center and arrange the three valence electrons around it, with two on one side and one on the other side. Lewis symbol for Aluminium (Al) is: \[ \chemfig{Al(-[:0,.5]-[,0.5]H)(-[:90,.5]-[,0.5]H)(-[:180,.5]-[,0.5]H)} \] (b) Bromine (Br)
03

Determine the number of valence electrons

Bromine is in group 17 of the periodic table, which means it has 7 valence electrons.
04

Write the Lewis symbol

To write the Lewis symbol, place the chemical symbol in the center and arrange the seven valence electrons around it, with four placed on one side and three on the other side. Lewis symbol for Bromine (Br) is: \[ \chemfig{Br(-[:0,.5]-[,0.5]H)(-[:90,.5]-[,0.5]H)(-[:180,.5]-[,0.5]H)(-[:225,.5]-[,0.5]H)(-[:270,.5]-[,0.5]H)(-[:315,.5]-[,0.5]H)} \] (c) Argon (Ar)
05

Determine the number of valence electrons

Argon is in group 18 of the periodic table, which means it has 8 valence electrons.
06

Write the Lewis symbol

To write the Lewis symbol, place the chemical symbol in the center and arrange the eight valence electrons around it. As Argon is a noble gas, it has a full octet and doesn't usually form bonds with other elements. Lewis symbol for Argon (Ar) is: \[ \chemfig{Ar(-[:0,.5]-[,0.5]H)(-[:45,.5]-[,0.5]H)(-[:90,.5]-[,0.5]H)(-[:135,.5]-[,0.5]H)(-[:180,.5]-[,0.5]H)(-[:225,.5]-[,0.5]H)(-[:270,.5]-[,0.5]H)(-[:315,.5]-[,0.5]H)} \] (d) Strontium (Sr)
07

Determine the number of valence electrons

Strontium is in group 2 of the periodic table, which means it has 2 valence electrons.
08

Write the Lewis symbol

To write the Lewis symbol, place the chemical symbol in the center and arrange the two valence electrons on one side. Lewis symbol for Strontium (Sr) is: \[ \chemfig{Sr(-[:0,.5]-[,0.5]H)(-[:45,.5]-[,0.5]H)} \]

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Most popular questions from this chapter

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity \(=k(I-E A)\), where \(k\) is a proportionality constant. (a) How does this definition explain why the electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{Cl}\) even though \(\mathrm{Cl}\) has the greater electron affinity? (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity? (c) By using data in Chapter 7 , determine the value of \(k\) that would lead to an electronegativity of \(4.0\) for \(F\) under this definition. (d) Use your result from part (c) to determine the electronegativities of \(\mathrm{Cl}\) and \(\mathrm{O}\) using this scale. (e) Another scale for electronegativity defines electronegativity as the average of an atom's first ionization energy and its electron affinity. Using this scale, calculate the electronegativities for the halogens, and scale them so fluorine has an electronegativity of 4.0. On this scale, what is Br's electronegativity?

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround these atoms: (a) \(\mathrm{PH}_{3}\), (b) \(\mathrm{AlH}_{3}\), (c) \(\mathrm{N}_{3}^{-}\), (d) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), (e) \(\mathrm{SnF}_{6-}\)

Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\); (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P}\); (c) \(\mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\).

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+}\), and energy is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-}\). Yet \(\mathrm{CaO}\) is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of \(\mathrm{CaO}\) is large enough to overcome these processes. (b) \(\mathrm{CaO}\) is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or \(\mathrm{O}\). (d) The enthalpy of formation of \(\mathrm{CaO}\) is small. (e) \(\mathrm{CaO}\) is stable to atmospheric conditions.

(a) Write the electron configuration for the element titanium, Ti. How many valence electrons does this atom possess? (b) Hafnium, Hf, is also found in group 4B. Write the electron configuration for Hf. (c) Ti and Hf behave as though they possess the same number of valence electrons. Which of the subshells in the electron configuration of Hf behave as valence orbitals? Which behave as core orbitals?
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