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Chapter 4: Problem 165

Arrange in order of increasing valence-shell size: \(\mathrm{Sr}, \mathrm{Mg}, \mathrm{Ba}, \mathrm{Ca}, \mathrm{Be}\).

Short Answer

Expert verified
The elements from Group 2 in the periodic table increase in valence-shell size as we go down the group. Their positions within the group are as follows: Be (2nd period), Mg (3rd period), Ca (4th period), Sr (5th period), and Ba (6th period). Therefore, the order of increasing valence-shell size for the given elements is: \(\mathrm{Be} < \mathrm{Mg} < \mathrm{Ca} < \mathrm{Sr} < \mathrm{Ba}\).

Step by step solution

01

Find the placement of each element in the periodic table

Locate the positions of each of the given elements from Group 2 in the periodic table. This will aid in determining which ones are placed higher and which ones are lower within the group.
02

Compare the elements within the group

Remember that as we go down a group, the valence-shell size increases. Compare the elements within Group 2 based on their placement and order them accordingly.
03

Arrange the elements in order of increasing valence-shell size

Organize the given elements in ascending order based on their valence-shell size, as gathered from observing their positions within Group 2 in the periodic table. From our analysis, we know that as we go down Group 2, the valence-shell size will increase. The placement of each element within Group 2 is as follows: - Be (Beryllium): 2nd period - Mg (Magnesium): 3rd period - Ca (Calcium): 4th period - Sr (Strontium): 5th period - Ba (Barium): 6th period Considering this information, we can now arrange the given elements in order of increasing valence-shell size: \(\mathrm{Be} < \mathrm{Mg} < \mathrm{Ca} < \mathrm{Sr} < \mathrm{Ba}\)

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Most popular questions from this chapter

What is the total number of \(p\) -subshell electrons for each of the following atoms: \(\mathrm{P}, \mathrm{Mg}\), Se, \(\mathrm{Zn} ?\)

Predict the formula of the compound formed by the reaction between: (a) \(\mathrm{Ca}\) and \(\mathrm{Br}\) (b) \(\mathrm{K}\) and \(\mathrm{N}\) (c) \(\mathrm{Al}\) and \(\mathrm{S}\) (d) \(\mathrm{Na}\) and \(\mathrm{I}\) (e) \(\mathrm{Mg}\) and \(\mathrm{O}\)

Circle the correct choice to indicate how many electrons each element must gain or lose to form an octet: \(\begin{array}{llll}\text { (a) } \mathrm{Mg} & \text { gains, loses } & 1,2,3 & \text { electrons }\end{array}\) \(\begin{array}{lll}\text { (b) Se gains, loses } 1,2,3 & \text { electrons }\end{array}\) (c) Al gains, loses \(1,2,3\) electrons \(\begin{array}{llll}\text { (d) Sr gains, loses } & 1,2,3 & \text { electrons }\end{array}\) (e) Br gains, loses \(1,2,3\) electrons (f) P gains, loses \(1,2,3\) electrons

Identify the period 2 element that is described by the ionization data below. \(\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+} 1 \mathrm{e}^{-} \quad \mathrm{IE}(1)=1.40 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{+}(\mathrm{g}) \rightarrow \mathrm{M}^{2+} 1 \mathrm{e}^{-} \mathrm{IE}(2)=2.86 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{2+}(\mathrm{g}) \rightarrow \mathrm{M}^{3+} 1 \mathrm{e}^{-} \mathrm{IE}(3)=4.58 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{3+}(\mathrm{g}) \rightarrow \mathrm{M}^{4+} 1 \mathrm{e}^{-} \mathrm{IE}(4)=7.48 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{4+}(\mathrm{g}) \rightarrow \mathrm{M}^{5+} 1 \mathrm{e}^{-} \mathrm{IE}(5)=9.44 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{5+}(\mathrm{g}) \rightarrow \mathrm{M}^{6+} 1 \mathrm{e}^{-} \mathrm{IE}(6)=5.33 \times 10^{4} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{6+}(\mathrm{g}) \rightarrow \mathrm{M}^{7+} 1 \mathrm{e}^{-} \mathrm{IE}(7)=6.44 \times 10^{4} \mathrm{~J} / \mathrm{mol}\)

Predict the formula of the compound that forms when lithium atoms react with nitrogen atoms. Completely explain your reasoning.
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