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9 questions about energies and periodic trends

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1. Which process is energetically favorable: a) adding an electron to K to form K-; or b) losing an electron to form K+? Explain.

2. Using Slater's rules, determine the effective nuclear charge for a 4s and 3d electrons in vanadium and V2+. Discuss the relative sizes of the atomic/ionic radii based on your result.

3. Compare the electron affinity for group 17 elements. Explain the lack of trend.

4. Which of Na or Cl will have the greatest first ionization energy? Which will have the greatest second ionization energy? Explain.

5. Why is the second ionization energy of Cr higher than that of Mn?

6. Give the ground state electron configuration of a)Ca; b) Ga3+ c) Bi and d) Pb2+. Identify each as paramaganetic or diamagnetic.

7. Why do niobium and tantalum have the same atomic radii?

8. Write the chemical equation illustrating the second electron affinity for oxygen. Do you expect this process to be exothermic or endothermic. Explain.

9. Put the following atoms or ions in order of increasing size F-, Cl, P, S, Pb and Sn. Explain briefly.

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Solution Summary

In this solution we consider the solutions to 9 different problems related to periodic trends, including atomic size, radius, electron affinities, and ionization energies. Tricky questions regarding niobium and tantalum are also answered in depth. Slater's rules are also discussed. Don't miss this one!

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1. Which process is energetically favorable: a) adding an electron to K to form K-; or b) losing an electron to form K+? Explain.

B, since it is easier to lose an electron to form an octet than to gain electrons (since you need 7 to make an octet)

2. Using Slater's rules, determine the effective nuclear charge for a 4s and 3d electrons in vanadium and V2+. Discuss the relative sizes of the atomic/ionic radii based on your result.

See this reference for slater's rules: http://intro.chem.okstate.edu/WorkshopFolder/SlaterRule.html

We can use them to calculate vanadium

(1s2)(2s2,2p6)(3s2,3p6,3d3)(4s2)
For 4s2, 1 in the same group, 1*0.35. None to the right. 11 one group below, 11*0.85. 10 further below 10*1.00
23-(0.35+11*0.85+10) = 23-19.7 = 3.3 nuclear charge felt by 4s

For 3d, ignore the 4s electrons since they do not contribute.
10 electrons in same group, 10*0.35. 10 electrons to the left, 10*1.

23-(10*0.35+10) = 23-13.5 = 9.5

For V2+, there are no 4s electrons since it lost ...

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