1- Consider the following ionization energies for aluminum: Al(g) ? Al+(g) + e I

Question

1- Consider the following ionization energies for aluminum:
Al(g) ? Al+(g) + e IE1 = 580 kJ/mol
Al+(g) ? Al2+(g) + e IE2 = 1815 kJ/mol
Al2+(g) ? Al3+(g) + e IE3 = 2740 kJ/mol
Al3+(g) ? Al4+(g) + e IE4 = 11,600 kJ/mol

a.Account for the trend in the values of the ionization energies.

b.Explain the large increase between IE3 and IE4.

c.Which one of the four ions has the greatest electron affinity? Explain.

d.List the four aluminum ions in order of increasing size

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2- The electron affinity for sulfur is higher than that for oxygen, and the electron affinity for chlorine is higher than that for fluorine. How do you account for this?

Explanation / Answer

c.Which one of the four ions has the greatest electron affinity? Explain.

d.List the four aluminum ions in order of increasing siz

c. Al3+(g) ? Al4+(g) + e ion is the answer; the reason why is because it will have the electron configuration of a halogen, making it very easy for this particular ion to gain an electron to complete the octet, which means it will have the greatest electron affinity.

d. In order of increasing size: Al3+(g) ? Al4+(g) + e, Al2+(g) ? Al3+(g) + e, Al+(g) ? Al2+(g) + e , Al(g) ? Al+(g) + e

The reason why is because as successive electrons are removed, the ions become more positive, allowing the nucleus to have a tighter grip on the remaining electrons, which means that the fewer electrons, the smaller the ion will be. Hence, the ions having the least electrons will be the smallest ions and the ones with the most electrons will be the largest ions.

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