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 cwhich one of the four ions has the greatest electron affinity? explaindlist the four aluminum ions given in order of increasing size, and explain your ordering.
Ionization occurs whilst sufficient electric powered potential is utilized to an atom, that no rely if it is helpful, it somewhat is going to tension the electron, it somewhat is the damaging particle that flies in orbit around the nucleus, out of its orbit, it somewhat is termed a valence shellIf damaging potential is utilized, the nucleus will capture the unfastened electron into its valence shell/orbitWikipedia has greater special information, and mentions that the potential required to ionize an atom would desire to be greater suitable than the potential conserving the electron in orbityet another important part of point out is that an ionized atom would desire to have one greater or one much less electron than it may many times have.
cAl3+(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 affinitydIn 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 beHence, the ions having the least electrons will be the smallest ions and the ones with the most electrons will be the largest ions.
