The following are ionization energies for a third period element in kJ/mol: IE1:

Question

The following are ionization energies for a third period element in kJ/mol: IE1: 590, IE2: 1820, IE3L 2880, IE4: 12800. a. Which element could this be based on these ionization energies? Justify your answer b. Draw the PES diagram for this element c. Compare the size of the neutral version of this atom to its ion form. Which is larger and why? d. Arrange the following atoms and ions in order of increasing radius: S^2-, Ar and Ne. Explain your ranking in terms of nuclear charge and electronic structure.

Explanation / Answer

a) the element is Al . the reason is that first three ionisation energies are lower and not much difference but forth ionisation energy is very high . so the element can loose only 3 electrons in its outmost shell . so it is Al

c) neutral atom is Al and ion is Al+3 . Al size is larger than Al+3 size. the reason is that in cation numnber of protons are more than number of electrons . these protons strongly hold the less numnber of electrons . so the size of cation always smaller than neutral atom

size: Al > Al+3

d) increase in order of radius :   S^-2 > Ar > Ne .

the nuclear charge is very less in anion compared to cation and neutral . in S^-2 there are more electrons than protons so its nuclear charge is lesser than other two . so it is greater size. for S^-2 electrons are more than protons. less number of protons cannot hold good these electrons so size is large for anion

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