Explain the following variations in atomic or ionic radii: Fe>Fe2+>Fe3+ 1. The 4

Question

Explain the following variations in atomic or ionic radii:

Fe>Fe2+>Fe3+

1. The 4s valence electrons in Fe are on average                             [ Select ]                       ["closer", "farther"]          from the nucleus than the 3d electrons, so Fe is larger than Fe2+.

2. Because there are five 3d orbitals, in Fe2+ at least                             [ Select ]                       ["three orbitals", "two orbitals", "five orbitals", "four orbitals", "one orbital"]          must contain a pair of electrons.

3. Removing one more electron from Fe2+ to form Fe3+ significantly                            [ Select ]                       ["reduces", "increases"]          repulsion, increasing the nuclear charge experienced by each of the other d electrons and decreasing the size of the ion.

Explanation / Answer

1. The 4s electrons in Fe are the valence electrons which is farthest from the nucleus. On the other hand, Fe2+ is formed by removing the two 4s valence electrons from Fe. So, the 3d electrons become the valence electrons in Fe2+ which is closer than the 4s valence electrons in Fe. Hence, answer is:

The 4s valence electrons in Fe are on average farther then the 3d electrons, so Fe is larger than Fe2+.

2. There are six 3d electrons in Fe2+. So, atleast one of the five 3d orbitals in Fe2+ will contain a pair of electrons. Hence, the answer is:

Because there are five 3d orbitals, in Fe2+ atleast one orbital must contain a pair of electrons.

3. When an electron is removed from Fe2+, Fe3+ is formed. This results in decreasing the number of electrons in 3d orbitals and increases the nuclear charge. So, the internuclear repulsion is reduced and the attraction of the nucleus with the 3d electrons increases and decreases the radius of Fe3+ ion. Hence, the answer is:

Removing one more electron from Fe2+ to form Fe3+reduces repulsion, increasing the nuclear charge experienced by each of the other d electrons and decreasing the size of the ion.

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