In general, ionization energies increase across a period from left to right. Explain why the second ionization energy of Cr is higher, not lower, than that of Mn.

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Chemistry

Melissa Yoder

8 September, 18:03

In general, ionization energies increase across a period from left to right. Explain why the second ionization energy of Cr is higher, not lower, than that of Mn.

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Bennett · Answer 1

Answer:So this leads to the fact that second ionization energy of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

Explanation:

Ionization energy is the amount of energy that we require to remove an electron form an isolated gaseous atom.

As we move from left to right across a period electrons are added to the same outermost shell therefore the attraction between the electrons and nucleus increases since more number of negatively charged electron are attracted to the positively charged nucleus. This attraction leads to the decrease in atomic radii across a period and increase in ionization energy.

The increase in ionization energy occurs due to the fact that as the attraction between the nucleus and outermost electrons increases so the electrons are more tightly bound to the nucleus hence more amount of energy is required to ionize the electron which leads to increase in ionization energy.

The electronic configuration of Cr and Mn are:

Cr:[Ar]3d⁵4S¹

Mn:[Ar]3d⁵4S²

The electronic configuration of Cr and Mn after 1st ionization:

Cr:[Ar]3d⁵4S⁰

Mn:[Ar]3d⁵4S¹

The electronic configuration of Cr and Mn after 2nd ionization:

Cr:[Ar]3d⁴4S⁰

Mn:[Ar]3d⁵4S⁰

As we can see that that 3d orbital of Cr (Chromium) is half filled with 5 electrons in it and 4s orbital of Cr is also half-filled.

So when Cr is ionized for the first time then the electron from the half-filled 4s orbital will be removed. As the 1 electron present in outer most 4s orbital is removed so the 4s orbital now is completely vacant.

Now for the second ionization energy an electron ahs to be removed from half-filled 3d⁵ orbital. Hunds rule of maximum multiplicity states that the fully-filled or half-filled orbitals have maximum stability on account of symmetry and exchange energy.

So half-filled 3d⁵ orbital of Cr is very stable and hence to remove an electron from this would be require a lot of energy and hence the second ionization energy of chromium is higher than that of Manganese.

In case of Mn the 3d orbital is also half - filled as chromium but the 4s orbital contains two electrons. when we remove the first electron from this orbital then also there is 1 electron present in the 4s orbital. So for the second ionization of Mn the only electron left in 4s orbital will be removed as the removal of electron from a 4s orbital is much easier as it requires less amount of energy as compared to removal of a electron from stable half filled 3d orbital.

So this leads to the fact that second ionization energy of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.