Select the statement that explains this exception to the trend in ionization energy. O Boron has an electron in a 2p orbital, whereas beryllium only has electrons in the 2s orbital. Both 2s electrons must be
removed at the same time, so it takes less energy to remove the one electron from the 2p orbital in boron than to remove
both electrons from the 2s orbital in beryllium.
O Beryllium has a noble gas electron configuration, whereas boron is one electron away from achieving a noble gas
configuration. Consequently, it takes much less energy to remove one electron from boron and achieve a noble gas
configuration than it does to remove an electron from beryllium which is already in a noble gas configuration.
The 2p electron in boron experience a greater effective nuclear charge than the 2s electrons in beryllium, meaning it is
easier to remove the 2p electron from boron than it is to remove a 2s electron from beryllium.
Boron has an electron in a 2p orbital which is higher in energy than the 2s orbital. Consequently, it is easier to remove
an electron from the 2p orbital in boron than from the 2s orbital in beryllium.