The first ionzation energy generally increases from left to right across a period. However, there are some exceptions to this trend. For example, boron has a smaller ionization energy than beryllium even though it lies to the right of it in the same period. Select the statement that explains this exception to the trend in ionization energy. Boron has an electron in a 2 p 2p orbital, whereas beryllium only has electrons in the 2 s 2s orbital. Both 2 s 2s electrons must be removed at the same time, so it takes less energy to remove the one electron from the 2 p 2p orbital in boron than to remove both electrons from the 2 s 2s orbital in beryllium. 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. Boron has an electron in a 2 p 2p orbital which is higher in energy than the 2 s 2s orbital. Consequently, it is easier to remove an electron from the 2 p 2p orbital in boron than from the 2 s 2s orbital in beryllium. The 2 p 2p electron in boron experience a greater effective nuclear charge than the 2 s 2s electrons in beryllium, meaning it is easier to remove the 2 p 2p electron from boron than it is to remove a 2 s 2s electron from beryllium.