Question

Question 10 4 pts Which of the following explanations best explains, from first principles, the fact that boron has a lower first ionization energy than beryllium? 0 Beryllium has more core electrons than boron. 0 Boron is more boring and hence it will be easier to remove the bored electrons from the shell. Electrons are more closely bound in beryllium because God invented the Octet Rule, such that beryllium is closer to an octet than boron. 0 In beryllium, an unpaired electron must be removed whereas in boron a paired electron will be removed. 0 In beryllium, an electron is removed from the 2s orbital, which is closer (and hence has a lower energy) than in boron, whereas electrons are removed from the 2p orbital.

Answer 1

Let's first write the elctronic configuration of Beryllium and Boron

Beryllium: 1s² 2s²

Boron :1s² 2s² 2p¹

Now, looking at general periodic trends it says that  boron should have a higher ionization energy than beryllium

BUT, since Beryllium has two electrons in its outermost orbital which is 2s orbital. The electrons in the 2s orbital are already paired and hence it is quite a stable configuration for the beryllium atom. Hence the first ionization enthalpy of beryllium is higher than that for boron. Also, we can say that Boron's valence electron (2p¹) is shielded by the 2s electrons, less energy is required to remove the the 2p electron(s) from a boron atom than is requried to remove the 2s electron from a beryllium atom.

Hence, option (5) would be the best explaination for the quesition.