Question

Which has larger (IE) in each pair?
A.$N$ or ${O^ + }$
B.$B{e^ + }$ or ${B^ + }$
C.$Be$ or $B$
D.$N{a^ + }$ or $Ne$
Give your answer as a four digit number, example- $ 1212/2122$ and so on ($ 1$ for first element and $ 2$ for second)

Answer 1

Hint: Let’s first discuss what ionization energy is. Ionization energy refers to the amount of energy needed to remove an electron from an atom. Ionization energy decreases as we go down a group. Ionization energy increases from left to right across the periodic table. It can be used to help predict the strength of chemical bonds.

Complete answer:
There are two electrons in the first orbital of the $ O$ sublevel, there is greater electronic repulsion in the $ 2p$ sublevel for $ O$ than $ N$ . Therefore, it is easier to remove an electron from the $ O$ than the $ N$ , and the ionization energy of $ O$ is lower than $ N$ . Also the effective nuclear charge of $ {O^ + }$ . Therefore, the correct order is $ N < {O^ + }$ .
When an electron is removed from $ {B^ + }$ , stable $ 2{s^2}$ electronic configuration is broken. so, the correct order is $ {B^ + } < B{e^ + }$
The first ionization energy of beryllium is greater than that of boron because beryllium has a stable complete electronic configuration so it requires more energy to remove the first electron from it. So, the correct order is $ Be > B$
The ionization enthalpy of $ N{a^ + }$ ion is more than twice that of $ Ne$ . The $ N{a^ + }$ ion has a higher effective nuclear charge than NE. So, the electrons in the outermost shell are more tightly bound to the nucleus in $ N{a^ + }$ ion than in the neutral $ Ne$ atom. higher will be the ionization energy, so the correct order is $ N{a^ + } > Ne$
Hence, the correct answer is $ 2211$

Note:
$ N$ and $ {O^ + }$ are isoelectronic species. Isoelectronic species are known as atoms or ions that have the same number of electrons. In isoelectronic species, there the number of electrons would be equal but the elements would be different.