Which of the following is incorrect order of property as indicated?

Correct option is D

Ionization potential (I.P): It is the energy required to take out the outermost electron from an isolated gaseous atom. The energy required to remove the first electron is called the first ionization potential. The more stable the atom, the higher is its I.P.
Smaller the size of the atoms, the higher is the I.P value.
It is harder to remove an electron from a stable electronic configuration- fully filled and half-filled.
Trends in the periodic table:
• Along the period (horizontally): Electrons keep adding in the same shell. Increased nuclear charge and number of electrons lead to a greater force of attraction between the nucleus and outermost electrons. Hence it requires more energy to remove the electrons from the outermost shell. Therefore, ionization energy increases along the period.
• Down the group (vertically): The electrons start getting placed in the higher energy shells. The distance between the outermost electrons and the nucleus increases and the force of attraction between them decreases. Owing to the easier removal of electrons from the outermost shell, the ionization energy decreases down the group.
Si, Al, Mg, Na all belong to the same period 3.
The trend in I.P value should be Si > Al > Mg > Na.However, it is seen that Mg has higher ionization potential than Al because:
Mg has electronic configuration 1s22s22p63s2 that has fully filled 3s orbital. Al has electronic configuration 1s22s22p63s23p1 and its easier to remove the 3p electron. Hence Mg has a higher ionisation potential than Al. Thus, the true order of the first ionisation potential is Na < Mg > Al < Si.
Electron Affinity: It is defined as the amount of energy released when an electron is added to a neutral atom to form a negatively charged ion. On moving down the group, the electron affinity generally decreases. This is because as you go down the group in a periodic table, new valence shells are added and therefore the atomic radius increases. The new orbital or the shell is far away from the nucleus which means the attraction between the positively charged nucleus and the new electron decreases. As the size of fluorine is smaller electron repulsion is more and in Br and I, nuclear influence is less. The increasing order of electron affinity is I < Br < F < Cl.
Ionic radii follow similar trends to atomic radii with one critical difference. Cations have very different radii than anions. Cations are all smaller than their neutral analogs while anions are all larger. This is easy to understand since cations have lost electrons. As a result they have both fewer electrons in the highest energy atomic orbitals farthest from the nuclei and the remaining electrons feel a stronger pull from the nucleus. Look at Na+ which is isoelectronic with Ne. It goes from being one of the "largest" atoms on the left-hand side of the periodic table to effectively one of the smallest with an electron configuration that is the same a neon (all the way on the right-hand side). Moreover, Ne has a nuclear charge of Z=10 and Na+ has a Z=11. Thus, the Na+ should be smaller than Ne. However as atomic radii and ionic radii are often defined differently than this comparison is difficult to make.
Conversely adding an electron to F to make F- also generates an ion that isoelectronic with Ne. However, now you have added electrons and kept the number of protons constant. Thus F- will be larger than Ne (and larger than Na+).
The general trends will continue to hold. From top to bottom of the periodic table ions will increase in radii. However, now left to right the radius is more of a function of the number of electrons. O2- will be larger than F- as both have 10 electrons but Z=8 for oxygen and Z=9 for fluorine.