In the given nuclear reaction, the element X is:

${}_{11}^{22}\mathrm{Na}$ → X + e⁺ + v

CONCEPT:

Beta decay is the radioactive decay process in which the proton is transformed into a neutron or vice versa. There are two types of beta decay processes that follow;

1. Beta-Plus decay
2. Beta-Negative decay.

1. Beta-Plus decay - In this process proton disintegrates into the neutron which results in a decrease in the atomic number of the given sample. It is written as,

 ${}_Z^A\mathrm{X}$→ ${}_{Z-1}^A\mathrm{Y}$​​ + e+ + v

Here, A is the mass number, Z is the atomic number and v is the electron neutrino.

2. Beta-Negative decay - In this process proton disintegrates into the neutron which results in an increase in the atomic number of the given sample. It is written as,

${}_Z^A\mathrm{X}$ → ${}_{Z+1}^A\mathrm{Y}$ + e^- + $\overline{v}$

Here, A is the mass number, Z is the atomic number, and​ $\overline{v}$​ is antineutrino.

CALCULATION:

Given nuclear reaction is 

${}_{11}^{22}\mathrm{Na}$ → X + e+ + v

Here we can see that it is the Beta-Plus decay because we are given e⁺ and v on the product side and according to the Beta-Plus decay, the proton disintegrates into the neutron which results in a decrease in the atomic number. Therefore in X the atomic number decrease we have;

As there is a decrease in Atomic Number it turns from (Z = 11 i.e. Na to Z = 10 i.e. Ne)

Following is the reaction.

${}_{11}^{22}\mathrm{Na}$ → ${}_{10}^{22}\mathrm{Ne}$+ e+ + v

Therefore, X = ${}_{10}^{22}\mathrm{Ne}$

Hence option 4) is the correct answer.