The ionization energies (IE₁ to IE₅) of ‘s’ and/or ‘p’ block elements (X, Y and Z) are given below.

The number of valence electrons in X, Y and Z are​

Correct option is C

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron of an isolated gaseous atom, positive ion, or molecule. The first ionization energy is quantitatively expressed as

where X is any atom or molecule, X⁺ is the resultant ion when the original atom was stripped of a single electron, and e^- is the removed electron. Roughly speaking, the closer the outermost electrons are to the nucleus of the atom, the higher the atom's ionization energy.
The nth ionization energy refers to the amount of energy required to remove the most loosely bound electron from the species having a positive charge of (n − 1). For example, the first three ionization energies are defined as follows:

​Ionization energy generally increases from left to right within a given period (that is, row). As the nuclear charge of the nucleus increases across the period, the electrostatic attraction increases between electrons and protons, hence the atomic radius decreases, and the electron cloud comes closer to the nucleus because the electrons, especially the outermost one, are held more tightly by the higher effective nuclear charge.
X= 1s²2s²2p² (Four valence electrons)
Fifth electron will be removed from 1s², resulting in very high IE₅ value
Y= 1s² 2s² 2p¹ (Three valence electrons)
Fourth electron will be removed from 1s², resulting in very high IE₄ value
Z= 1s² 2s¹ (One valence electron)
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