Rb has the least
Element Rb (Rubidium) has the lowest ionization energy among Rb, Na, C, and F. This is because as you move down a group on the periodic table, the ionization energy typically decreases due to the increase in atomic size. Rubidium is located below sodium (Na) in the same group, so it has a lower ionization energy.
Potassium (K) has a lower ionization energy than sodium (Na).
The ionization energy increase from sodium to fluorine.
K (lowest) Na Li B N (highest)
Na, sodium, should have the lowest first ionization energy of those four elements.
The element with the smallest first ionization energy is Francium, as it is located in Group 1 of the periodic table and has the largest atomic size. Among the elements listed, lithium (Li) would have the smallest first ionization energy as it is closer to the upper right of the periodic table compared to sodium (Na), potassium (K), and rubidium (Rb).
The amount of energy needed to remove the most loosely held electron is referred to as the ionization energy. It is the energy required to remove an electron from a gaseous atom or ion.
Sodium (Na) has the lowest first ionization energy in period 3.
The order from highest to lowest ionization energy is Cl > Al > Si > Na > S. This is because ionization energy generally increases from left to right and from bottom to top in the periodic table.
As we move from left to right across Period 3 from Na to Cl, electronegativity and first ionization energy generally increase. This is due to the increasing effective nuclear charge as electrons are added, causing a stronger attraction between the nucleus and outer electrons. Chlorine, being closer to the right of the period, has a higher electronegativity and first ionization energy compared to sodium.
I'm assuming you're referring to the problem Na(g)+Cl(g) -> Na+(g)+Cl-(g)Since, ionization energy is basically the amount of energy it takes to dislodge one electron from a neutral atom, Cl has to dislodge one electron to stabilize NaThus,Na -> Na++e-
Rb