
Which of following item has the highest first ionization energy:
A. \[Na\]
B. \[K\]
C. \[Sc\]
D. \[Rb\]
Answer
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Hint: The first ionization energy alludes to the energy it takes to eliminate one electron from an atom. It basically relies upon the electrostatic fascination between the positive protons in the core and negative electrons—on the off chance that protons and electrons are more pulled in to one another, at that point the main ionization energy would be higher, and the other way around.
Complete step by step answer:
Alkali metals have very low ionization energies because they have only one valence electron to dose before attaining noble gas configuration.
There is a difference between the two: a neutral \[K\] atom has 19 protons and 19 electrons, while a neutral \[Sc\] atom has 21 protons and 21 electrons.
So, the magnitude of attraction between the positively charged protons and negatively charged electrons is greater for \[Sc\] than it is for \[K\]. As such, it will take more energy to remove one electron-which is why it has a higher first ionization energy.
This fits with the trend of ionization energy increasing as we move right, across a period, too!
Option 1st: \[Na\], This Option is incorrect
Option 2nd: \[K\], This Option is incorrect
Option 3rd: \[Sc\], This Option is correct. So, scandium is the only element that is not in the first group and has the highest I.E among all.
Option 4th: \[Rb\], This Option is incorrect
So, the correct option is C. $Sc$
Note:
The energy expected to eliminate at least one electron from an impartial iota to shape a decidedly charged particle is a physical property that impacts the synthetic conduct of the molecule. By definition, the primary ionization energy of a component is the energy expected to eliminate the furthest, or most noteworthy energy, electron from a nonpartisan molecule in the gas stage.
Complete step by step answer:
Alkali metals have very low ionization energies because they have only one valence electron to dose before attaining noble gas configuration.
There is a difference between the two: a neutral \[K\] atom has 19 protons and 19 electrons, while a neutral \[Sc\] atom has 21 protons and 21 electrons.
So, the magnitude of attraction between the positively charged protons and negatively charged electrons is greater for \[Sc\] than it is for \[K\]. As such, it will take more energy to remove one electron-which is why it has a higher first ionization energy.
This fits with the trend of ionization energy increasing as we move right, across a period, too!
Option 1st: \[Na\], This Option is incorrect
Option 2nd: \[K\], This Option is incorrect
Option 3rd: \[Sc\], This Option is correct. So, scandium is the only element that is not in the first group and has the highest I.E among all.
Option 4th: \[Rb\], This Option is incorrect
So, the correct option is C. $Sc$
Note:
The energy expected to eliminate at least one electron from an impartial iota to shape a decidedly charged particle is a physical property that impacts the synthetic conduct of the molecule. By definition, the primary ionization energy of a component is the energy expected to eliminate the furthest, or most noteworthy energy, electron from a nonpartisan molecule in the gas stage.
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