\(\frac{12 \mathrm{mg}}{24 \mathrm{g}}=0.5 \times 10^{-3} \mathrm{moles}\)
Energy per atom \(=7.646+15.035=22.68 \mathrm{eV}\)
\(22.68 \times 96.48=21.88 \times 10^{2} \mathrm{KJ} / \mathrm{mole}\)
E(needed) \(21.88 \times 10^{2} \times 0.5 \times 10^{-3}\)
\(10.94 \times 10^{-1}=1.094 \mathrm{KJ}\)
The ionization potential is the minimum amount of energy required to remove one electron from each atom in a mole of atoms in the gaseous state. The first ionization energy is the energy required to remove two, the ionization energy is the energy required to remove the atom's nth electron, after the (n −1) electrons before it have been removed. Trend-wise, ionization energy tends to increase while one progresses across a period because the greater number of protons (higher nuclear charge) attract the orbiting electrons more strongly, thereby increasing the energy required to remove one of the electrons. Ionization energy and ionization potentials are completely different. The potential is an intensive property and it is measured by "volt" ; whereas the energy is an extensive property expressed by "eV" or "kJ/mole".
As one progresses down a group on the periodic table, the ionization energy will likely decrease since the valence electrons are farther away from the nucleus and experience a weaker attraction to the nucleus's positive charge. There will be an increase of ionization energy from left to right of a given period and a decrease from top to bottom. As a rule, it requires far less energy to remove an outer-shell electron than an inner-shell electron. As a result, the ionization energies for a given element will increase steadily within a given shell, and when starting on the next shell down will show a drastic jump in ionization energy. Simply put, the lower the principal quantum number, the higher the ionization energy for the electrons within that shell. The exceptions are the elements in the boron and oxygen family, which require slightly less energy than the general trend. Helium has the highest ionization energy while Francium has the lowest.
Factor Influencing Ionization energy
Variation in ionization energies in a period and group may or may not be regular and can be influenced by the following factor.
(A)Size of the Atom :ionization energy decreases with increase in atomic size.
As the distance between the outer most electrons and the nucleus increases, the force of attraction between the valence shell electrons and the nucleus decreases. As a result, outer most electrons are held less firmly and a lesser amount of energy is required to knock them out.
For example, ionization energy decreases in a group from top to bottom with an increase in atomic size.
(B)Magnitude of Nuclear Charge:ionization energy increases with an increase in nuclear charge. This is due to the fact that with an increase in nuclear charge, the electrons of the outer most shell are more firmly held by the nucleus and thus greater amount of energy is required to pull out an electron from the atom.
For example, ionization energy increases as we move from left to right along a period due to the increase in nuclear charge.
(C)Shielding or screening effect :
The inner electronic shells act as a screen between the nucleus and the outer shell. This reduces the inward pull of the outer shell towards the nucleus. In other words, the electrons that lie between the nucleus and the valence (i.e. outermost) shell have a shielding or screening effect on the electrons of valence (i.e. outermost) shell. This is called the shielding effect. The larger the number of electrons in the inner shells, the greater will be the screening effect and lesser will be the nuclear attraction of the outermost electrons. Thus increase in the number of inner electrons will tend to decrease ionization energy. It should be carefully noted that electrons in the same or similar type of orbital do not shield each other. Thus the electrons in s-orbital will not shield each other. Similaly, electrons in px, py and pz orbital of the same subshell do not shield each other. It has also been found that the screening effect of
s-electrons is more than that of p-electron shoes effect is in turn more than that of a d-electron and so on.
(D)Penetration effect of the electron :
The ionization energy also depends on the type of electron which is removed. s, p, d and f electrons have orbitals with different shapes. An s electron penetrates close to the nucleus, and is therefore more tightly held than a p electron. Similarly a p-orbital electron is more tightly held than a d-orbital electron and a d-orbital electron is more tightly held than an f-orbital electron. All other factors being equal, ionization energies are in the order s>p>d>f .
For example, the ionization energy of aluminium is comparatively less than magnesium because the outer most electron is to be removed from a 3p-orbital (having lesser penetration effect) in aluminium where as in magnesium it will be removed from a 3s-orbital (having larger penetration effect) of the same energy level.
(E)Electronic Configuration :
If an atom has exactly half-filled or completely filled orbitals, then such an arrangement has extra stability.
The removal of an electron from such an atom requires more energy then expected. For example, the first ionization energy of beryllium is greater than boron because beryllium has an extra stable completely filled outer most 2s orbital while boron has a partially filled less stable outer most 2p-orbital.
Be \((Z=4) 1 s^{2}, 2 s^{2}\)
\(\mathbf{B}(\mathbf{Z}=5) 1 \mathbf{s}^{2}, 2 \mathbf{s}^{2}, 2 \mathbf{p}^{1}\)
Hence, the correct option is (A)
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