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Hint: The ionization enthalpy or energy is the minimum amount of energy that needs to be supplied to an isolated gaseous atom in order to remove the most loosely bound electron to convert it into a gaseous cation. It is also called ionization potential.
Complete answer:
First let us understand ionization enthalpy. The ionization enthalpy or energy is the minimum amount of energy that needs to be supplied to an isolated gaseous atom in order to remove the most loosely bound electron to convert it into a gaseous cation. It will be called the first ionization enthalpy since only one electron will be removed. Similarly the enthalpy for removing a second electron from a mono-positive gaseous cation will be called second ionization enthalpy.
There are certain factors that govern the ionization enthalpy:
(i) Nuclear charge: Greater the nuclear charge, greater will be the ionization enthalpy since with more nuclear charge the electrons in the outermost shell will be more tightly bound. Hence the ionization enthalpy increases across a period.
(ii) Atomic radius: As the atomic radius increases the ionization enthalpy decreases since the electrostatic attractive force on the electrons present in the outermost shell will decrease as the distance will increase. Hence the ionization enthalpy decreases down a group.
(iii) Penetration effect of the orbitals: It is more difficult to remove an electron from an orbital that has more penetration effect than an orbital that has less penetration effect if both of them belong to the same shell. For the same shell, the penetration effect of orbitals is in the order: $s>p>d>f$
(iv) Screening effect of the inner shell electrons: With the increase in the screening effect of the inner electrons, the ionization enthalpy decreases. This is because the effective nuclear charge on the valence electrons decreases with increasing screening effect.
(v) Electronic configuration: The half-filled or completely filled orbitals have extra stability. Due to this, if we want to remove a valence electron which is present in a half-filled or completely filled orbital, then we need to supply more energy.
Keeping all these points in mind we will solve the question. Mg, Al, Si and P all belong to the third period. The atomic radius across a period decreases while the nuclear charge increases. Therefore P has the highest first ionization enthalpy.
So, the correct answer is “Option D”.
Note: Mg has greater first ionization enthalpy than Al since in Mg the electron needs to be removed from a fully filled s-orbital while in Al it is removed from the 3p orbital which is neither half-filled or fully filled.
Complete answer:
First let us understand ionization enthalpy. The ionization enthalpy or energy is the minimum amount of energy that needs to be supplied to an isolated gaseous atom in order to remove the most loosely bound electron to convert it into a gaseous cation. It will be called the first ionization enthalpy since only one electron will be removed. Similarly the enthalpy for removing a second electron from a mono-positive gaseous cation will be called second ionization enthalpy.
There are certain factors that govern the ionization enthalpy:
(i) Nuclear charge: Greater the nuclear charge, greater will be the ionization enthalpy since with more nuclear charge the electrons in the outermost shell will be more tightly bound. Hence the ionization enthalpy increases across a period.
(ii) Atomic radius: As the atomic radius increases the ionization enthalpy decreases since the electrostatic attractive force on the electrons present in the outermost shell will decrease as the distance will increase. Hence the ionization enthalpy decreases down a group.
(iii) Penetration effect of the orbitals: It is more difficult to remove an electron from an orbital that has more penetration effect than an orbital that has less penetration effect if both of them belong to the same shell. For the same shell, the penetration effect of orbitals is in the order: $s>p>d>f$
(iv) Screening effect of the inner shell electrons: With the increase in the screening effect of the inner electrons, the ionization enthalpy decreases. This is because the effective nuclear charge on the valence electrons decreases with increasing screening effect.
(v) Electronic configuration: The half-filled or completely filled orbitals have extra stability. Due to this, if we want to remove a valence electron which is present in a half-filled or completely filled orbital, then we need to supply more energy.
Keeping all these points in mind we will solve the question. Mg, Al, Si and P all belong to the third period. The atomic radius across a period decreases while the nuclear charge increases. Therefore P has the highest first ionization enthalpy.
So, the correct answer is “Option D”.
Note: Mg has greater first ionization enthalpy than Al since in Mg the electron needs to be removed from a fully filled s-orbital while in Al it is removed from the 3p orbital which is neither half-filled or fully filled.
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