
What is the order of ionization enthalpy for $ B,\text{ }Al $ and $ Ga $ ?
Answer
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Hint :We know that we know that ionization energy increases with the decrease in atomic radius across the period from left to right. But there are some exceptions: Due to higher penetration power of s-orbitals than p-orbitals, electron removal from s-orbital requires more energy.
Complete Step By Step Answer:
First let us understand what ionization energy is. Ionization energy is the minimum amount of energy required to remove an electron from an isolated neutral gaseous atom or molecule. Electrons in half-filled orbitals are stable; they face less repulsion and cannot be easily removed. Penetration is the ability of an electron to get close to the nucleus. It is the ability of the orbital to attract electrons. The penetration effect of s-orbital electrons is maximum due to its closeness to the nucleus than p-orbital.
Ionization energy increases with the decrease in size across the period. All the atoms given in the options belong to the second period. Their size decreases in the order, Due to higher penetration power of s-orbitals than p-orbitals, electron removal from s-orbital requires more energy. Therefore, ionization energy of electrons in s-orbital is greater than ionization energy of electrons in p-orbital. Electrons in half-filled orbitals are stable; they face less repulsion and cannot be easily removed. Therefore, ionization energy of electrons in half-filled configurations is greater.
$ B\text{ }<\text{ }Al\text{ }<\text{ }Ga $ The trend in ionization enthalpy is because there is decrease in Ionization enthalpy from $ B $ to $ Al $ due to increase in size and from $ Al $ to $ Ga. $ $ Ga $ has more ionization energy than $ Al $ due to ineffective screening effect.
Note :
Remember that it is very important to know the exceptions in periodic trends. Although in general ionization energy increases with decrease in atomic size across period as we move from left to right, the exceptions due to penetration effect of electrons in s-orbital and stability of half-filled configuration.
Complete Step By Step Answer:
First let us understand what ionization energy is. Ionization energy is the minimum amount of energy required to remove an electron from an isolated neutral gaseous atom or molecule. Electrons in half-filled orbitals are stable; they face less repulsion and cannot be easily removed. Penetration is the ability of an electron to get close to the nucleus. It is the ability of the orbital to attract electrons. The penetration effect of s-orbital electrons is maximum due to its closeness to the nucleus than p-orbital.
Ionization energy increases with the decrease in size across the period. All the atoms given in the options belong to the second period. Their size decreases in the order, Due to higher penetration power of s-orbitals than p-orbitals, electron removal from s-orbital requires more energy. Therefore, ionization energy of electrons in s-orbital is greater than ionization energy of electrons in p-orbital. Electrons in half-filled orbitals are stable; they face less repulsion and cannot be easily removed. Therefore, ionization energy of electrons in half-filled configurations is greater.
$ B\text{ }<\text{ }Al\text{ }<\text{ }Ga $ The trend in ionization enthalpy is because there is decrease in Ionization enthalpy from $ B $ to $ Al $ due to increase in size and from $ Al $ to $ Ga. $ $ Ga $ has more ionization energy than $ Al $ due to ineffective screening effect.
Note :
Remember that it is very important to know the exceptions in periodic trends. Although in general ionization energy increases with decrease in atomic size across period as we move from left to right, the exceptions due to penetration effect of electrons in s-orbital and stability of half-filled configuration.
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