
Why does fluorine have higher ionization energy than bromine?
Answer
480.3k+ views
Hint: We need to understand the definition of ionization energy and compare the ionization of both fluorine and bromine. Ionization energy is the amount of energy necessary to remove a valence electron from a gaseous atom or ion, causing it to obtain a positive charge when the electron is removed.
Complete answer:
Fluorine is in period two of the periodic table, which means it only has electrons in the first two energy levels in the ground state. Fluorine has greater ionization energy than bromine because its valence electrons are more strongly attracted to its positively charged atomic nuclei. This is due to the fact that fluorine atoms only have two energy levels, therefore there is no substantial shielding effect. As a result, removing a valence electron from fluorine necessitates greater ionization energy.
Bromine belongs to Period 4 of the periodic table. It has electrons in the first four energy levels in its ground state. To protect valence electrons from positively charged atomic nuclei, electrons have three energy states. This indicates that bromine maintains its valence electrons more loosely than fluorine, necessitating lower ionization energy to remove a valence electron.
Note:
It must be noted that when the ionization energies of atoms in the periodic table are compared, two periodic trends emerge that obey the laws of Coulombic attraction: Within a particular time period, ionization energy typically rises from left to right (that is, row). In general, ionization energy falls from top to bottom in a particular group (that is, column). The latter trend is caused by the outer electron shell moving away from the nucleus, with one inner shell added to each row as one travels down the column.
Complete answer:
Fluorine is in period two of the periodic table, which means it only has electrons in the first two energy levels in the ground state. Fluorine has greater ionization energy than bromine because its valence electrons are more strongly attracted to its positively charged atomic nuclei. This is due to the fact that fluorine atoms only have two energy levels, therefore there is no substantial shielding effect. As a result, removing a valence electron from fluorine necessitates greater ionization energy.
Bromine belongs to Period 4 of the periodic table. It has electrons in the first four energy levels in its ground state. To protect valence electrons from positively charged atomic nuclei, electrons have three energy states. This indicates that bromine maintains its valence electrons more loosely than fluorine, necessitating lower ionization energy to remove a valence electron.
Note:
It must be noted that when the ionization energies of atoms in the periodic table are compared, two periodic trends emerge that obey the laws of Coulombic attraction: Within a particular time period, ionization energy typically rises from left to right (that is, row). In general, ionization energy falls from top to bottom in a particular group (that is, column). The latter trend is caused by the outer electron shell moving away from the nucleus, with one inner shell added to each row as one travels down the column.
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