
The first ionization for phosphorus is \[{\text{3}}{{\text{p}}^3}\], and that for sulfur is \[1005\,kJ/mol\].
Which is the best explanation for this difference in first ionization energy.
A. Sulphur has a larger nuclear charge than phosphorus
B. Sulphur has a larger number of inner electrons that create a greater shielding effect.
C. Electron repulsion in the \[{\text{3}}{{\text{p}}^{\text{4}}}\] in sulfur makes the electron easier to remove than the \[{\text{3}}{{\text{p}}^3}\] electron in phosphorus.
D. A sulfur atom is larger than a phosphorus atom.
Answer
557.1k+ views
Hint: In simple language, ionization can be defined as the amount of energy which is required to remove an electron from its valence shell. The energy required to remove the first electron from the valence shell is called the first ionization energy, energy required to remove the second electron from the valence shell is called the second ionization energy, energy required to remove the third electron from the valence shell is called the third ionization energy and so on.
Complete step by step solution:
Given that the first ionization energy for phosphorus is \[{\text{3}}{{\text{p}}^3}\], and the first ionization energy for sulfur is \[1005\,kJ/mol\]. We know that the phosphorus half-filled subshell of \[{\text{3}}{{\text{p}}^3}\] is a stable paramagnetic ally. Sulphur on the other hand has a configuration of \[{\text{3}}{{\text{p}}^{\text{4}}}\] due to removal of an electron is easier due to the repulsion.
Therefore, C is the correct option..
Additional information:
There are various factors which affect the ionization energy. These are:
The electrons are strongly attracted to it if the nucleus is positively charged. If an electron is near to the nucleus then there will be greater attraction than the electrons far. When there exists two electrons in the same orbital, the electrons experience repulsion. This repulsion creates some disturbances in the attraction of the nucleus. The ionization energy is less in the paired electrons than the unpaired electrons because in paired electrons, it can be easily removed.
Note: In the periodic table, the elements placed to the left have a much lower ionization energy. The elements placed to the right have a higher value of ionization energy.
Complete step by step solution:
Given that the first ionization energy for phosphorus is \[{\text{3}}{{\text{p}}^3}\], and the first ionization energy for sulfur is \[1005\,kJ/mol\]. We know that the phosphorus half-filled subshell of \[{\text{3}}{{\text{p}}^3}\] is a stable paramagnetic ally. Sulphur on the other hand has a configuration of \[{\text{3}}{{\text{p}}^{\text{4}}}\] due to removal of an electron is easier due to the repulsion.
Therefore, C is the correct option..
Additional information:
There are various factors which affect the ionization energy. These are:
The electrons are strongly attracted to it if the nucleus is positively charged. If an electron is near to the nucleus then there will be greater attraction than the electrons far. When there exists two electrons in the same orbital, the electrons experience repulsion. This repulsion creates some disturbances in the attraction of the nucleus. The ionization energy is less in the paired electrons than the unpaired electrons because in paired electrons, it can be easily removed.
Note: In the periodic table, the elements placed to the left have a much lower ionization energy. The elements placed to the right have a higher value of ionization energy.
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