
The $ I{P_1} $ of $ O,S,F $ and $ Cl $ are in the order of:
(a) $ F > O > Cl > S $
(b) $ S > Cl > O > F $
(c) $ Cl > S > O > F $
(d) $ F > Cl > O > S $
Answer
533.4k+ views
Hint: Ionization potential is the minimum energy required to remove the electron which is loosely bonded, from an isolated neutral gaseous atom. It is also known as ionization energy or ionization enthalpy.
Complete step-by-step answer:
We will start with the definition of the first ionization potential:
First Ionization potential: First Ionization potential is the minimum energy required to remove the first loosely bonded electron from an isolated neutral gaseous atom. It is also known as ionization energy or ionization enthalpy.
Now, we will talk about how the ionization potential or ionization energy or ionization enthalpy varies:
In period: Along the period, as the size decreases, the removal of electrons ccea is more difficult. So, along the period, left to right, the ionization potential increases.
In group: Along the group, as the size increases, the removal of electrons is easier. So, along the group, top to bottom, the ionization potential decreases.
Now, comparing the ionization potential of the given elements, $ O,S,F $ and $ Cl $ .
From the above discussions of the trend, the higher ionization potential is of fluorine, as having the smallest size, the ionization potential of the chlorine is lower than fluorine and also, the ionization potential of the sulphur is lower than oxygen, because along the period, left to right, the ionization potential increases. As along the group, top to bottom, the ionization potential decreases, the ionization potential of oxygen is higher than sulphur and also the ionization potential of fluorine is higher than chlorine.
So, the correct order is $ F > O > Cl > S $ .
Hence, the correct option is (a) $ F > O > Cl > S $ .
Note:
Ionization potential is dependent on the atomic size, as the size of the atom is small, it has higher nuclear charge. So, due to the higher nuclear charge, it is difficult to remove an electron from the atom of small size, and for the larger size, nuclear charge is low. Hence, easier to remove an electron.
Complete step-by-step answer:
We will start with the definition of the first ionization potential:
First Ionization potential: First Ionization potential is the minimum energy required to remove the first loosely bonded electron from an isolated neutral gaseous atom. It is also known as ionization energy or ionization enthalpy.
Now, we will talk about how the ionization potential or ionization energy or ionization enthalpy varies:
In period: Along the period, as the size decreases, the removal of electrons ccea is more difficult. So, along the period, left to right, the ionization potential increases.
In group: Along the group, as the size increases, the removal of electrons is easier. So, along the group, top to bottom, the ionization potential decreases.
Now, comparing the ionization potential of the given elements, $ O,S,F $ and $ Cl $ .
From the above discussions of the trend, the higher ionization potential is of fluorine, as having the smallest size, the ionization potential of the chlorine is lower than fluorine and also, the ionization potential of the sulphur is lower than oxygen, because along the period, left to right, the ionization potential increases. As along the group, top to bottom, the ionization potential decreases, the ionization potential of oxygen is higher than sulphur and also the ionization potential of fluorine is higher than chlorine.
So, the correct order is $ F > O > Cl > S $ .
Hence, the correct option is (a) $ F > O > Cl > S $ .
Note:
Ionization potential is dependent on the atomic size, as the size of the atom is small, it has higher nuclear charge. So, due to the higher nuclear charge, it is difficult to remove an electron from the atom of small size, and for the larger size, nuclear charge is low. Hence, easier to remove an electron.
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