
Which of the following elements will have the highest ionization energy?
A) $1s^2\;2s^2\;2p^6\;3s^1$
B) $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$
C) $1s^2\;2s^2\;2p^6\;3s^2\;3p^4$
D) $1s^2\;2s^2\;2p^6\;3s^2\;3p^1$
Answer
574.2k+ views
Hint: At first we have to know that ionization energy is defined as the minimum amount of energy is required to remove an loosely bound valence electron from an isolated gaseous system.
Complete answer:
Now we have to look forward to the answer-
> From the first electronic configuration $1s^2\;2s^2\;2p^6\;3s^1$, we see that one electron
of ‘3s’ orbital is easily removed as its preceding ‘p’ orbital is fulfilled [p- orbital contains maximum six electrons]. It becomes like a stable noble gas configuration which has all fulfilled orbitals.
> From the other configuration $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$ we see that ionization energy will be higher as the valence shell becomes a half filled which is a stable electronic configuration for the element.
> The third configuration $1s^2\;2s^2\;2p^6\;3s^2\;3p^4$ we can conclude that lower energy will be required to remove one valence electron that is ‘3p’ orbital electron because the configuration is not half filled or fully filled so one electron can be removed easily.
> From the last electronic configuration $1s^2\;2s^2\;2p^6\;3s^2\;3p^1$ we can say that, lower amount of energy is required to remove the last electron that is ‘3p’ orbital electron.
So from the above fact we have to clearly understand that the highest ionization energy is required for the electronic configuration - $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$
Therefore the correct answer is (B).
Note: This should be noted that from the electronic configuration we can find out the element.
Here the elements from their respective electronic configuration are:-
A) $1s^2\;2s^2\;2p^6\;3s^1$ - Sodium
B) $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$ - Phosphorous
C) $1s^2\;2s^2\;2p^6\;3s^2\;3p^4$ - Sulphur
D) $1s^2\;2s^2\;2p^6\;3s^2\;3p^1$ - Aluminium
From the above element increasing order of their first ionization energy as-
Sodium < Aluminum < Sulphur < Phosphorus
Complete answer:
Now we have to look forward to the answer-
> From the first electronic configuration $1s^2\;2s^2\;2p^6\;3s^1$, we see that one electron
of ‘3s’ orbital is easily removed as its preceding ‘p’ orbital is fulfilled [p- orbital contains maximum six electrons]. It becomes like a stable noble gas configuration which has all fulfilled orbitals.
> From the other configuration $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$ we see that ionization energy will be higher as the valence shell becomes a half filled which is a stable electronic configuration for the element.
> The third configuration $1s^2\;2s^2\;2p^6\;3s^2\;3p^4$ we can conclude that lower energy will be required to remove one valence electron that is ‘3p’ orbital electron because the configuration is not half filled or fully filled so one electron can be removed easily.
> From the last electronic configuration $1s^2\;2s^2\;2p^6\;3s^2\;3p^1$ we can say that, lower amount of energy is required to remove the last electron that is ‘3p’ orbital electron.
So from the above fact we have to clearly understand that the highest ionization energy is required for the electronic configuration - $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$
Therefore the correct answer is (B).
Note: This should be noted that from the electronic configuration we can find out the element.
Here the elements from their respective electronic configuration are:-
A) $1s^2\;2s^2\;2p^6\;3s^1$ - Sodium
B) $1s^2\;2s^2\;2p^6\;3s^2\;3p^3$ - Phosphorous
C) $1s^2\;2s^2\;2p^6\;3s^2\;3p^4$ - Sulphur
D) $1s^2\;2s^2\;2p^6\;3s^2\;3p^1$ - Aluminium
From the above element increasing order of their first ionization energy as-
Sodium < Aluminum < Sulphur < Phosphorus
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