Chemistry, asked by apple1426, 1 year ago

the answer is option 4 but explain how. ..clearly. . pls guys ..help me. .. fast ...

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CoolestCat015: sorry am not sure of answer but i can add your query to priority group
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Answers

Answered by Moony15
2

In Option 1 and 4, it can be seen clearly that the outermost electrons are closer to the nucleus, so they will require more ionization energy than rest. Now, in option 1 the 1st ionization energy (I.E) will be higher than option 4 because s subshell is fully filled in option 1 but not in option 4. But as 2nd I.E is asked so in option 4 after the removal of one electron ( 1st ionization) the next electron is in another subshell, unlike in the 1st option where we can still get the last electron from s subshell. Note here that the 3s subshell is farther than the 2p so removal of an electron from a 2p subshell will need more energy. Moreover the 2p subshell is fully filled so extra energy will be required. So 2nd I.E of option 4 will be the greatest. I hope you could follow.


apple1426: can you say that in simple sentence
apple1426: plsss
Moony15: The energy required to remove one electron is IE 1 and the energy required to remove another electron is IE 2.
Moony15: According to your question IE 2 is asked so we can say that one electron has already been removed from the element. In option 4 the last electron is now in a new subshell ( I hope you know what that is) and it is closer to the nucleus by aufbau memory map (refer net for this)
Moony15: so the nucleus will try to not to give away the electron as its force of attraction on the electron is more ( than the force of attraction by the nucleus of the other options on their respective outermost electron because their electrons are farther away from them) . So more energy would have to be given that is more IE.
apple1426: tq
apple1426: ...
Moony15: You're welcome
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