Solution:- Atomic number of F is 9 and it has 9 electrons. It's electron configuration is:-
[tex]1s^2s2^22p^5[/tex]
Atomic number of [tex]F^-[/tex] is 9 and it has 10 electrons. This ion is formed when F accepts one electron to gain nearest noble gas electron configuration.
[tex]1s^2s2^22p^6[/tex]
Atomic number of O is 8 and it has 8 electrons. It's electron configuration is:-
[tex]1s^2s2^22p^4[/tex]
Atomic number for [tex]O^2^-[/tex] is 8 and it has 10 electrons. This ion is formed when O accepts two electrons to gain nearest noble gas electron configuration.
[tex]1s^2s2^22p^6[/tex]
Atomic number for Na is 11 and it has 11 electrons. It's electron configuration is:-
[tex]1s^2s2^22p^63s^1[/tex]
Atomic number for [tex]Na^+[/tex] is 11 and it has 10 electrons. It loses one electron from it's outer most shell to look like it's nearest noble gas.(Ne)
It's electron configuration is:-
[tex]1s^2s2^22p^6[/tex]
Atomic number of Ca is 20 and it has 20 electrons. It's electron configuration is:-
[tex]1s^22s^22p^63s^23p^64s^2[/tex]
Atomic number of [tex]Ca^2^+[/tex] is 20 and it has 18 electrons. It loses two electrons from it's outer most shell to look like it's nearest noble gas.(Ar)
It's electron configuration is:-
[tex]1s^22s^22p^63s^23p^6[/tex]
The electronic configuration is done based on the octet rule. The are 2 electrons in s orbital followed by 6 electrons in p orbital. The -ve charge imparts the gain of electron, and +ve charge imparts the loss of electrons.
(a) F , atomic number 9
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^5[/tex]
(b) The ion of [tex]\rm F^-[/tex] gains and electron and has 10 electrons.
The electronic configuration will be: [tex]\rm 1s^2\;2s^2\;2p^6[/tex]
(c) The atomic number of oxygen is 8, It has 8 electrons.
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^4[/tex]
(d) The ion form accepts 2 electrons. The total number of eelctrons will be 10.
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^6[/tex]
(e) The atomic number of Na is 11.
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^6\;3s^1[/tex]
(f) The ion of Na gains 1 electron. Total of 12 electrons are there.
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^6\;3s^2[/tex]
(g) The Ca has 20 electrons in its ground sate.
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^6\;3s^2\;3p^6\;4s^2[/tex]
(h) The ion of Ca loses 2 electrons and has 18 electrons in its octet.
Electronic configuration = [tex]\rm 1s^2\;2s^2\;2p^6\;3s^2\;3p^6[/tex]
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