Question

How do you write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each of the following?
(a) \[N\] (b) \[Rb\] (c) \[Br\]

Answer 1

Hint: The electronic configuration of an element is the portrayal of the course of action of the electrons in the electronic shells. Under every huge orbital there are various subshells and these subshells have degenerate orbitals in which the electrons are masterminded.

Complete step by step answer:
The electronic setup of an element is done based on the quantity of electrons present in each orbital and in the subshells of each orbital.
(a) \[N\]
Nitrogen has a place with Group \[5A\] . It implies that nitrogen has \[5\] valence electrons. It needs to finish its octet electron or have similar arrangement of a novel gas to turn into a stable particle. It would be liable to acknowledge \[3\] electrons to finish its octet. Thus, its particle structure would be \[{N^{3 - }}\] .
Since it surrendered an electron, it will have similar configuration with the previous noble gas which is Neon:1{s^2}2{s^2}2{p^6}.
(b) \[Rb\]
Rubidium has a place with Group \[1A\] . It implies that rubidium has \[1\] valence electrons. It needs to finish its octet electron or have a similar setup of a novel gas to turn into a stable particle. It would be liable to deliver \[1\] electron to finish its octet than acknowledge \[7\] more. Hence, its particle structure would be \[R{b^ + }\] .
Since it surrendered an electron, it will have similar arrangement with the past noble gas which is krypton:1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}
(c) \[Br\]
Bromine has a place with Group \[7A\] . It implies that bromine has \[7\] valence electrons. It needs to finish its octet electron or have similar configuration of a novel gas to turn into a stable particle. It would be prone to acknowledge \[1\] electrons to finish its octet. Along these lines, its particle structure would be \[B{r^ + }\] .
Since it surrendered an electron, it will have similar arrangement with the next noble gas which is krypton: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}$

Note: The electrons are topped off in the orbitals and in the individual orbitals as per their greatest limit. The main shell can have a limit of two electrons, the subsequent shell can have most extreme eight electrons, while the third one can have greatest eighteen electrons, etc. These electrons are masterminded in the distinctive subshells as per their most extreme limit. The \[s\] subshell can have \[2\] electrons, \[p\] subshells can have \[6\] electrons \[d\] can have \[10\] electrons, and \[f\] orbitals can have most extreme \[14\] electrons.