Question

Which has the same electronic configuration as the neon atom?
A.$N{a^ + }$
B. $Al$
C. $F$
D. $Ti$
E. $B$

Answer 1

Inert gases are not reactive because their outermost shell is complete. So they do not donate, accept or share electrons during chemical reaction. Therefore, they are called inert gases.

Complete step by step answer:
Let us write electronic configuration of given elements one by one.
(1) Neon $[Ne]$
It’s atomic number is $10.$
Electronic configuration, $1{s^2}2{s^2}2{p^6}$
(2) Sodium $[Na]$
Atomic number of sodium is$11.$
$\therefore$Number of electrons$= 11$
Sodium ion $[N{a^ + }],$is formed by losing one electron.
$\therefore$Number of electron in $N{a^ + } = 10.$
It’s electronic configuration, $1{s^2},2{s^2}2{p^6}$
(3) Aluminum $(Al)$
Atomic number is $13.$
It’s electronic configuration, $1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}.$
(4) Fluorine $(F)$
Atomic number is $9$.
It’s electronic configuration is, $1{s^2}2{s^2}2{p_x}^22{p_y}^22{p^1}.$
(5) Titanium $(Ti)$
Atomic number is $22.$
Total number of electrons$= 22.$
It’s electronic configuration is, $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^2}.$
(6) Boron [B]$\to$Atomic number is $5$.
Total number of electrons $= 5$
Its electronic configuration is, $1{s^2}2{s^2}2{p^1}$
From the above discussion we see that the number of electrons in $N{a^ + }$ and $Ne$ are same.
Therefore, the correct option is $N{a^{ + .}}$
Electronic configuration is the distribution of electrons in discrete orbitals. Electrons get filled according to the increasing order of their energies. Orbitals with less energy fills first. This is called Aufbau principle.

Therefore, by the above explanation, the correct option is [A]

Note: Except Helium, all inert gases have complete octet. Helium has $2$ electrons in its outermost orbit. Species with the same number of electrons are called isoelectronic species.
$Ne$ and $N{a^ + }$ are isoelectronic species.
$A{l^{ + + + }}$ is also isoelectronic species with $Ne,N{a^ + }.$
Similarly, $M{g^{2 + }},C{l^ - }$ are isoelectronic species.