Question

Which of the following ions has the same number of electrons as a krypton atom with atomic number 36?
A.Chlorine
B.Sodium
C.Rubidium
D.Xenon

Answer 1

Hint: Ions are charged atomic or molecular species. The amount of positive charge on an ion is equal to the number of electrons it has lost and the amount of negative charge on an ion is equal to the number of electrons it has gained.

Complete step by step answer:
A positively charged ion is called a cation, while a negatively charged ion is called an anion.
To find out the number of electrons in an ion, we need to add the amount of negative charge on an anion to its atomic number and subtract the amount of positive charge on a cation from its atomic number.
Ionic species of chlorine $({}^{17}Cl)$ is chloride ion $C{l^ - }$.
$Cl = [Ne]3{s^2}3{p^5}$
$C{l^ - } = [Ne]3{s^2}3{p^6}$
Number of electrons in $C{l^ - } = 17 + 1 = 18$.
Ionic species of sodium $({}^{11}Na)$ is $N{a^ + }$.
$Na = [Ne]3{s^1}$
$N{a^ + } = [Ne]3{s^0}$
Number of electrons in $N{a^ + } = 11 - 1 = 10$.
Ionic species of rubidium $({}^{37}Rb)$ is $R{b^ + }$.
$Rb = [Kr]5{s^1}$
$R{b^ + } = [Kr]5{s^0}$
Number of electrons in $R{b^ + } = 37 - 1 = 36$.
Xenon $(Xe)$ is a noble gas and has a very stable electronic configuration with complete octet. Though at times it forms a number of ions in +2, +4, +6 and +8 states. But mostly it occurs as $Xe$.
$Xe = [Kr]4{d^{10}}5{s^2}5{p^6}$
Number of electrons in $Xe = 54$.
Thus, rubidium ion $(R{b^ + })$ has the same number of electrons as a krypton atom.

Hence option C is correct.

Note:
 Atomic number is the number of protons present in the nucleus of an atom. In a neutral atom the number of protons is equal to the number of electrons, therefore, the number of electrons is also represented by the atomic number.