Question

The species Ne, $N{a^ + }$ and ${F^ - }$ have the same:
(A) Number of protons
(B) Number of nucleons
(C) Number of electrons
(D) Mass number

Answer 1

Hint: We need to find the number of electrons of the given species. We will write the electronic configuration of the given species. Calculating the number of electrons, mass number and number of nucleons for the given species we will get the correct answer.


Complete solution:
 We know that the number of electrons of a given species is equal to the effective atomic number. We will calculate it for the species in the question.
For Neon which is Ne, the atomic number is ten. So the number of electrons in neon is $10$.
Now for sodium which is Na, the atomic number is eleven($11$). Writing the electronic configuration for this sodium atom we get
 $Na \to 1{s^2}2{s^2}2{p^6}3{s^1}$
 it is the electronic configuration of the sodium atom.
We have been given sodium ions. When it loses one outermost electron, it forms $N{a^ + }$ . The cation has ten electrons. The electronic configuration will look like as follows
$N{a^ + } \to 1{s^2}2{s^2}2{p^6}$ which can be written as $[Ne]$.
For fluorine which is F, the atomic number is nine($9$). Writing the electronic configuration for fluorine atom we get,
 $F \to 1{s^2}2{s^2}2{p^5}$
it is the electronic configuration of the fluorine atom.
We have been given fluorine ions. When it accepts one electron it forms ${F^ - }$. This anion has ten($10$) electrons.
Thus all the species $Ne,N{a^ + },{F^ - }$ have the same number of electrons.
Now the mass number of sodium is twenty-three, neon is twenty, fluorine is twenty so option d is incorrect
Now $N{a^ + }$ has eleven protons, Ne has ten protons and ${F^ - }$ has nine protons.So option a is incorrect
The number of nucleons is equal to the mass number so option b is incorrect.


Note: The species having the same number of electrons are termed as isoelectronic species. The total number of nucleons is equal to the sum of neutrons and protons for a particular species.