Question

The size of isoelectronic species${F^ - },Ne$ and $N{a^ + }$ is affected by
(i) Principal Quantum Number $\left( n \right)$
(ii) Electron-electron interaction in the outer orbital
(iii) Nuclear Charge $\left( Z \right)$
(iv) None as all have same size

Answer 1

First write down the Atomic Number of the neutral atoms. Then find out the number of electrons and protons present in each of the neutral atoms. Then check whether any electron is gained or lost and do the necessary changes to the number of electrons present in the ions. Determine the common factor in the three given ions. After eliminating the common factor determine whether the other factor is affecting the size of the given ions.

Complete step-by-step answer: The neutral atoms corresponding to the given ions are $F,\,Ne$ and $Na$respectively.-For $F$, the Atomic Number $= \,\,9$
Therefore, the number of protons present in the atom $= \,\,9$
Also, the number of electrons present in the atom $= \,\,9$ (since the atom is neutral the number of protons and electrons are equal).The charge present in its ion is $\left( { - 1} \right)$ which means the neutral atom $F$ accepts one electron to form${F^ - }$.
So the number of electrons present in the ${F^ - }$ ion $= \,\,9 + 1\,\, = 10$.
The number of protons present in the ion is the same as that of the atom, as only an electron is gained.
For $Ne$, the Atomic Number $= \,\,10$
Therefore, the number of protons present in the atom $= \,\,10$
Also, the number of electrons present in the atom $= \,\,10$ (since the atom is neutral the number of protons and electrons are equal).Since the neutral atom $Ne$ has not gained or lost an electron, it remains in its neutral form.For $Na$, the Atomic Number $= \,\,11$
Therefore, the number of protons present in the atom $= \,\,11$
Also, the number of electrons present in the atom $= \,\,11$ (since the atom is neutral the number of protons and electrons are equal).The charge present in its ion is $\left( { + 1} \right)$ which means the neutral atom $Na$ loses one electron to form $N{a^ + }$.
So the number of electrons present in the $N{a^ + }$ ion $= \,\,11 + 1\,\, = 10$.
The number of protons present in the ion is the same as that of the atom, as only an electron is lost.
So we can see the number of electrons present in all the three ions are the same and these are known as isoelectronic species.So the only guiding factor that determines the size of the ions is the number of protons present in that ion. More is the number of protons present in the nucleus of the ion, more they can pull the electrons towards themselves which results in the decrease in size of the ion.
Since the number of protons present in $N{a^ + } > Ne > {F^ - }$, the size of the ions follows the order
${F^ - } > Ne > N{a^ + }$.
Since the nucleus consists of protons and neutrons and neutrons are neutral in charge, the nuclear charge fully depends on the charge of the protons. So we can say the nuclear charge is equal to the charge of the protons. Hence the size of the ions is affected by the nuclear charge only.

Hence the correct answer is (iii) Nuclear Charge $\left( Z \right)$.

Note: You should know the periodic table properly in order to do this kind of question. If the periodic table is known you can easily know the atomic number and hence you can find out the number of protons and electrons present in the given ion. Also check properly whether the ion is negatively or positively charged for proper calculation of the number of electrons present in the ion.