Question

The size of isoelectronic species; ${{F}^{-}}$,$Ne$, $N{{a}^{+}}$ is affected by
(A)- nuclear charge(Z)
(B)- valence principal quantum number
(C)- electron-electron interaction in the outer orbitals
(D)- none of the factors because their size is the same

Answer 1

Atoms and ions that have the same electron configuration are said to be isoelectronic. The term means "equal electric" or "equal charge". Isoelectronic chemical species typically display similar chemical properties. For atoms or ions that are isoelectronic, the quantity of protons determines the radius of it.

Complete step by step answer:
Nuclear charge is a measure of the effect of the number of protons in the nucleus and their ability to attract the negative electrons in orbits around the nucleus.
- Higher is the nuclear charge, the higher will be the attraction between the nucleus and the valence electrons and lower will be the ionic size.
- no. of electrons in ${{F}^{-}}$ is 10 and protons are 9.
no. of electrons in $Ne$ is 10 and protons is 10.
no. of electrons in $N{{a}^{+}}$ is 10 and protons is 11.

-The number of protons in the nucleus remains the same and only one electron is getting added to the valence shell, so the pulling effect of protons decreases and so nuclear charge will also decrease which will result in larger size of fluoride ion.
- Lesser the number of protons, and higher the number of electrons, nuclear charge decreases and atomic/ionic size increases.
-With the increase in the nuclear charge, the force of attraction will increase. Hence the size of the species decreases.
-Here, in the isoelectronic species the no. of protons is different. So, the size of the isoelectronic species will vary.
-Electron-electron interaction in the outer orbitals does not affect the size as the no. of electrons is the same in all isoelectronic species.
-The valence electrons are determined by how many electrons are in the outermost shell of an atom. It also does not affect size.
So, the size of the isoelectronic species ${{F}^{-}}$,$Ne$, $N{{a}^{+}}$ is affected by nuclear charge.
So, the correct answer is “Option A”.

Additional Information:
According to Coulomb's law, the attraction of an electron to a nucleus depends only on three factors: the charge of the nucleus (+Z), the charge of the electron (-1), and the distance between the two (r). Each electron in a multi-electron atom experiences a different magnitude of (and attraction to) the nuclear charge depending on what specific subshell the electron occupies. The amount of positive charge experienced by any individual electron is the effective nuclear charge $({{Z}_{eff}})$.

Note: As isoelectronic series is a sequence of species all having the same no. of electrons which means there will be the same amount of electron-electron repulsion but they differ from each other in nuclear charge. The effect of increasing nuclear charge on the radius is decreasing.