Question

In case of isoelectronic species the size of F^– , Ne and Na⁺ is affected by:

• A. Principal quantum number (n)
• B. None of the factors because their size is the same
• C. Electron-electron interaction in the outer orbitals
• D. Nuclear charge (z)

Answer 1

The Correct Option is D

The question asks to identify the primary factor that affects the size of the isoelectronic species F⁻, Ne, and Na⁺.

Concept Used:

The key concepts for determining the size of isoelectronic species are:

1. Isoelectronic Species: These are atoms or ions that have the same number of electrons.
2. Nuclear Charge (Z): This is the total positive charge of the nucleus, which is equal to the number of protons in the atom.
3. Effective Nuclear Charge (Z_eff): This is the net positive charge experienced by an electron in a multi-electron atom. The electrons in the outer shells are shielded from the full nuclear charge by the inner-shell electrons. The effective nuclear charge is approximately the nuclear charge minus the screening constant (\(\sigma\)), \(Z_{eff} = Z - \sigma\).

For an isoelectronic series, the number of electrons is constant. Therefore, the screening effect from the inner electrons is also roughly constant. The size of the species is then primarily determined by the magnitude of the nuclear charge (number of protons). A higher nuclear charge exerts a stronger electrostatic pull on the same number of electrons, causing the electron cloud to contract and resulting in a smaller atomic or ionic radius.

Step-by-Step Solution:

Step 1: Determine the number of protons (Nuclear Charge, Z) and electrons for each species.

• For Fluoride ion (F⁻):
  • The atomic number of Fluorine (F) is Z = 9. So, it has 9 protons.
  • The charge is -1, meaning it has gained one electron.
  • Number of electrons = 9 (protons) + 1 = 10 electrons.
• For Neon atom (Ne):
  • The atomic number of Neon (Ne) is Z = 10. So, it has 10 protons.
  • It is a neutral atom.
  • Number of electrons = 10 (protons) = 10 electrons.
• For Sodium ion (Na⁺):
  • The atomic number of Sodium (Na) is Z = 11. So, it has 11 protons.
  • The charge is +1, meaning it has lost one electron.
  • Number of electrons = 11 (protons) - 1 = 10 electrons.

Since all three species (F⁻, Ne, Na⁺) have 10 electrons, they are isoelectronic.

Step 2: Compare the nuclear charge of the species.

The number of protons (nuclear charge) for each species is:

• F⁻: Z = 9
• Ne: Z = 10
• Na⁺: Z = 11

The nuclear charge increases in the order F⁻ < Ne < Na⁺.

Step 3: Relate nuclear charge to ionic/atomic size.

All three species have the same number of electrons (10) arranged in the same electronic configuration (1s²2s²2p⁶). However, they have different numbers of protons in their nuclei.

The Na⁺ ion, with 11 protons, has the highest nuclear charge. It will exert the strongest electrostatic attraction on its 10 electrons, pulling them closer to the nucleus. This results in the smallest radius.

The F⁻ ion, with only 9 protons, has the lowest nuclear charge. Its nucleus exerts the weakest attraction on the 10 electrons, resulting in the largest radius.

The Ne atom is intermediate with 10 protons and 10 electrons.

Thus, the order of size is: Na⁺ < Ne < F⁻.

Final Result:

For the isoelectronic species F⁻, Ne, and Na⁺, the number of electrons is the same, but the nuclear charge is different. The electrostatic attraction between the nucleus and the electrons increases with the increase in nuclear charge. Therefore, the size of these species is primarily affected by the nuclear charge (number of protons).