Question

The effective nuclear charge experienced by the outermost electron of Na is different than the effective nuclear charge experienced by the outermost electron of Ne. This difference best accounts for which of the following?

(A) Na has a greater density at standard conditions than Ne.
(B) Na has a lower first ionization energy than Ne.
(C) Na has a higher melting point than Ne.
(D) Na has a higher neutron-to-proton ratio than Ne.
(E) Na has fewer naturally occurring isotopes than Ne.

Answer 1

(B) Na has a lower first ionization energy than Ne.

Step-by-step explanation:

As we know that ionization potential is the energy required to remove one electron from outermost shell. So if we found the ionization potential of Na and then ionization potential of Ne we will find the difference.

Experimentally it is found that Na has smaller value of ionization potential compare to Ne. So we can say the effective charge of Nucleus of Na must be less than the effective charge of nucleus of Ne.

Because if effective charge of nucleus is more than it will have strong attraction on outer electron and it must have more ionization potential.

So here correct answer will be

(B) Na has a lower first ionization energy than Ne.

Answer 2

B) Na has a lower first ionization energy than Ne.

Step-by-step explanation:

The atomic number¹ for Na has a value of 11 while in the case of Ne this value is 10. That means that Sodium (Na) has a total number of 11 protons, 11 neutrons and 11 electrons (since it is electrically neutral²). For the case of Neon (Ne) it has 10 protons, 10 neutrons and 10 electrons.

As the atomic number increases, the atomic radius³ shrinks (the orbitals are closer to the nucleus) as a consequence of the electric force. For the case of sodium (Na) the electron in the outermost orbital will experience a lower electric force than the electron placed in the outermost orbital in the atom of Neon (Ne).

Although, the sodium’s atom has more protons and therefore electrons, these eleven electrons will be organized according with the electronic configuration⁴ in the different shells (orbitals) of probabilities of their positions around the atom.

The electronic configuration for Na is:

1s²2s²2p⁶3s¹

The electronic configuration for Ne is:

1s²2s²2p⁶

Since Na needs another orbital to placed its outermost electron, the atomic radius will have a greater value than Ne. The electric force is inversely proportional to the square of the distance between two charged particles, as is established in Coulomb’s law:

`F = \kappa_(0) (q1q2)/(r^(2))` (1)

Where q1 and q2 are the charges,
`\kappa_(0)` is the proportionality constant and r is the distance between the two charges.

Hence, the electron in the outermost orbital of Ne is submitted to a greater electric force according with equation 1, the required energy to remove it (ionization energy⁵) will be greater than in the case of Na (for that case will be the first ionization energy).

¹Atomic number: The number of protons or electrons in an atom.

²Electricaly neutral: All the charges are balanced (same number of positive charges and negative charges).

³Atomic radius: Distance between the center of the nucleus and an electron placed in the outermost orbital for a specific atom.

⁴Electronic configuration: Show how the electrons of an atom will be arranged in different orbitals according with the fact that each orbital has a specific number of electrons that can be held.

⁵Ionization energy: Energy required to remove an electron from an atom.

Key values:

First ionization energy of Na: 495 kJ/mol

First ionization energy of Ne: 2080 kJ/mol

Atomic radius of Na: 2.27 Å

Atomic radius of Ne: 1.54 Å

Atomic number of Na: 11

Atomic number of Ne: 10