Question

Consider the following nuclides : $^{12}_6\text{C}$, $^{198}_{80}\text{Hg}$, $^{14}_6\text{C}$, $^{197}_{79}\text{Au}$. Group them into isotopes and isotones.

Hint:

A brilliant mnemonic trick: Isotopes have the same number of Protons. Isotones have the same number of Neutrons. Isobars have the same mass number (A, resembles Both combined).

Answer 1

Concept:
• Isotopes are distinct chemical nuclides that share the exact same atomic number $Z$ (identical number of positive protons) but purposely possess entirely different mass numbers $A$ (different number of neutrons).
• Isotones are completely different chemical nuclides that coincidentally contain the exact same physical number of internal neutrons $N$, which is mathematically calculated actively as $N = A - Z$.

Step 1: Identify and group the Isotopes
We meticulously examine the lower subscript index (atomic number $Z$) strictly provided for each individual nuclide in the list.
For $^{12}_6\text{C}$, the atomic number is $Z = 6$.
For $^{14}_6\text{C}$, the atomic number is also $Z = 6$.
Because they both rigorously share the exact same atomic number ($Z=6$), but definitively have vastly different top mass numbers ($12$ vs $14$), they chemically belong to the same specific element.
Therefore, $^{12}_6\text{C}$ and $^{14}_6\text{C}$ solidly form a perfect pair of isotopes.

Step 2: Identify and group the Isotones
To firmly hunt for isotones, we must first mathematically calculate the hidden internal neutron count ($N = A - Z$) specifically for the remaining heavy nuclides.
For the heavy Mercury nuclide $^{198}_{80}\text{Hg}$, the neutron count calculates to:
\[ N = 198 - 80 = 118 \]
For the heavy Gold nuclide $^{197}_{79}\text{Au}$, the neutron count similarly calculates to:
\[ N = 197 - 79 = 118 \]
Because both highly distinct heavy nuclei amazingly share the exact same internal neutron count of exactly $118$, they fit the definition perfectly.
Therefore, $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$ solidly form a perfect pair of isotones.