Subatomic Particles, Isotopes, Isobars and Isoelectronic Species

The three particles and the two numbers that describe a nuclide:

• Proton — charge +1, mass about 1 u, in the nucleus. Its count is the atomic number.
• Neutron — charge 0, mass about 1 u, in the nucleus. Protons and neutrons together are nucleons.
• Electron — charge −1, mass about 1/1836 of a proton (nearly massless), in shells outside the nucleus.
• The nuclide is written $^{A}_{Z}\text{X}$: mass number $A$ (top) and atomic number $Z$ (bottom), with neutrons $= A - Z$.

The counts, straight from $Z$, $A$ and the charge:

• Protons $= Z$; neutrons $= A - Z$.
• Electrons in a neutral atom $= Z$.
• Electrons in an ion $= Z - (\text{charge})$: subtract the charge, so a $+$ ion has fewer electrons and a $-$ ion has more.
• Example — $\text{Ca}$ has $Z = 20$, so a neutral calcium atom has exactly 20 electrons, whereas $\text{K}^{+}$ ($Z = 19$) has $19 - 1 = 18$.

The four families, sorted by which count is held equal:

• Isotopes — same protons (same element, same $Z$), different neutrons and mass number. Same chemical properties, same periodic-table position. Example: $^{35}\text{Cl}$ and $^{37}\text{Cl}$.
• Isobars — same mass number $A$, different elements (different $Z$). Example: $^{40}\text{Ar}$ and $^{40}\text{Ca}$.
• Isotones — same number of neutrons $(A - Z)$, different $Z$ and $A$. Example: $^{12}_{6}\text{C}$ and $^{11}_{5}\text{B}$ (both $6$ neutrons).
• Isoelectronic species — same number of electrons, regardless of element or charge. Example: $\text{Na}^{+}$, $\text{F}^{-}$, $\text{O}^{2-}$ and $\text{Ne}$ (all $10$).

Two species are isoelectronic when they have the same number of electrons:

• Compute electrons for each: $e^- = Z - (\text{charge})$.
• The $10$-electron set (neon core) is the one PYQs use most: $\text{O}^{2-}$, $\text{F}^{-}$, $\text{Ne}$, $\text{Na}^{+}$, $\text{Mg}^{2+}$, $\text{Al}^{3+}$ all have $10$.
• The odd-one-out is usually a neutral atom slipped in beside its ions — e.g. neutral $\text{Na}$ has $11$, not $10$.

The weighted-average mass, used both ways:

• Forwards — average $= \sum (\text{isotope mass} \times \text{fraction})$, where the fractions sum to $1$.
• Backwards — let the abundance of one isotope be $x\%$ and solve. For chlorine: $\dfrac{35x + 37(100 - x)}{100} = 35.5$ gives $x = 75$, so the ratio $^{35}\text{Cl} : {}^{37}\text{Cl} = 75 : 25 = 3 : 1$.

The recall facts the bank tests here:

• Number of isotopes — nitrogen has 2 natural isotopes ($^{14}\text{N}$, $^{15}\text{N}$); hydrogen has $3$ (protium, deuterium, tritium); carbon has $3$ ($^{12}\text{C}, {}^{13}\text{C}, {}^{14}\text{C}$).
• Hydrogen-like (hydrogenic) species have exactly one electron: $\text{H}$, $\text{He}^{+}$, $\text{Li}^{2+}$, $\text{Be}^{3+}$. Neutral $\text{He}$ has $2$ electrons, so it is not hydrogen-like.
• Radioactivity — heavy elements such as $\text{At}$ (astatine), $\text{Po}$ (polonium) and $\text{Rn}$ (radon) are radioactive; the noble gas $\text{Ar}$ (argon) is stable and not radioactive.