Exercise 2.41
Symbols 79Br and 79Br can be written, whereas symbols 35Br and 35Br are not acceptable. Answer briefly.
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NCERT’s answer
For a given element the number of prontons is the same for the isotopes, whereas the mass number can be different for the given atomic number.
The atomic number is fixed by which element the symbol names, so it can be dropped — the mass number cannot, because it changes from isotope to isotope, and swapping the two numbers' positions changes what the symbol is claiming.The full isotope symbol is written as
\[{}^{A}_{Z}\text{X}
\]
where \(\displaystyle A\) is the mass number (protons + neutrons, written as a left superscript) and \(\displaystyle Z\) is the atomic number (protons alone, written as a left subscript). For bromine, \(\displaystyle Z = 35\) always — that is what makes an atom "bromine" — while the two naturally occurring isotopes have \(\displaystyle A = 79\) and \(\displaystyle A = 81\).Why \(\displaystyle {}^{79}_{35}\text{Br}\) and \(\displaystyle {}^{79}\text{Br}\) are fine.
Writing the atomic number is optional, not required, because it carries no new information: the symbol "Br" already tells you \(\displaystyle Z=35\) — no other element has that atomic number, so stating it again is repetition, not identification. The mass number, on the other hand, is genuinely needed, because it is the one thing that changes between isotopes of the same element. So \(\displaystyle {}^{79}\text{Br}\) is a complete, unambiguous symbol: the "$\displaystyle 79$" sits in the correct upper-left slot reserved for the mass number, and \(\displaystyle Z=35\) is understood from "Br" itself.Why \(\displaystyle {}^{35}_{79}\text{Br}\) is not acceptable.
Here the numbers have been swapped into the wrong slots: $\displaystyle 35$ now sits where the mass number belongs, and $\displaystyle 79$ sits where the atomic number belongs. That symbol is asserting \(\displaystyle Z = 79\) for bromine — but atomic number $\displaystyle 79$ belongs to gold, not bromine, so it directly contradicts the element named by the symbol "Br". It also fails on a physical count: the number of neutrons is
\[N = A - Z
\]
Substituting the swapped values, \(\displaystyle N = 35 - 79 = -44\), a negative neutron count, which cannot exist. This is the mistake people make when they treat superscript and subscript as interchangeable labels instead of two specific quantities tied to specific positions.Why \(\displaystyle {}^{35}\text{Br}\) is not acceptable.
With no subscript shown, the lone number "$\displaystyle 35$" is read as sitting in the mass-number position — so this symbol claims bromine has mass number \(\displaystyle A=35\). Combined with the correct \(\displaystyle Z=35\) that "Br" always implies, that would give
\[N = A - Z = 35 - 35 = 0
\]
a bromine atom with zero neutrons — which is not one of bromine's real isotopes (only \(\displaystyle A=79\) and \(\displaystyle A=81\) occur). Dropping the atomic-number subscript is allowed only because it duplicates information already fixed by the element symbol; it is never allowed to drop or misplace the mass number, since that is the one variable piece of information the symbol exists to convey.**Answer: \(\displaystyle {}^{79}_{35}\text{Br}\) and \(\displaystyle {}^{79}\text{Br}\) are valid because the atomic number ($\displaystyle 35$) is redundant once "Br" is written and can be omitted, while the mass number ($\displaystyle 79$) must always be shown in its correct upper-left position. \(\displaystyle {}^{35}_{79}\text{Br}\) is invalid because it misplaces the numbers and asserts \(\displaystyle Z=79\) (gold's atomic number, not bromine's), giving an impossible negative neutron count. \(\displaystyle {}^{35}\text{Br}\) is invalid because it misstates the mass number as $\displaystyle 35$, which is not a real isotope of bromine (only $\displaystyle 79$ and $\displaystyle 81$ are).