Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A) (i) Z = 17, A = 35

Exercises · 2.4

Q. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(i) Z = 17, A = 35.

(ii) Z = 92, A = 233.

(iii) Z = 4, A = 9.

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Concept understanding — Atomic Notation

Atomic Notation: The ID Card of an Atom

Imagine you're at a huge stadium filled with people. To identify any one person, you'd need more than just a name — you'd need their jersey number, their team, maybe their position. An atom is the same way. Just saying "carbon" or "oxygen" tells you the kind of atom, but not the full story. Atomic notation is the shorthand that gives you the complete identity.

The Intuition: Three Numbers, One Symbol

Every atom is built from three key particles:

Protons (positive charge) — these define what element it is.
Neutrons (no charge) — these add mass and stability.
Electrons (negative charge) — these orbit the nucleus and determine chemical behaviour.

Atomic notation packs all this information around the element's symbol. Think of it as a label with two numbers attached:

^{A}_{Z}\text{X}

Where:

X is the element symbol (e.g., C for carbon, O for oxygen).
Z is the atomic number (number of protons).
A is the mass number (protons + neutrons).

Note

The atomic number $Z$ is what makes an atom that element. Change $Z$ , and you change the element entirely. Carbon always has $Z=6$ ; if you change it to $7$ , it's nitrogen.

The Precise Statement

For any atom represented as $^{A}_{Z}\text{X}$ :

Atomic number $Z$ = number of protons = number of electrons (in a neutral atom).
Mass number $A$ = number of protons + number of neutrons.
Number of neutrons $N = A - Z$ .

A neutral atom has equal protons and electrons. If the atom gains or loses electrons, it becomes an ion — but the notation for the nucleus stays the same.

Example: Carbon-12

The most common carbon atom is written as:

^{12}_{6}\text{C}

From this:

$Z=6$ → 6 protons, and in a neutral atom, 6 electrons.
$A=12$ → total nucleons (protons + neutrons) = 12.
Neutrons $N = 12 - 6 = 6$ .

Why Two Numbers? Why Not Just One?

If you only knew the mass number $A$ , you wouldn't know the element — because different elements can have the same mass number (e.g., $^{14}\text{C}$ and $^{14}\text{N}$ both have $A=14$ ). If you only knew the atomic number $Z$ , you wouldn't know how many neutrons are present — and that matters for isotopes.

Important

Isotopes are atoms of the same element (same $Z$ ) with different numbers of neutrons (different $A$ ). For example:

$^{12}_{6}\text{C}$ (6 neutrons)
$^{13}_{6}\text{C}$ (7 neutrons)
$^{14}_{6}\text{C}$ (8 neutrons) All are carbon, but they have different masses and some are radioactive.

A Quick Table for Clarity

Notation	Element	Protons ( $Z$ )	Neutrons ( $N$ )	Mass Number ( $A$ )
$^{1}_{1}\text{H}$	Hydrogen	1	0	1
$^{4}_{2}\text{He}$	Helium	2	2	4
$^{23}_{11}\text{Na}$	Sodium	11	12	23

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The standard notation for a nuclide is $\ce{^{A}_{Z}X}$ , where $X$ is the chemical symbol. For the given pairs, the symbols are $\ce{^{35}_{17}Cl}$ , $\ce{^{233}_{92}U}$ , and $\ce{^{9}_{4}Be}$ .

The notation $\ce{^{A}_{Z}X}$ is the universal shorthand for a specific atomic nucleus. The subscript $Z$ is the atomic number — it tells you how many protons are in the nucleus, and it uniquely identifies the element. The superscript $A$ is the mass number — the total number of protons plus neutrons. Once you know $Z$ , you look up the element’s symbol on the periodic table; then you just write $A$ above it and $Z$ below it, left of the symbol.

A common mistake is to swap the positions of $Z$ and $A$ , or to forget that $Z$ determines the element, not the other way around. Let’s apply this cleanly to each case.

$Z = 17$ , $A = 35$

Atomic number 17 is chlorine (Cl). So the symbol is $\ce{^{35}_{17}Cl}$ . This is the most abundant isotope of chlorine.
$Z = 92$ , $A = 233$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set D-41 markMCQ
$Q. The number of orbitals associated with ‘N’ shell of an atom is (A) 16 (B) 32 (C) 3 (D) 4$
›Reveal solutionSolution

N shell $\Rightarrow n = 4$ ; the number of orbitals in a shell is $n^2 = 16$ (the number of electrons would be $2n^2 = 32$ , which is the trap).

Step 1 — Identify the shell.

The spectroscopic shell letters map to principal quantum numbers in order:
$K \to n=1,\quad L \to n=2,\quad M \to n=3,\quad \mathbf{N \to n=4}$
Step 2 — List the subshells allowed for $n = 4$ .

The azimuthal quantum number obeys $l = 0, 1, \dots, (n-1)$ , so for $n=4$ : $l = 0, 1, 2, 3$ — that is $4s, 4p, 4d, 4f$ .

Step 3 — Count the orbitals in each subshell.

For a given $l$ the magnetic quantum number takes $m_l = -l \dots 0 \dots +l$ , which is $(2l+1)$ values — and each value of $m_l$ is one orbital:

Subshell $l$ Orbitals $(2l+1)$
$4s$ 0 1
$4p$ 1 3
$4d$ 2 5
$4f$ 3 7

Step 4 — Add them up.
$1 + 3 + 5 + 7 = 16$
This reproduces the general result …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. An Anion formed by an atom carries one unit of negative charge and has a mass number equal to 37u. If this Anion contains 11.1% more neutrons than electrons, what is the correct number of protons, neutrons and electrons in it? (A) protons=17 electrons=18 neutrons=20 (B) protons=16 electrons=16 neutrons=21 (C) protons=19 electrons=19 neutrons=18 (D) protons=18 electrons=18 neutrons=19$
›Reveal solutionSolution

Solving the neutron-excess condition for a singly-charged anion of mass 37 gives 17 protons, 18 electrons, 20 neutrons.

Set up (anion charge $-1$ $\Rightarrow$ electrons $= p+1$ ; mass number $37 \Rightarrow$ neutrons $= 37-p$ ):

11.1% more neutrons than electrons:
$37-p = 1.111\,(p+1)$
$37 - 1.111 = 1.111p + p \Rightarrow 35.889 = 2.111p \Rightarrow p = 17$ …
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Subshell	$l$	Orbitals $(2l+1)$
$4s$	0	1
$4p$	1	3
$4d$	2	5
$4f$	3	7

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Q.Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

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