An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than

Exercises · 2.43

Q. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

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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 key is to relate the mass number (protons + neutrons) to the charge (electrons = protons + 1) and the given neutron excess (neutrons = 1.111 × electrons). Solving gives 17 protons, 20 neutrons, and 18 electrons — the ion is chloride, $\ce{^{37}_{17}Cl^{-}}$ .

Let’s unpack this step by step. The problem gives you three pieces of information about an ion: its mass number (37), its charge (−1), and a percentage relationship between its neutrons and electrons. The goal is to identify the element and write its symbol with mass number and charge.

The core idea is simple: an ion’s mass number is the sum of protons and neutrons. Its charge tells you how many extra or missing electrons there are relative to protons. And the percentage condition gives you a direct equation linking neutrons to electrons. Solve for the number of protons — that’s the atomic number, which identifies the element.

Set up the variables. Let $p$ = number of protons, $n$ = number of neutrons, $e$ = number of electrons. For a neutral atom, $p = e$ . But this is an ion with one unit of negative charge, meaning it has one extra electron:

e = p + 1

Use the mass number. Mass number = protons + neutrons = 37:

p + n = 37

Translate the percentage condition. “11.1% more neutrons than electrons” means:

n = e + 0.111 \times e = 1.111 \, e

(11.1% = 0.111 as a decimal; “more than” means add that fraction of the base quantity.)

Tip

A quick check: 11.1% is exactly $\frac{1}{9}$ . So $n = e + \frac{e}{9} = \frac{10}{9}e$ . This fraction will make the algebra cleaner.

So we can write:

n = \frac{10}{9} e

Substitute $e$ in terms of $p$ . From step 1, $e = p + 1$ . Therefore:

n = \frac{10}{9}(p + 1)

Now use the mass number equation. From step 2: $p + n = 37$ . Substitute $n$ :

p + \frac{10}{9}(p + 1) = 37

Multiply through by 9 to clear the denominator: …

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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.An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

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