Atomic number Z, mass number A, isotopes, isobars, isotones: Exercises

Before we count anything, let's make the reading of the nuclear symbol completely mechanical with one picture.

Figure — Atomic number Z, mass number A, isotopes, isobars, isotones

Level 1 — Recognition

(Read the symbol, name the number. No arithmetic beyond one subtraction.)

L1.1 In the symbol $_{11}^{23} Na$ , state $Z$ , $A$ , and $N$ .

Recall Solution L1.1

WHAT we do: read the two corners, then subtract for neutrons.

$Z = 11$ (bottom-left) — this is the proton count.
$A = 23$ (top-left) — this is the nucleon count.
$N = A - Z = 23 - 11 = 12$ neutrons. Why subtract? $A$ counts all nucleons; removing the $Z$ protons leaves only neutrons.

L1.2 For a neutral atom of $_{20}^{40} Ca$ , how many electrons does it have?

Recall Solution L1.2

Neutral means charge $q = 0$ , so electrons $= Z - q = 20 - 0 = 20$ . Why: a neutral atom balances every positive proton with one negative electron, so electrons $= Z$ .

L1.3 Which single number in $_{Z}^{A} X$ decides which element it is — $A$ or $Z$ ?

Recall Solution L1.3

$Z$ , the atomic number. Change $Z$ and you change the element itself (the iso-tope "same place" idea lives here). $A$ only tells you how heavy that element's particular atom is; it does not name the element.

Level 2 — Application

(Use $A = Z + N$ and electrons $= Z - q$ directly.)

L2.1 An atom has $17$ protons and $18$ neutrons. Write its full nuclear symbol (element: chlorine, symbol Cl) and give $A$ .

Recall Solution L2.1

WHAT: build $A$ from its two parts. $A = Z + N = 17 + 18 = 35$ , and $Z = 17$ . Symbol: $_{17}^{35} Cl$ . Why add? $A$ is defined as the total nucleon count = protons + neutrons.

L2.2 How many neutrons are in $_{26}^{56} Fe$ ?

Recall Solution L2.1

$N = A - Z = 56 - 26 = 30$ neutrons.

L2.3 The ion $_{8}^{16} O^{2 -}$ : give protons, neutrons, and electrons.

Recall Solution L2.3

Protons $= Z = 8$ — ionization never touches the nucleus.
Neutrons $= A - Z = 16 - 8 = 8$ — also untouched.
Electrons $= Z - q = 8 - (- 2) = 10$ . The charge is $- 2$ , meaning 2 extra electrons were gained. Why the minus-minus becomes plus: $q = - 2$ , so $Z - q = 8 - (- 2) = 8 + 2 = 10$ . An anion has more electrons than protons.

L2.4 A cation $_{13}^{27} Al^{3 +}$ : how many electrons?

Recall Solution L2.4

Electrons $= Z - q = 13 - (+ 3) = 10$ . Why fewer: a $3 +$ charge means the atom lost 3 electrons, so it now has $13 - 3 = 10$ .

Level 3 — Analysis

(Classify pairs. Always compute the trio $(Z, A, N)$ first, then see which one matches.)

The picture below is the decision you run every time.

L3.1 Classify: $_{17}^{35} Cl$ and $_{17}^{37} Cl$ .

Recall Solution L3.1

Trio for each:

Cl-35: $Z = 17, A = 35, N = 18$ .
Cl-37: $Z = 17, A = 37, N = 20$ . Same $Z = 17$ , different $A$ ⇒ isotopes (same element, different mass). iso-tope = same place = same $Z$ .

L3.2 Classify: $_{18}^{40} Ar$ , $_{19}^{40} K$ , $_{20}^{40} Ca$ .

Recall Solution L3.2

Ar: $Z = 18, A = 40, N = 22$ .
K: $Z = 19, A = 40, N = 21$ .
Ca: $Z = 20, A = 40, N = 20$ . Same $A = 40$ , different $Z$ ⇒ isobars (different elements, same mass). iso-bar = same weight = same $A$ .

L3.3 Classify: $_{19}^{39} K$ and $_{20}^{40} Ca$ .

Recall Solution L3.3

K: $N = 39 - 19 = 20$ .
Ca: $N = 40 - 20 = 20$ . Different $Z$ , different $A$ , but same $N = 20$ ⇒ isotones. iso-tone = same Neutrons.

L3.4 Classify: $_{6}^{14} C$ and $_{7}^{14} N$ . (This is the classic trap.)

Recall Solution L3.4

C: $Z = 6, A = 14, N = 8$ .
N: $Z = 7, A = 14, N = 7$ . Same $A = 14$ , different $Z$ ⇒ isobars. Note $N$ differs ( $8$ vs $7$ ), so they are not isotones.

Level 4 — Synthesis

(Work backwards, or combine two clues.)

L4.1 An atom has mass number $A = 39$ and contains $20$ neutrons. Find $Z$ , and use $Z$ to name the element from the list Ar( $18$ ), K( $19$ ), Ca( $20$ ).

Recall Solution L4.1

WHAT: invert $A = Z + N$ to get $Z = A - N$ . $Z = A - N = 39 - 20 = 19$ . $Z = 19$ ⇒ potassium (K). Symbol $_{19}^{39} K$ . Why this rearrangement: the same conservation $A = Z + N$ can be solved for any one of the three when the other two are known.

L4.2 Species X has $18$ electrons and charge $+ 2$ . Find its $Z$ . Then, if $A = 40$ , how many neutrons?

Recall Solution L4.2

From electrons $= Z - q$ : rearrange to $Z = electrons + q = 18 + (+ 2) = 20$ . So $Z = 20$ (calcium). Neutrons $N = A - Z = 40 - 20 = 20$ . Why add the charge back: a $2 +$ ion is missing 2 electrons, so the neutral proton count is $2$ more than the electrons you see.

L4.3 Build an isotone partner: give a nucleus that is an isotone of $_{8}^{16} O$ but is the element carbon ( $Z = 6$ ).

Recall Solution L4.3

Isotone means same $N$ . For O-16: $N = 16 - 8 = 8$ . We need carbon ( $Z = 6$ ) with $N = 8$ , so $A = Z + N = 6 + 8 = 14$ . Answer: $_{6}^{14} C$ . Why: fix the target $N$ , then rebuild $A$ from the new element's $Z$ .

L4.4 Two neutral atoms are isotopes. Atom 1 is $_{29}^{63} Cu$ . Atom 2 has $36$ neutrons. Find atom 2's $A$ and write its symbol.

Recall Solution L4.4

Isotopes share $Z$ , so atom 2 also has $Z = 29$ (copper). $A = Z + N = 29 + 36 = 65$ . Symbol: $_{29}^{65} Cu$ . Why $Z$ is locked: "isotope" means same element, and same element means same proton number.

Level 5 — Mastery

(Multi-step reasoning; several families or ions at once.)

L5.1 A neutral atom Y has $30$ neutrons and its mass number is $56$ . (a) Find $Z$ and name it from Fe( $26$ ), Ni( $28$ ), Zn( $30$ ). (b) A different atom, $_{28}^{58} Ni$ , is compared to Y: are they isotopes, isobars, or isotones? (c) State the neutron numbers you used.

Recall Solution L5.1

(a) $Z = A - N = 56 - 30 = 26$ ⇒ iron, $_{26}^{56} Fe$ . (b) Trio for each:

Y (Fe-56): $Z = 26, A = 56, N = 30$ .
Ni-58: $Z = 28, A = 58, N = 58 - 28 = 30$ . Different $Z$ , different $A$ , same $N = 30$ ⇒ isotones. (c) $N (Fe-56) = 30$ , $N (Ni-58) = 30$ .

L5.2 The ion $_{15}^{31} P^{3 -}$ : give protons, neutrons, and electrons. Then name a neutral atom that has the same electron count as this ion (choose from Ne( $10$ ), Ar( $18$ ), Kr( $36$ )).

Recall Solution L5.2

Protons $= Z = 15$ .
Neutrons $= A - Z = 31 - 15 = 16$ .
Electrons $= Z - q = 15 - (- 3) = 18$ . An electron count of $18$ matches neutral argon ( $Z = 18$ ). (This is why $P^{3 -}$ is isoelectronic with Ar — same electrons, though different nuclei.) Why they match: both have $18$ electrons; the ion reached $18$ by gaining $3$ , argon starts at $18$ neutral.

L5.3 You are told: element M has an isotope with $A = 24$ and $N = 12$ , and another isotope with $N = 14$ . (a) Find $Z$ of M. (b) Give the mass number of the second isotope. (c) Verify the two are genuinely isotopes.

Recall Solution L5.3

(a) From the first isotope: $Z = A - N = 24 - 12 = 12$ (magnesium). (b) Isotopes share $Z = 12$ ; second isotope $A = Z + N = 12 + 14 = 26$ . So $_{12}^{26} Mg$ . (c) Both have $Z = 12$ but $A = 24$ vs $A = 26$ (different $N$ : $12$ vs $14$ ). Same $Z$ , different $A$ ⇒ genuine isotopes. ✓

L5.4 Full audit of $_{8}^{18} O^{2 -}$ and classify it against $_{9}^{18} F$ : (a) protons/neutrons/electrons of the ion; (b) trio $(Z, A, N)$ of both; (c) family relationship between the two nuclei (ignore charge).

Recall Solution L5.4

(a) For $_{8}^{18} O^{2 -}$ : protons $= 8$ ; neutrons $= 18 - 8 = 10$ ; electrons $= 8 - (- 2) = 10$ . (b) Trios:

O-18: $Z = 8, A = 18, N = 10$ .
F-18: $Z = 9, A = 18, N = 18 - 9 = 9$ . (c) Same $A = 18$ , different $Z$ ⇒ isobars (their $N$ differ, $10$ vs $9$ , so not isotones). Why charge is ignored for the family test: isotope/isobar/isotone classify nuclei ( $Z$ , $A$ , $N$ ). Electrons/charge play no role in these three families.

Recall One-line self-check before you leave

For any nucleus, can you produce $(Z, A, N)$ and, for an ion, electrons $= Z - q$ , in under five seconds? If yes ::: you own D4. If a step stalls, re-open figure s01 and rebuild the trio from the two corners.

Connections

Parent topic (Hinglish) — the definitions these exercises drill.
Discovery of Protons, Neutrons and Electrons — the particles $Z$ , $N$ , and electrons count.
Average Atomic Mass and Isotopic Abundance — where isotope counting leads next.
Radioactivity and Nuclear Stability — why the $N$ you compute here governs stability.
Periodic Table and Atomic Number — elements ordered by the $Z$ you extract.