Noble Gases Worked ExamplesIGCSE Chemistry 0620

(a) The noble gases are in the last column, Group VIII: E and F.

(b) F is in Period 1, so it has only one shell, which is full with 2 electrons — it is helium. Answer: F.

(c) G is in Group VI, Period 2 (oxygen, $\text{O}_2$) and D is in Group VII, Period 3 (chlorine, $\text{Cl}_2$): D and G. The noble gases are monatomic, not diatomic.

(d) E is neon, 2,8. A (sodium, 2,8,1) loses one electron to form $\text{Na}^+$, 2,8; G (oxygen, 2,6) gains two electrons to form $\text{O}^{2-}$, 2,8. Answer: A and G. (D forms $\text{Cl}^-$, which is 2,8,8 — like argon, not neon.)

(a) Count the electrons shell by shell: X 2,8,8; Y 2,8,7; Z 2,8,1.

(b) X (argon) is the noble gas. It has a full outer shell of electrons, so it has no tendency to lose, gain or share electrons — it is unreactive.

(c) Because its outer shell is already full, an X atom does not need to share electrons with another atom, so it does not form bonds and exists as single atoms (it is monatomic).

Y (chlorine) has 7 outer electrons; two Y atoms each share one electron to form a covalent bond, giving each a full outer shell — so Y exists as $\text{Y}_2$ molecules.

Statement 1 is correct: neon is monatomic. Statement 2 is correct: neon is 2,8, a full outer shell.

Because the outer shell is full, a neon atom has no need to share electrons, so no bonds form between neon atoms — statement 2 explains statement 1. Answer: A.

B is the tempting wrong answer for candidates who know both facts separately but not the link. C and D each reject a correct fact.

(a) The boiling point increases down the group (from helium to radon).

(b) Argon lies between neon ($-246\ ^{\circ}\text{C}$) and krypton ($-153\ ^{\circ}\text{C}$). Halfway between them gives an estimate:

Any value between about $-210$ and $-180\ ^{\circ}\text{C}$ follows the trend. (The actual value is $-186\ ^{\circ}\text{C}$.)

(c) $-150\ ^{\circ}\text{C}$ is above the boiling point of argon, so argon is a gas at that temperature.

(a) Noble gases are monatomic, so the mass of one mole of gas is simply $A_r$ in grams. Density = mass of one mole ÷ volume of one mole.

(b) $1.67 = \dfrac{40}{24}$: the student used a molar mass of 40 for neon, i.e. treated it as a diatomic molecule, $\text{Ne}_2$. Neon exists as single atoms, so its molar mass is 20.

(c) Convert to $\text{dm}^3$ and find the moles, then multiply by $A_r(\text{Xe}) = 131$.

Electrons in the ion = proton number − charge (a negative charge adds electrons).

(a) $\text{O}^{2-}$: $8 + 2 = 10$ → 2,8, neon. $\text{Ca}^{2+}$: $20 - 2 = 18$ → 2,8,8, argon.

$\text{Li}^+$: $3 - 1 = 2$ → 2, helium. $\text{Al}^{3+}$: $13 - 3 = 10$ → 2,8, neon.

(b) Neon has 10 electrons. $\text{X}^{3-}$ has gained 3 electrons, so X has $10 - 3 = 7$ protons: nitrogen.

(c) Argon already has a full outer shell (2,8,8), so it has no tendency to lose or gain electrons.

Krypton is below argon, and density at r.t.p. increases down the group, so A and B can be ruled out straight away.

Krypton is monatomic, so one mole of the gas has a mass of $A_r = 84$ g and occupies $24 \text{ dm}^3$. Answer: C.

D is the trap: it doubles $A_r$ as though krypton were diatomic ($168 \div 24 = 7.00$). A comes from halving the density of argon instead of scaling it up.

(a) The pink-brown copper turns black as copper(II) oxide forms.

(b) The decrease in volume is the oxygen removed.

(c) Gases expand when heated, so a reading taken while the gas is hot would be too large.

(d) Argon is a noble gas with a full outer shell of electrons, so it does not lose, gain or share electrons — it is unreactive.

(e) $0.93\%$ of $100 \text{ cm}^3$ is $0.93 \text{ cm}^3$ of argon, all still present in the $79 \text{ cm}^3$.

(a) L and U: they exist as single atoms (monatomic), they do not react, and both are gases at room temperature (very low boiling points).

M and T are diatomic and reactive (halogen-like and oxygen-like); N is a metal.

(b) Boiling point increases down Group VIII. U ($-153\ ^{\circ}\text{C}$) has a higher boiling point than L ($-186\ ^{\circ}\text{C}$), so U is lower in the group.

(c) Density at r.t.p. also increases down the group (larger $A_r$), so U is denser.

(a) One dot and one cross sit in the overlap: two electrons (one shared pair — a single covalent bond).

(b) A chlorine atom has 7 outer electrons — one short of a full shell. By sharing one pair of electrons, each chlorine atom gains a full outer shell of 8, so chlorine exists as molecules of two atoms.

An argon atom already has a full outer shell of 8, so it gains nothing by sharing, and argon exists as single atoms.

(c) Chlorine: $\text{Cl}_2$. Argon: Ar.

(a) Equal volumes of gases at the same temperature and pressure contain equal numbers of moles of particles:

(b) Helium and argon are monatomic, so use $A_r$: He = 4, Ar = 40.

(c) Argon particles are single atoms:

Chlorine particles are $\text{Cl}_2$ molecules, each containing two atoms:

Both balloons hold the same number of particles, but argon is monatomic while chlorine is diatomic, so the chlorine balloon has twice as many atoms.

(a) Subtract the two readings:

(b) Moles of gas in $250 \text{ cm}^3$ at r.t.p.:

Molar mass = mass ÷ moles (keep the unrounded value of $n$):

Because Q is monatomic, $A_r = 84$. From the Periodic Table this is krypton, a noble gas — consistent with Q being unreactive.

(c) Weighing the flask full of air would include the mass of the air, so the difference would not be the mass of Q alone.

(d) The mass of gas is less than 1 g, so a precise balance is needed to keep the percentage error small.