Atomic Foundations Of Matter Class 9 NCERT Solutions

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Atomic Foundations of Matter Class 9 NCERT Solutions is simple, accurate, and step-by-step solutions to all the NCERT textbook questions. These solutions explain important concepts such as the Law of Conservation of Mass, Law of Constant Proportions, Dalton’s Atomic Theory, atoms, molecules, ions, valency, chemical formulae, molecular mass, and chemical bonding in easy-to-understand language.

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Atomic Foundations of Matter Class 9 NCERT Solutions

Q1. Water can be obtained from various sources. Are all these samples of water chemically identical?

Answer: Yes, all samples of pure water are chemically the same.

Water always has the formula H₂O → 2 hydrogen atoms + 1 oxygen atom.
Whether it comes from a river, well, or ocean, after purification it is identical.
The difference is only due to impurities (like salts or dirt), not the chemical substance itself.

Pure water from any source is chemically identical (H₂O).

Q2. Oxygen is sometimes represented as O and sometimes as O₂. What is the difference between these symbols?

Answer:

Symbol O → means one atom of oxygen.
Symbol O₂ → means two oxygen atoms joined together to form a molecule.
Oxygen in nature is found as O₂ molecules, not as single atoms, because single atoms are unstable.

Q3. Why does dissolved salt in water conduct electricity, but sugar does not?

Answer: Salt solution conducts electricity because salt breaks into ions (Na⁺ and Cl⁻) in water. These ions move and carry electric current. Sugar solution does not conduct electricity because sugar dissolves only as neutral molecules, without forming ions. So, no charged particles are present to carry current.

Q4. A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left in the beaker. Does this mean the Law of Conservation of Mass is violated? Explain.

Answer: No, the Law of Conservation of Mass is not violated. When ethanol burns in an open beaker, it changes into gases like carbon dioxide and water vapour. These gases escape into the air, so nothing is left in the beaker. The mass is still conserved, but it is spread into the surroundings.

Q5. When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation of Mass?

Answer: According to the Law of Conservation of Mass, the total mass of reactants = total mass of products. Here, hydrogen = 20 g and oxygen = 160 g. So, the total mass of reactants = 20 + 160 = 180 g.

This means the mass of water formed will also be 180 g.

Q6. A compound consists of 40% sulfur and 60% oxygen by mass. In a sample of the same compound containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?

Answer: The compound has 40% sulfur and 60% oxygen. This means for every 40 g of sulfur, there must be 60 g of oxygen. If the sample has 20 g of sulfur, then oxygen must be in the same ratio.

\[ \text{Oxygen mass} = \frac{60}{40} \times 20 = 30 \, \text{g} \]

Q7. Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3:4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?

Answer: Carbon monoxide has carbon and oxygen in the mass ratio 3:4. This means for every 3 g of carbon, 4 g of oxygen is needed. If we have 9 g of carbon, then oxygen required will be:

\[ \frac{4}{3} \times 9 = 12 \, \text{g} \]

Q8. The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.

Answer:

The law works for compounds because their elements always join in a fixed ratio (like water is always 1:8 by mass).
In mixtures, substances are just mixed, so the ratio can change (like salt water can be strong or weak).

Compounds = fixed ratio, Mixtures = variable ratio.

Q9. Students X and Y, both prepared an oxide of copper by combining copper and oxygen in the ratios of 4:1 and 8:2, espectively. Do their results justify the Law of Constant Proportions? Explain.

Answer: Yes, their results justify the Law of Constant Proportions. Student X used copper and oxygen in the ratio 4:1, while Student Y used 8:2. If we simplify 8:2, it also becomes 4:1. This shows that both students got the same fixed ratio of copper to oxygen.

Q10. Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.
Reason (R): According to Dalton’s Atomic Theory, atoms combine in a simple whole number
ratio by mass to form compounds.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Answer: (i) Both A and R are true, and R is the correct explanation of A.

Q11. Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (N₂).

Answer:

Q12. The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).

Answer: Fluorine has an atomic number of 9, so each atom has 9 electrons — with 7 in the outermost shell. To complete its octet, each fluorine atom needs 1 more electron. When two fluorine atoms come close, they share one pair of electrons, forming a single covalent bond (F–F). This sharing helps both atoms get 8 electrons in their outer shell, making the molecule stable.

Q13. Show the formation of the following molecules:
(i) Carbon dioxide (CO₂)
(ii) Hydrogen sulfide (H₂S)
(iii) Ammonia (NH₃)

Answer:

Q14. Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.

Answer: Neon has 10 electrons, with 8 in its outermost shell. This means its outer shell is already full. Because it is stable, neon does not need to gain, lose, or share electrons.

Q15. What kind of ion will oxygen (O) form?

Answer: Oxygen has 6 valence electrons. It needs 2 more to complete its outer shell. So, oxygen gains 2 electrons and forms a negative ion (O²⁻) called an oxide ion.

Q16. Fill in the blanks.
Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become _. Now, _ ion of magnesium and __ ions of chlorine combine to give magnesium chloride.

Answer: Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become Cl⁻. Now, one ion of magnesium and two ions of chlorine combine to give magnesium chloride.

Q17. Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.

Answer:

(i) Potassium (K → K⁺)

Potassium has 1 valence electron.
It loses 1 electron to form K⁺ cation.
Chlorine has 7 valence electrons. It gains 1 electron to form Cl⁻ anion.
Together, K⁺ + Cl⁻ → KCl (ionic bond).

(ii) Calcium (Ca → Ca²⁺)

Calcium has 2 valence electrons.
It loses 2 electrons to form Ca²⁺ cation.
Each chlorine atom needs 1 electron, so two Cl atoms take these electrons.
Together, Ca²⁺ + 2Cl⁻ → CaCl₂ (ionic bond).

Q18. Illustrate how sodium sulfide (Na₂S) is formed.

Answer: Formation of Sodium Sulfide (Na₂S):

Sodium (Na): Each sodium atom has 1 valence electron. It loses this electron to form a Na⁺ cation.
Sulfur (S): Sulfur has 6 valence electrons. It needs 2 more electrons to complete its octet.
So, two sodium atoms each give 1 electron to sulfur.
Sulfur gains these 2 electrons and becomes S²⁻ anion.
Finally, 2 Na⁺ ions + 1 S²⁻ ion → Na₂S (ionic compound).”

Q19. Name the following:
(i) CO₂
(ii) NO₂
(iii) SF₆
(iv) PCl₃

Answer: Here are the names of the given compounds:

CO₂ → Carbon dioxide
NO₂ → Nitrogen dioxide
SF₆ → Sulfur hexafluoride
PCl₃ → Phosphorus trichloride

Q20. Write the formula for the following:
(i) Sodium hydrogencarbonate .
(ii) Sulfur dioxide .
(iii) Ferric chloride .
(iv) Cuprous oxide _.

Answer: Here are the formulas for the given compounds:

Sodium hydrogencarbonate → NaHCO₃
Sulfur dioxide → SO₂
Ferric chloride → FeCl₃
Cuprous oxide → Cu₂O

Q21. Write the formulae for the compounds formed from the following pairs of ions:
(i) Fe³⁺ and OH^‒
(ii) K⁺ and CO2-/3

Answer:

(i) Fe³⁺ and OH⁻

Iron (Fe³⁺) has a charge of +3.
Hydroxide (OH⁻) has a charge of –1.
To balance charges, 3 OH⁻ ions are needed for 1 Fe³⁺ ion.
Formula: Fe(OH)₃

(ii) K⁺ and CO₃²⁻

Potassium (K⁺) has a charge of +1.
Carbonate (CO₃²⁻) has a charge of –2.
To balance charges, 2 K⁺ ions are needed for 1 CO₃²⁻ ion.
Formula: K₂CO₃

Q22. What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?

Answer: A solid compound that does not conduct electricity in solid state but does conduct when dissolved in water has an ionic bond.

In solid form, ions are fixed in a lattice → no free movement.
In water, ions separate and move freely → they carry electric current.

Q23. Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its:
(i) formula
(ii) type of bond
(iii) electrical conductivity of its aqueous solution.

Answer:

1. Identify the formula

Metal M loses 2 electrons, oxygen gains 2 electrons.
M → M²⁺ (cation)
O → O²⁻ (anion)
Combine in 1:1 ratio → MO

2. Determine bond type

Electrons are transferred from M to O.
Transfer of electrons → Ionic bond
Strong electrostatic attraction holds ions together

3. Predict conductivity

Check behavior in solid and aqueous states.
Solid MO: ions fixed, no conduction
Aqueous MO: ions free, conducts electricity
Slight solubility → weak but present conductivity

(i) Formula: MO
(ii) Type of bond: Ionic bond
(iii) Electrical conductivity: Conducts electricity in aqueous solution (due to free ions).

Q24. Find the molecular mass of nitric acid (HNO₃).
Atomic mass — H = 1 u; N = 14 u; O = 16 u.

Answer: Step‑by‑step calculation of molecular mass of nitric acid (HNO₃):

Hydrogen (H): Atomic mass = 1 u → contributes 1 u
Nitrogen (N): Atomic mass = 14 u → contributes 14 u
Oxygen (O): Atomic mass = 16 u × 3 atoms = 48 u

Total molecular mass = 1 + 14 + 48 = 63 u

Molecular mass of HNO₃ = 63 u

Q25. Find the molecular mass of methane (CH₄).
Atomic mass — C = 12 u; H = 1 u.

Answer: Step‑by‑step calculation of molecular mass of methane (CH₄):

Carbon (C): Atomic mass = 12 u → contributes 12 u
Hydrogen (H): Atomic mass = 1 u × 4 atoms = 4 u

Total molecular mass = 12 + 4 = 16 u

Molecular mass of CH₄ = 16 u

Q26. Find the formula unit mass of potassium chloride (KCl).
Atomic mass — K = 39 u; Cl = 35.5 u.

Answer: Step‑by‑step calculation of formula unit mass of potassium chloride (KCl):

Potassium (K): Atomic mass = 39 u → contributes 39 u
Chlorine (Cl): Atomic mass = 35.5 u → contributes 35.5 u

Total formula unit mass = 39 + 35.5 = 74.5 u

Formula unit mass of KCl = 74.5 u

Q27. Find the formula unit mass of magnesium hydroxide, Mg(OH)₂.
Atomic mass — Mg = 24 u; O = 16 u; H = 1 u.

Answer: Step‑by‑step calculation of formula unit mass of magnesium hydroxide, Mg(OH)₂:

Magnesium (Mg): Atomic mass = 24 u → contributes 24 u
Oxygen (O): Atomic mass = 16 u × 2 atoms = 32 u
Hydrogen (H): Atomic mass = 1 u × 2 atoms = 2 u

Total formula unit mass = 24 + 32 + 2 = 58 u

Formula unit mass of Mg(OH)₂ = 58 u

Q28. A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell.
(i) How many electrons does A tend to give or take to become stable?
(ii) What kind of ion would it form?
(iii) How many electrons does B tend to give or take to become stable?
(iv) What kind of ion would it form?
(v) If A and B were to combine, what kind of bond would be formed?
(vi) What would be the formula for the compound thus formed?

Answer:

Element A

Has 1 electron in its 3rd shell → like sodium (Na).

(i) It will give 1 electron to become stable.
(ii) It forms a positive ion (A⁺ cation).

Element B

Has 6 electrons in its 2nd shell → like oxygen (O).

(iii) It will take 2 electrons to complete its octet.
(iv) It forms a negative ion (B²⁻ anion).

When A and B combine

(v) Ionic bond is formed (transfer of electrons).
(vi) To balance charges: 2 A⁺ ions + 1 B²⁻ ion → A₂B

Final Answer:

A gives 1 electron → A⁺
B takes 2 electrons → B²⁻
Bond: Ionic
Formula: A₂B

Q29. An element X has six electrons in its outer shell and forms a diatomic molecule.
(i) Why would that be so?
(ii) What kind of bond would it form?
(iii) Draw the structure of the molecule it would form.
(iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.

Answer:

(i) Why would X form a diatomic molecule?

Element X has 6 valence electrons (like oxygen).
To complete its octet, it needs 2 more electrons.
Two X atoms share 2 pairs of electrons → each gets 8 electrons.
So, X forms a diatomic molecule (X₂).

(ii) What kind of bond would it form?

By sharing electrons, X forms a covalent bond.

(iii) Structure of X₂ molecule

Two X atoms share two pairs of electrons → a double bond (X = X).

(iv) Structure of molecule formed with Y

Element Y has 2 valence electrons (like magnesium).
Y loses 2 electrons → forms Y²⁺ cation.
X gains 2 electrons → forms X²⁻ anion.
Together, they form an ionic bond.
Formula: YX

Q30. You want to design a new ionic compound, where the total positive charge is 6+ and the total negative charge is 6 –. Which of the following combinations gives the correct number of ions?
(i) 2 Al³⁺ and 3 Cl^–
(ii) 3 Mg²⁺ and 1 PO3-/4
(iii) 2 Fe³⁺ and 3 O^2-
(iv) 3 Ca²⁺ and 2 SO2-/4

Answer:

(i) 2 Al³⁺ and 3 Cl⁻

2 Al³⁺ → total +6
3 Cl⁻ → total –3

Not balanced

(ii) 3 Mg²⁺ and 1 PO₄³⁻

3 Mg²⁺ → total +6
1 PO₄³⁻ → total –3

Not balanced

(iii) 2 Fe³⁺ and 3 O²⁻

2 Fe³⁺ → total +6
3 O²⁻ → total –6

Balanced → Correct combination, Formula: Fe₂O₃

(iv) 3 Ca²⁺ and 2 SO₄²⁻

3 Ca²⁺ → total +6
2 SO₄²⁻ → total –4

Not balanced

Q31. Choose the correct statement(s) and correct the false statement(s).
(i) Elements are made up of molecules and compounds are made up of atoms.
(ii) The molecule of a compound is always made up of two or more atoms of the same kind.
(iii) One molecule of nitrogen gas contains three nitrogen atoms.
(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.

Answer: (iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.

Q32. Write the chemical formulae for the following compounds.
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide

Answer:

Aluminium nitrate → Al(NO₃)₃
Calcium oxide → CaO
Ferric oxide → Fe₂O₃

Q33. Write the formulae of the compounds formed from the following pairs of ions.
(i) Ca²⁺ and Br^–
(ii) Al³⁺ and CO2-/3
(iii) K⁺ and SO2-/4
(iv) NH+4 and Cl^–

Answer:

(i) Ca²⁺ and Br⁻

Calcium ion = +2
Bromide ion = –1
Need 2 Br⁻ ions to balance 1 Ca²⁺.
Formula: CaBr₂

(ii) Al³⁺ and CO₃²⁻

Aluminium ion = +3
Carbonate ion = –2
Balance by criss‑cross: 2 Al³⁺ and 3 CO₃²⁻.
Formula: Al₂(CO₃)₃

(iii) K⁺ and SO₄²⁻

Potassium ion = +1
Sulfate ion = –2
Need 2 K⁺ ions for 1 SO₄²⁻.
Formula: K₂SO₄

(iv) NH₄⁺ and Cl⁻

Ammonium ion = +1
Chloride ion = –1
Balance: 1 NH₄⁺ with 1 Cl⁻.
Formula: NH₄Cl

Q34. Which of the following, in Fig. 9.18, correctly represents Cl– ion (Atomic number of chlorine = 17).

Answer:

Chlorine atom (Z = 17) has configuration 2, 8, 7 → shown in diagram (iii).
Chloride ion (Cl⁻) gains 1 electron → configuration becomes 2, 8, 8 → shown in diagram (iv).

So, diagram (iv) correctly represents Cl⁻ ion.

Q35. Determine the formula unit mass of the following substances.
(i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser, which is essential for plant growth.
(ii) Phosphoric acid (H³PO₄), used to make phosphate fertiliser and detergents.
(iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity and helps in digestion.

Answer:

(i) Ammonium nitrate (NH₄NO₃)

N = 14 u × 2 atoms = 28 u
H = 1 u × 4 atoms = 4 u
O = 16 u × 3 atoms = 48 u
Total = 28 + 4 + 48 = 80 u
Formula unit mass = 80 u

(ii) Phosphoric acid (H₃PO₄)

H = 1 u × 3 atoms = 3 u
P = 31 u × 1 atom = 31 u
O = 16 u × 4 atoms = 64 u
Total = 3 + 31 + 64 = 98 u
Formula unit mass = 98 u

(iii) Sodium hydrogencarbonate (NaHCO₃)

Na = 23 u × 1 atom = 23 u
H = 1 u × 1 atom = 1 u
C = 12 u × 1 atom = 12 u
O = 16 u × 3 atoms = 48 u
Total = 23 + 1 + 12 + 48 = 84 u
Formula unit mass = 84 u

Q36. Write the formulae for the compounds formed by the reaction of:
(i) Magnesium and nitrogen
(ii) Lithium and nitrogen
(iii) Sodium and sulfur
(iv) Aluminium and oxygen

Answer:

(i) Magnesium and nitrogen

Mg → Mg²⁺
N → N³⁻
Balance charges: 3 Mg²⁺ (total +6) with 2 N³⁻ (total –6).
Formula: Mg₃N₂

(ii) Lithium and nitrogen

Li → Li⁺
N → N³⁻
Balance charges: 3 Li⁺ (total +3) with 1 N³⁻ (total –3).
Formula: Li₃N

(iii) Sodium and sulfur

Na → Na⁺
S → S²⁻
Balance charges: 2 Na⁺ (total +2) with 1 S²⁻ (total –2).
Formula: Na₂S

(iv) Aluminium and oxygen

Al → Al³⁺
O → O²⁻
Balance charges: 2 Al³⁺ (total +6) with 3 O²⁻ (total –6).
Formula: Al₂O₃

Q37. Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiNO₃ is given as an example.

Answer:

Cation / Anion	NO₃⁻	SO₄²⁻	PO₄³⁻
NH₄⁺	NH₄NO₃	(NH₄)₂SO₄	(NH₄)₃PO₄
Li⁺	LiNO₃ (given)	Li₂SO₄	Li₃PO₄
Al³⁺	Al(NO₃)₃	Al₂(SO₄)₃	AlPO₄
Cu²⁺	Cu(NO₃)₂	CuSO₄	Cu₃(PO₄)₂

Q38. 5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.

Answer:

Reactants (before reaction)

Sodium carbonate = 5.3 g
Acetic acid = 6.0 g
Total mass of reactants = 5.3 + 6.0 = 11.3 g

Products (after reaction)

Carbon dioxide = 2.2 g
Water = 0.9 g
Sodium acetate = 8.2 g
Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g

Verification

Mass of reactants = 11.3 g
Mass of products = 11.3 g
Both are equal.

The Law of Conservation of Mass is valid in this reaction, because the total mass before and after the reaction is the same.

Q39. If a species has 11 protons, 12 neutrons and 10 electrons then
(i) what is its atomic number and mass number?
(ii) is it neutral, a cation or an anion? Explain.
(iii) write its electronic configuration.
(iv) name the species.

Answer:

(i) Atomic number and mass number

Protons = 11 → Atomic number = 11
Protons + Neutrons = 11 + 12 = 23 → Mass number = 23

(ii) Neutral, cation or anion?

Neutral atom of sodium would have 11 protons and 11 electrons.
Here, electrons = 10 (one less).
So, it is a cation (Na⁺) because it has lost 1 electron.

(iii) Electronic configuration

For Na atom (11 e⁻): 2, 8, 1
For Na⁺ (10 e⁻): 2, 8
Electronic configuration = 2, 8

(iv) Name of the species

Atomic number 11 = Sodium (Na).
Since it is a cation with +1 charge → Sodium ion (Na⁺)

Final Answer:

Atomic number = 11, Mass number = 23
Species is a cation (Na⁺)
Electronic configuration = 2, 8
Name = Sodium ion (Na⁺)

Q40. Two elements, A and B, have the following configurations —
A: 2, 8, 5
B: 2, 8, 7
(i) Which element is more reactive?
(ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing.
(iii) Predict the formula of the compound they would form.

Answer:

(i) Which element is more reactive?

A: 2, 8, 5 → Atomic number 15 (Phosphorus)
B: 2, 8, 7 → Atomic number 17 (Chlorine)
Reactivity depends on how easily atoms achieve stability.
Chlorine (B) needs only 1 electron to complete its octet, so it is more reactive than phosphorus (A).
Correct: Element B (Chlorine) is more reactive.

(ii) Type of bond when A and B combine

A (Phosphorus) has 5 valence electrons → needs 3 more.
B (Chlorine) has 7 valence electrons → needs 1 more.
Both are non‑metals, so they will share electrons.
They form a covalent bond by sharing electrons.

(iii) Formula of the compound formed

Phosphorus (A) needs 3 electrons → bonds with 3 chlorine atoms (each providing 1 electron).
Compound formed = PCl₃ (phosphorus trichloride).

Final Answer:

More reactive element = B (Chlorine)
Bond type = Covalent bond (electron sharing)
Formula = PCl₃

Q41. Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state.
Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Answer: (iii) A is true, but R is false.

Q42. The species ²⁷Al, ⁸⁰Br^– and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?

Answer:

Species 1: ²⁷Al

Protons = 13 → Atomic number = 13
Neutrons = 27 − 13 = 14
Electrons = 13 (neutral atom) → 13
²⁷Al has 13 protons, 14 neutrons, 13 electrons

Species 2: ⁸⁰Br⁻

Protons = 35 → Atomic number = 35
Neutrons = 80 − 35 = 45
Electrons = 35 + 1 (extra electron for negative charge) = 36
⁸⁰Br⁻ has 35 protons, 45 neutrons, 36 electrons

Species 3: ²⁰¹Hg²⁺

Protons = 80 → Atomic number = 80
Neutrons = 201 − 80 = 121
Electrons = 80 − 2 (lost two for positive charge) = 78
²⁰¹Hg²⁺ has 80 protons, 121 neutrons, 78 electrons

Species	Protons	Neutrons	Electrons
²⁷Al	13	14	13
⁸⁰Br⁻	35	45	36
²⁰¹Hg²⁺	80	121	78

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