CBSE Class 7 Science Advanced Chemistry Atomic Structure and Transformation of Matter Long Answer Questions Set 01

Long Answer Questions

Question 1. Explain in detail the modern structure of atom.
Answer:
1. An atom has three fundamental particles: protons (+1, 1 amu), neutrons (no charge, 1 amu) and electrons (−1, \( \frac{1}{1837} \) amu).
2. Protons and neutrons are packed in a small central region, the nucleus. They are called nucleons.
3. The nucleus is very small compared with the atom, so most of the atom is empty space.
4. Electrons revolve at high speed round the nucleus in fixed paths called orbits or shells, which are named K, L, M, N or 1, 2, 3, 4.
5. The shells hold at most 2, 8, 18, 32 electrons (\( 2n^2 \)), and the energy of a shell rises as its distance from the nucleus increases.
6. The number of electrons equals the number of protons, so an atom is electrically neutral.
7. The outermost shell is the valence shell, and it can have at most 8 electrons.
8. The atomic number Z is the number of protons, and the mass number A is protons + neutrons.
In simple words: A tiny, heavy nucleus sits at the centre and light electrons move around it in shells.

Teacher's Note:
1. Cover nucleus, shells, charge and capacity.
2. A neat labelled sketch will earn extra marks.

Question 2. Give the electronic configuration and geometrical representation of the following elements: (a) fluorine (b) sodium (c) chlorine (d) carbon
Answer:
(a) Fluorine (Z = 9): electronic configuration 2, 7; nucleus has 9 protons and 10 neutrons.

(b) Sodium (Z = 11): electronic configuration 2, 8, 1; nucleus has 11 protons and 12 neutrons.

(c) Chlorine (Z = 17): electronic configuration 2, 8, 7; nucleus has 17 protons and 18 neutrons.

(d) Carbon (Z = 6): electronic configuration 2, 4; nucleus has 6 protons and 6 neutrons.

In simple words: Put the protons and neutrons in the nucleus and the electrons in the shells.

Teacher's Note:
1. Neutrons = mass number − atomic number.
2. Check each total number of electrons.

Question 3. Classify the chemical reactions into various types depending on the way the products are formed by giving suitable examples.
Answer:
Chemical reactions are of four types.
1. Chemical combination (\(\ce{X + Y -> XY}\)): two substances join to form one compound. Examples: \(\ce{H2 + Cl2 -> 2HCl}\) (element + element), \(\ce{2SO2 + O2 -> 2SO3}\) (element + compound), \(\ce{CaO + H2O -> Ca(OH)2}\) (compound + compound).
2. Chemical decomposition (\(\ce{AB -> A + B}\)): one compound breaks into more than one substance by heat, light or electricity. Examples: \(\ce{2Pb(NO3)2 -> 2PbO + 4NO2 + O2}\) (thermal), \(\ce{2HOCl -> 2HCl + O2}\) (photolytic), \(\ce{2H2O -> 2H2 + O2}\) (electrolytic).
3. Chemical displacement (\(\ce{X + YZ -> XZ + Y}\)): a more reactive element displaces a less reactive one from its compound. Example: \(\ce{Zn + FeSO4 -> ZnSO4 + Fe}\). But \(\ce{Cu + FeSO4}\) gives no reaction.
4. Double decomposition (double displacement) (\(\ce{AB + XY -> AY + XB}\)): two compounds exchange their radicals. Example: \(\ce{2KNO3 + FeSO4 -> K2SO4 + Fe(NO3)2}\). If one product is an insoluble solid, it is a precipitation reaction, for example \(\ce{NaCl + AgNO3 -> NaNO3 + AgCl v}\).
In simple words: Reactions either join substances, split a substance, swap an element, or swap radicals.

Teacher's Note:
1. Learn the general form of each type.
2. Give one balanced example for each.

Question 4. Mention the postulates and limitations of Dalton's atomic theory.
Answer:
Postulates
1. Matter is made of tiny indivisible particles called atoms, which can be neither created nor destroyed nor changed into atoms of another element.
2. Atoms of one element are identical in all respects.
3. Atoms of different elements differ in all respects.
4. Atoms of different elements combine in simple whole-number ratios to form compounds, and atoms take part in chemical reactions.
5. Atoms may combine in more than one simple whole-number ratio.
Limitations
1. Atoms are divisible, since electrons, protons and neutrons exist.
2. Atoms of the same element may differ in mass (isotopes), for example protium, deuterium and tritium of hydrogen.
3. Atoms of different elements may have the same mass number, for example calcium and argon.
In simple words: The theory said atoms are indivisible and identical, and later discoveries disproved this.

Teacher's Note:
1. Give five postulates and three limitations.
2. Name the isotopes of hydrogen.

Question 5. Write the steps in balancing a chemical equation with an example.
Answer:
Steps
Step I: Count how often each element occurs in the equation and balance first the element that appears the fewest times.
Step II: Among elements that appear equally often, balance the metal before the nonmetal.
Step III: Balance hydrogen and oxygen last.
Example: \(\ce{Fe(OH)3 + H2SO4 -> Fe2(SO4)3 + H2O}\)
Iron appears least often: put 2 before \(\ce{Fe(OH)3}\). Sulphur then needs 3 before \(\ce{H2SO4}\). Hydrogen: 6 + 6 = 12 on the left, so put 6 before \(\ce{H2O}\).
Balanced equation: \(\ce{2Fe(OH)3 + 3H2SO4 -> Fe2(SO4)3 + 6H2O}\) (2 Fe, 3 S, 12 H, 18 O on each side).
In simple words: Balance by adding numbers in front of formulae, one element at a time.

Teacher's Note:
1. Never change subscripts.
2. Verify every element at the end.

Question 6. What is valency? Write a short note on different types of valencies of cations with examples.
Answer:
Valency is the number of electrons an atom loses or gains to attain a stable octet or duplet configuration.
Cations are positive ions made by metals losing electrons. According to the charge:
1. Monovalent (+1): \(\ce{Na+}\), \(\ce{K+}\), \(\ce{H+}\).
2. Bivalent (+2): \(\ce{Mg^2+}\), \(\ce{Ca^2+}\), \(\ce{Zn^2+}\).
3. Trivalent (+3): \(\ce{Al^3+}\), \(\ce{Cr^3+}\).
4. Tetravalent (+4): \(\ce{Sn^4+}\).
Variable valency: some metals form ions with different charges, such as iron (\(\ce{Fe^2+}\) ferrous, \(\ce{Fe^3+}\) ferric) and copper (\(\ce{Cu+}\) cuprous, \(\ce{Cu^2+}\) cupric). The lower valency takes -ous and the higher takes -ic.
In simple words: Valency is the charge on an ion, and metals can give 1, 2, 3 or 4 electrons.

Teacher's Note:
1. Give at least one example per type.
2. Mention variable valency.

Teacher's Note:
1. Write the configuration first.
2. Argon forms no hydride.

Question 8. In an atom, the L and M shells have the same number of electrons. The N shell has one electron less than the K shell. Explain the formation of the compound by this element and oxygen atom.
Answer:
Same electrons in L and M means 8 and 8. The K shell has 2, so the N shell has 2 − 1 = 1 electron.
The configuration is 2, 8, 8, 1, so Z = 19 and the element is potassium.
Oxygen (Z = 8) is 2, 6 and needs 2 electrons to complete its octet.
Each potassium atom loses one electron and becomes \(\ce{K+}\). Oxygen gains two electrons and becomes \(\ce{O^2-}\). Two potassium atoms give their electrons to one oxygen atom.
The oppositely charged ions attract and form \(\ce{K2O}\) (potassium oxide).
In simple words: Potassium gives one electron each to oxygen, and two potassium atoms are needed per oxygen atom.

Teacher's Note:
1. Identify the element first.
2. Use criss-cross to confirm K2O.

Question 9. Explain the formation of the compound formed by two elements with Z = 20 and Z = 17.
Answer:
Z = 20 is calcium, 2, 8, 8, 2. Z = 17 is chlorine, 2, 8, 7.
Calcium loses two electrons to reach an octet and forms \(\ce{Ca^2+}\) (2, 8, 8). Chlorine gains one electron and forms \(\ce{Cl-}\) (2, 8, 8).
One calcium atom gives its two electrons to two chlorine atoms, one each.
The ions \(\ce{Ca^2+}\) and \(\ce{Cl-}\) attract each other and form \(\ce{CaCl2}\) (calcium chloride).
In simple words: One calcium atom gives one electron to each of two chlorine atoms.

Teacher's Note:
1. Show the configurations before and after.
2. Criss-cross gives CaCl₂.

Question 10. An atom of an element has the electrons distributed in four shells. If the ratio of the electrons present in the I and IV shells is 1 : 1, and the ratio of electrons in the II and III shells is also 1 : 1, predict its valency and justify. Write the formulae of (a) nitride (b) silicate (c) chloride.
Answer:
The K shell has 2 electrons, so shell IV also has 2. The L shell has 8, so shell III also has 8.
The configuration is 2, 8, 8, 2 (calcium, Z = 20, represented by M).
It loses 2 electrons to form \(\ce{M^2+}\) (2, 8, 8), so the valency is 2.
(a) Nitride: \(\ce{M^2+}\) and \(\ce{N^3-}\) give \(\ce{M3N2}\).
(b) Silicate: \(\ce{M^2+}\) and \(\ce{SiO3^2-}\) give \(\ce{MSiO3}\).
(c) Chloride: \(\ce{M^2+}\) and \(\ce{Cl-}\) give \(\ce{MCl2}\).
In simple words: Two outer electrons are lost, so the valency is 2, and the formulae follow by criss-cross.

Teacher's Note:
1. Work out the configuration from the ratios.
2. Valencies 2 and 2 cancel for silicate.

Question 11. In a chemical laboratory, a student broke the thermometer while performing an experiment. Immediately, the mercury which was spilled over was removed by the lab in-charge by sprinkling sulphur on it. The teacher asked the students to predict the type of change/reaction associated with the above process. Explain with appropriate reasons.
Answer:
Mercury and sulphur join to form mercuric sulphide: \(\ce{Hg + S -> HgS}\).
A new substance forms from two elements, so it is a chemical change, specifically a combination (synthesis) reaction.
The harmful, easily vaporised mercury is changed into a solid compound that is easy to remove, which is why sulphur is sprinkled on it.
In simple words: Sulphur turns the loose, harmful mercury into solid HgS by a combination reaction.

Teacher's Note:
1. Name the product.
2. Two elements giving one compound is synthesis.

Question 12. Asha's mother asked her daughter Asha to boil the milk and after some time, she told Usha, Asha's sister, to add a little curd in the same lukewarm milk. Next day, their father while teaching "transformation of matter" to his daughters wanted to relate the given activity. So he asked them "what type of change was involved in the activities performed by you yesterday?" Predict the answer given by them and justify.
Answer:
Boiling of milk is a physical change. On boiling, fat particles, being lighter than water, rise as water partly vaporises, and a layer of fat (cream) collects on the surface on cooling. The components of milk are only partly separated and no new substance forms.
Adding curd to lukewarm milk is a chemical change. The bacteria in curd convert the lactose (milk sugar) into lactic acid, which sets the milk into curd. A new substance forms, so the molecular composition changes.
In simple words: Boiling only separates and warms the milk, but setting curd makes new substances.

Teacher's Note:
1. Treat the two activities separately.
2. Say whether a new substance forms.

Question 13. In a science fair, a student took two test tubes labelled A and B filled with copper sulphate solution, then he put magnesium ribbon in test tube A and silver foil in test tube B. Predict the observation and justify.
Answer:
Test tube A: magnesium is more reactive than copper, so it displaces copper: \(\ce{CuSO4 + Mg -> MgSO4 + Cu}\). The blue colour fades slowly and the solution becomes colourless, and reddish-brown copper is deposited.
Test tube B: silver is less reactive than copper and cannot displace it, so there is no reaction and the solution stays blue.
In simple words: Magnesium pushes copper out, but silver cannot.

Teacher's Note:
1. Use the reactivity series.
2. Mention the colour of the solution.

Question 14. An atom of an element "X" has three completely filled shells. The ratio of electrons in the 1 and 4 shells is 1 : 1. Find the atomic number of "X". Also write the formulae of its bicarbonate, chromate and phosphide.
Answer:
Three completely filled shells: K = 2, L = 8 and M = 18. The 4th shell has the same number as the 1st, which is 2.
The configuration is 2, 8, 18, 2, so the atomic number is 2 + 8 + 18 + 2 = 30. X is zinc, which forms \(\ce{Zn^2+}\).
Bicarbonate: \(\ce{Zn^2+}\) and \(\ce{HCO3-}\) give \(\ce{Zn(HCO3)2}\).
Chromate: \(\ce{Zn^2+}\) and \(\ce{CrO4^2-}\) give \(\ce{ZnCrO4}\).
Phosphide: \(\ce{Zn^2+}\) and \(\ce{P^3-}\) give \(\ce{Zn3P2}\).
In simple words: Add up the electrons in all four shells to get Z, then criss-cross.

Teacher's Note:
1. The 1st shell holds 2 electrons.
2. Zinc has valency 2.

Question 15. A metal (M) and a nonmetal (X) form a compound with the formula \(\ce{M4X3}\). Predict the electronic arrangements of M and X assuming that M has three occupied shells and X has two occupied shells. Also derive the formulae of (a) sulphate, (b) dihydrogen phosphate and (c) hydroxide of ion of M.
Answer:
\(\ce{M4X3}\) means M has valency 3 and X has valency 4 (criss-cross: \(\ce{M^3+}\) and \(\ce{X^4-}\)).
M has 3 valence electrons in its third shell: 2, 8, 3, which is aluminium (Z = 13).
X needs 4 electrons and has two shells: 2, 4, which is carbon (Z = 6).
(a) Sulphate: \(\ce{Al2(SO4)3}\)
(b) Dihydrogen phosphate: \(\ce{Al(H2PO4)3}\)
(c) Hydroxide: \(\ce{Al(OH)3}\)
In simple words: The subscripts of M4X3 tell the valencies, from which the electrons in the outer shells follow.

Teacher's Note:
1. Swap the subscripts back to find valencies.
2. Use brackets for repeated radicals.

Teacher's Note:
1. There are two valid answers.
2. Check the nitrate and nitrite radicals carefully.

Question 17. The number of atoms on either side of a chemical equation is balanced by keeping appropriate coefficients and not by changing the subscripts of the elements in the formulae. Justify.
Answer:
The subscript in a formula shows how many atoms of an element are in one molecule of that substance. Changing it changes the molecular composition, so a different substance would be written. For example, \(\ce{H2O}\) is water, but changing the subscript to \(\ce{H2O2}\) gives hydrogen peroxide.
A coefficient only tells how many molecules of that same substance take part. It increases the number of atoms without changing the substance: \(\ce{2H2 + O2 -> 2H2O}\).
Therefore atoms are balanced only through coefficients.
In simple words: Subscripts decide what the substance is; coefficients decide how much of it is there.

Teacher's Note:
1. Give the H₂O and H₂O₂ example.
2. A balanced equation must keep the same substances.

Question 18. An atom of an element has two shells with electrons. The valence shell has 2n + 1 electrons where n is the number of electrons in the inner shell. This element can form a radical "X" with hydrogen and two radicals Y and Z with oxygen, Z has more number of oxygen than Y. Identify the element and the radicals. Write the reaction taking place between solutions of two salts, AgZ and XCl. Identify the type of reaction.
Answer:
The inner shell (K) has n = 2 electrons, so the valence shell has 2 × 2 + 1 = 5 electrons. The configuration is 2, 5 (Z = 7), so the element is nitrogen.
With hydrogen it forms the radical X = \(\ce{NH4+}\) (ammonium).
With oxygen it forms Y = \(\ce{NO2-}\) (nitrite) and Z = \(\ce{NO3-}\) (nitrate), where nitrate has more oxygen.
AgZ is \(\ce{AgNO3}\) and XCl is \(\ce{NH4Cl}\). The reaction is \(\ce{AgNO3 + NH4Cl -> AgCl v + NH4NO3}\).
Two compounds exchange radicals and insoluble silver chloride settles down, so it is a double decomposition reaction and, specifically, a precipitation reaction.
In simple words: Nitrogen (2, 5) gives ammonium, nitrite and nitrate, and the two salts swap radicals to give AgCl as a precipitate.

Teacher's Note:
1. First get the configuration 2, 5.
2. AgCl is the white precipitate.