Model Test – Solved ഉത്തരങ്ങൾ
Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated shell diagram or table picture for the longer questions, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.
In the shell diagrams the nucleus is a dark-red disc showing the number of protons (for example 15p), the shells K, L, M, N are orange rings and the electrons are green dots – not to scale. In the table pictures each cell is one element; rows are periods (പീരിയഡുകൾ) and columns are groups (ഗ്രൂപ്പുകൾ), and the colours show the families of Table 2.2:
Rules used everywhere: period = number of shells; group = outermost electrons (groups 1, 2) or outermost electrons + 10 (groups 13–18); helium is in group 18.
💡 Moseley or Mendeleev? The modern law is Moseley’s.
Henry Moseley’s X-ray diffraction experiments showed that the properties of elements depend on their atomic numbers, not on their atomic masses. This is the modern periodic law (ആധുനിക പീരിയോഡിക് നിയമം). Mendeleev’s law of 1869 used atomic masses.
Marking: 1 mark for (b).
📖 Review this on Page 4 · Modern periodic law, key box 1
💡 They show a gradual change between the two sides of the table.
The 10 groups 3–12 lie between the highly metallic groups 1–2 and the less metallic groups 13–18. They show the gradual change (transition) between them, so they are called transition elements.
Marking: 1 mark.
📖 Review this on Page 7 · Transition elements, key box 9
💡 Count the lanthanoids too.
Period 6 has 32 elements: 18 places in the main table, where one place (group 3) stands for the 15 lanthanoids (La 57 to Lu 71) of the row below – 17 + 15 = 32. (Periods: 2, 8, 8, 18, 18, 32, 32.)
Marking: 1 mark for (c).
📖 Review this on Page 8 · Period 6 and the lanthanoids · Page 5 · Periods
💡 Neon is in group 18.
Sodium (2, 8, 1) is in group 1 – the alkali metals. Neon (2, 8) is in group 18 – the noble gases (Table 2.2).
Marking: 1 mark.
📖 Review this on Page 5 · Table 2.2, families
💡 Count the numbers in the configuration, not their sum.
Period number = number of shells (ഷെല്ലുകളുടെ എണ്ണം). 2, 8, 18, 7 has 4 shells → period 4. (It is bromine, Z = 35; group 7 + 10 = 17.)
Marking: 1 mark.
📖 Review this on Page 6 · Table 2.7, period number
💡 Which of them is not in groups 3–12?
Iron (group 8), copper (group 11) and zinc (group 12) are transition elements. Magnesium (2, 8, 2) is in group 2 – a main group element (alkaline earth metal). All four are metals, so “metal” is not the reason.
Marking: ½ mark for magnesium, ½ mark for the reason.
📖 Review this on Page 7 · Transition elements · Page 6 · Main group elements
💡 Same group – which has the most shells?
All four are in group 2. Down a group the number of shells increases (Be 2, Mg 3, Ca 4, Sr 5), so the size of the atom increases. Strontium, the lowest of the four, is the largest.
Marking: 1 mark for (d).
📖 Review this on Page 9 · Size down a group, key box 14
💡 It is used to make computer chips.
Metalloids show both metallic and non-metallic properties: silicon, germanium, arsenic, antimony (and boron, tellurium). Sodium is a metal; sulphur and argon are non-metals.
Marking: 1 mark for (b).
📖 Review this on Page 6 · Metals, non-metals and metalloids · Page 1 · Metalloids
💡 More than 2 outermost electrons → add 10.
(a) 15 electrons: K takes 2, L takes 8 and M gets 15 − 10 = 5 → 2, 8, 5.
Given: configuration 2, 8, 5
Rule: period = number of shells; groups 13–18: group = outermost electrons + 10
Working: 3 shells; 5 outermost electrons → 5 + 10 = 15
Answer: (b) period 3, group 15
X is phosphorus, in the nitrogen family. (Adding 10 is needed because the 10 groups of transition elements come between group 2 and group 13.)
Marking: (a) 1 mark; (b) period ½, group ½ (the group must use + 10).
📖 Review this on Page 6 · Tables 2.5–2.7, group and period finder
💡 Compare 40 and 39, then 18 and 19.
(a) Mendeleev arranged the elements in the increasing order of atomic mass (ആരോഹണക്രമം). Argon (40) is heavier than potassium (39), yet it has to come first so that potassium can stand with the alkali metals – the order of atomic mass is not followed here.
(b) In the modern table the elements are arranged in the increasing order of atomic number. Argon is 18 and potassium is 19, so argon comes before potassium naturally – no exception. Argon falls in group 18 (noble gases) and potassium in group 1 (alkali metals).
Marking: (a) 1 mark; (b) 1 mark.
📖 Review this on Page 3 · Merits and limitations, card sort · Page 4 · Ar–K before and after
💡 Count the electrons in the last shell.
(a) Outermost electrons 8 → group 8 + 10 = 18; 4 shells → period 4. Krypton is a noble gas.
(b) Its outermost shell has 8 electrons. Such an arrangement is stable (സ്ഥിരത), so krypton, like the other group 18 elements, normally does not take part in chemical reactions (key box 8).
Marking: (a) ½ + ½; (b) 1 mark (8 outermost electrons ½, stable ½).
📖 Review this on Page 7 · Table 2.8, noble gases
💡 Each family is named after its first element – except groups 1, 2, 17 and 18.
(i) group 2 – alkaline earth metals ആൽക്കലൈൻ എർത്ത് ലോഹങ്ങൾ (Be, Mg, Ca …)
(ii) group 13 – boron family (B, Al, Ga …)
(iii) group 16 – oxygen family (O, S, Se …)
(iv) group 17 – halogens ഹാലൊജനുകൾ (F, Cl, Br, I …)
Marking: ½ mark for each.
📖 Review this on Page 5 · Table 2.2, families
💡 How many electrons can the K shell hold?
(a) Helium has only one shell, K, and K can hold at most 2 electrons. So helium’s outermost shell is full: it has a stable arrangement and, like the noble gases, does not normally react. Beryllium’s 2 outermost electrons are in the L shell, which is far from full – beryllium is a reactive metal of group 2.
(b) Period 1 elements have just one shell, K, which holds at most 2 electrons. So only hydrogen (1) and helium (2) fit in period 1.
Marking: (a) 1 mark; (b) 1 mark.
📖 Review this on Page 7 · Noble gases · Page 5 · Period 1
💡 Same number of outermost electrons → same group.
(a) P (Z = 6): 2, 4; Q (Z = 14): 2, 8, 4.
(b) Yes. All three have 4 outermost electrons → group 4 + 10 = 14, the carbon family (കാർബൺ കുടുംബം). (P, Q, R are carbon, silicon and germanium.)
(c) R. P has 2 shells, Q has 3 and R has 4. Down a group the number of shells increases, so the size of the atom increases – R, with the most shells, is the largest.
Marking: (a) ½ + ½; (b) group ½, family ½; (c) R ½, reason ½.
📖 Review this on Page 6 · Group number · Page 5 · Families · Page 9 · Size down a group
(a) Find their atomic numbers. (b) In which shell is the new electron added as we go from X to Y to W? What type of elements are these? (c) X, Y and W are in different groups, yet they show similar chemical properties. Why?
💡 Compare the last two numbers of each configuration.
(a) Atomic number = total electrons: X = 2 + 8 + 14 + 2 = 26, Y = 2 + 8 + 15 + 2 = 27, W = 2 + 8 + 16 + 2 = 28. (Iron, cobalt and nickel.)
(b) The third shell grows 14 → 15 → 16 while the outermost (fourth) shell stays at 2: the new electron enters the shell just inside the outermost shell (ബാഹ്യതമ ഷെല്ലിന് തൊട്ടുള്ളിലുള്ള ഷെൽ). Such elements are transition elements (key box 10).
(c) Chemical properties depend mainly on the outermost electrons. X, Y and W all have 2 outermost electrons, so they are similar along the period too (key box 11).
Marking: (a) 1 mark (two correct ½); (b) shell ½, transition ½; (c) 1 mark.
📖 Review this on Page 7 · Table 2.9, penultimate-shell walk
| Element | Atomic number | Electron configuration | Period | Group |
|---|---|---|---|---|
| Nitrogen | 7 | (i) …… | (ii) …… | 15 |
| Calcium | 20 | 2, 8, 8, 2 | 4 | (iii) …… |
| Gallium | (iv) …… | 2, 8, 18, 3 | (v) …… | (vi) …… |
💡 Period: count shells. Group: look at the last number.
| Element | Z | Configuration | Period | Group |
|---|---|---|---|---|
| Nitrogen | 7 | (i) 2, 5 | (ii) 2 (2 shells) | 15 |
| Calcium | 20 | 2, 8, 8, 2 | 4 | (iii) 2 (2 outermost electrons) |
| Gallium | (iv) 31 = 2 + 8 + 18 + 3 | 2, 8, 18, 3 | (v) 4 (4 shells) | (vi) 13 = 3 + 10 |
Period = number of shells. Group = outermost electrons for groups 1 and 2, outermost electrons + 10 for groups 13–18. Atomic number = sum of the electrons.
Marking: ½ mark for each blank.
📖 Review this on Page 6 · Group and period finder · Page 2 · Electron configuration
💡 Down a group: shells win. Across a period: nuclear charge wins.
(a) Potassium – it has 4 shells, lithium only 2 (same group 1).
(b) A larger nuclear charge (ന്യൂക്ലിയർ ചാർജ്) pulls the electrons in, but potassium has two more shells. Its outermost electron is much farther from the nucleus and the inner electrons screen it (shielding). The effect of the extra shells is greater than the effect of the extra nuclear charge, so the atom is larger (key box 14).
(c) Boron. Both are in period 2 with the same number of shells (2). Fluorine has the greater nuclear charge (+9 against +5), so it attracts its outermost electrons more strongly and is smaller (key box 15). (Approx. radii: B 82 pm, F 71 pm.)
Marking: 1 mark for each part (answer ½, reason ½).
📖 Review this on Page 9 · Tables 2.11–2.12, size lab
(b) Write the electron configuration of J. Show how its group number is found from it.
(c) Find the atomic number of Q. Why is Q called a transition element?
(d) G and M show similar chemical properties. Why?
(e) Arrange G, J and L in the increasing order of atomic size. Give the reason.
💡 Read each letter’s period (row) and group (column) first.
(a) Group 17 holds the halogens: E and L. D is in group 2 – the alkaline earth metals.
(b) J is in period 3 → 3 shells; group 16 → 16 − 10 = 6 outermost electrons. Configuration 2, 8, 6 (Z = 16). From the configuration: outermost electrons 6 → group 6 + 10 = 16.
(c) Period 4 starts with Z = 19 in group 1, then 20 (group 2), 21 (group 3), 22 (group 4), so Q in group 5 is Z = 23. Q is in groups 3–12; its new electron enters the shell just inside the outermost shell – a transition element.
(d) G and M are both in group 1: each has 1 outermost electron, and the outermost electrons decide the chemical properties.
(e) L < J < G. All three are in period 3 – same number of shells. From G (group 1) to J (16) to L (17) the nuclear charge increases, the pull on the outermost electrons increases, and the size decreases.
Real elements: D Be, E F, G Na, J S, L Cl, M K, Q V, R Kr, T He.
Marking: 1 mark for each part (½ + ½ where there are two answers).
📖 Review this on Page 11 · Partial-table question · Page 5 · The interactive table · Page 9 · Size across a period
(b) Potassium permanganate is purple and copper sulphate is blue. Which property of transition elements does this show? Lime is white – why?
(c) Calcium has the configuration 2, 8, 8, 2. Find its group number and name its family.
(d) Compare the configurations of titanium and manganese. In which shell are the extra electrons of manganese added? How many outermost electrons does each have?
(e) The rare earths obtained from monazite are lanthanoids. Which elements are the lanthanoids, and where are they placed in the periodic table?
💡 Count the shells; look at the second-last shell.
(a) Each configuration has 4 shells → period 4 (transition elements begin in period 4).
(b) Most compounds of transition elements are coloured (നിറമുള്ള സംയുക്തങ്ങൾ). Lime is a compound of calcium, a main group element (group 2), not a transition element – so it is white (like calcium carbonate in Table 2.10).
Given: calcium 2, 8, 8, 2
Rule: groups 1 and 2: group = number of outermost electrons
Working: 2 outermost electrons
Answer: (c) group 2, the alkaline earth metals
(d) Titanium 2, 8, 10, 2 → manganese 2, 8, 13, 2: the third shell (M, just inside the outermost shell) grows from 10 to 13, while the outermost shell has 2 electrons in both.
(e) Lanthanum (57) and the next 14 elements, up to lutetium (71) are the lanthanoids (ലാൻഥനോയ്ഡുകൾ, also called rare earths). They belong to period 6 but are placed in a separate row below the main table.
Marking: 1 mark for each part (½ + ½).
📖 Review this on Page 7 · Coloured compounds, Table 2.9 · Page 8 · Lanthanoids, rare earths of Chavara · Page 12 · Applications
Marks summary
| Section | Questions | Marks each | Total | Topics |
|---|---|---|---|---|
| A | 1–8 | 1 | 8 | modern periodic law, transition elements, period 6, families, period from the configuration, odd one out, size down a group, metalloid |
| B | 9–13 | 2 | 10 | group and period from Z, the Ar–K problem, krypton and stability, family names, helium and period 1 |
| C | 14–17 | 3 | 12 | the carbon family and size, the penultimate shell, completing a table, size with reasons |
| D | 18–19 | 5 | 10 | partial periodic table with letters, Kerala case study on transition elements and rare earths |
| Total | 40 | |||
- Cool-off (15 min) – read the whole paper. Mark the questions where you must find a group or period (Q5, Q9, Q11, Q14, Q16, Q18, Q19) and the size questions (Q7, Q14, Q17, Q18).
- Section A – about 10 minutes. One word or one letter; give the reason in Q6.
- Section B – about 20 minutes. Write the configuration first, then the period (shells) and the group (+10 only for 13–18).
- Section C – about 25 minutes. In every size answer write the reason: “more shells” down a group, “greater nuclear charge, same shells” across a period.
- Section D – about 25 minutes. In Q18 first write the period and group of every letter beside the figure; in Q19 underline the numbers in the passage.
- Checking – keep 10 minutes: did you add 10 for groups 13–18 and not for groups 1 and 2? Period = shells (not outermost electrons)? Helium in group 18?
Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5 to 9 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.