Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 1 Periodic Properties And Variation Of Properties

Chapter 1. Periodic Properties And Variation Of Properties

Solution 1:
Answer: Modern periodic law states that the physical and chemical properties of elements are a periodic function of their atomic numbers i.e., if the elements are arranged in the order of their atomic numbers, the elements with similar properties are repeated after definite regular intervals.
In simple words: When elements are arranged by their atomic number (number of protons), similar properties repeat in a pattern, just like how days of the week repeat every seven days.

📝 Teacher's Note: Use the analogy of a calendar to explain periodicity — just as Monday comes after every 7 days, similar properties appear after regular intervals in the periodic table. Emphasize that atomic number (not atomic mass) is the organizing principle.

🎯 Exam Tip: Always mention "atomic number" and "periodic function" in your answer. The examiner looks for the phrase "definite regular intervals" to award full marks.

Solution 2:
Answer: Modern periodic table consists of eighteen groups and seven periods.
In simple words: The modern periodic table is like a grid with 18 columns (groups) and 7 rows (periods), where each element has its own specific place.

📝 Teacher's Note: Draw a simple grid on the board to show 18 vertical columns and 7 horizontal rows. Students often confuse the numbers, so use the mnemonic "18 Groups, 7 Periods" repeatedly.

🎯 Exam Tip: Remember the exact numbers — 18 groups and 7 periods. Many students write 8 groups by mistake, which loses marks.

Solution 3:
Answer: The recurrence of similar properties of elements after certain regular intervals when they are arranged in the order of increasing atomic numbers is called periodicity.
In simple words: Periodicity is like a repeating pattern — when you arrange elements by atomic number, similar behaviors come back again and again at regular gaps.

📝 Teacher's Note: Connect this to students' daily life — like how seasons repeat every year or how school periods repeat daily. This helps them understand the concept of "recurrence" better.

🎯 Exam Tip: Use the keywords "recurrence," "regular intervals," and "increasing atomic numbers" — these are scoring points examiners specifically look for.

Solution 4:
Answer: In general, the elements belonging to a group have the same number of valence electrons. For example, all the group 1 elements have valency one since they have only one electron in their outermost shell. In general, the elements belonging to a period do not have same valency but their valence shell remains the same. For example, second period has 8 elements with atomic number 3 to 10 but in all of them the valence electrons are present in shell number two.
In simple words: Elements in the same column (group) have the same number of outer electrons, while elements in the same row (period) have their outer electrons in the same shell level.

📝 Teacher's Note: Use the analogy of apartment floors — elements in the same group live in different buildings but have the same number of people in their top floor, while elements in the same period live on the same floor level.

🎯 Exam Tip: Always give specific examples like "Group 1 elements have 1 valence electron" to support your explanation and earn full marks.

Solution 5:
Answer: Fluorine has lower electron affinity than chlorine because of the small size of fluorine which results in stronger repulsion between the electron and the electrons already present in the atom of fluorine. Hence the energy released in accepting an electron is lesser in fluorine than that of chlorine.
In simple words: Fluorine is like a small room that's already crowded — when you try to add one more person (electron), there's too much pushing and shoving, so less energy is released.

📝 Teacher's Note: This is a counterintuitive concept for students. Emphasize that smaller size doesn't always mean higher electron affinity due to electron-electron repulsion. Use visual aids showing crowded vs spacious atoms.

🎯 Exam Tip: Mention "electron-electron repulsion" and "small size of fluorine" as key points. This exception to the general trend often appears in exams.

Solution 6:
Answer: 1. The element with highest first ionization energy: Neon (Ne) 2. The element with highest electronegativity: Fluorine (F) 3. The element with largest atomic size: Lithium (Li) 4. The most reactive non-metal: Fluorine 5. The most reactive metal: Lithium
In simple words: In the second period, neon is hardest to remove electrons from, fluorine attracts electrons most strongly, lithium is the biggest atom, fluorine is the most eager non-metal, and lithium is the most eager metal to react.

📝 Teacher's Note: Create a visual map of the second period showing these trends. Students should understand that noble gases have highest ionization energy, halogens have highest electronegativity, and alkali metals have largest size and highest reactivity.

🎯 Exam Tip: Learn the second period trends by heart: Li (largest, most reactive metal), F (highest electronegativity, most reactive non-metal), Ne (highest ionization energy).

Solution 7:
Answer: 1. The most metallic element will be found at C. 2. The most non-metallic element will be found at D.
In simple words: The most metallic element is at the bottom-left corner (C) and the most non-metallic element is at the top-right corner (D) of the periodic table.

📝 Teacher's Note: Draw the periodic table and mark the metallic character trend — increases down the group and decreases across the period. Use arrows to show the direction of increasing metallic character.

🎯 Exam Tip: Remember that metallic character increases down and to the left, while non-metallic character increases up and to the right in the periodic table.

Solution 8:
Answer:

(i) Both A and C need one electron to get a stable configuration. Since A belongs to group 17 while C belongs to group 1, so size of atoms of A is greater than that of C hence Electron affinity of element A is more than that of element C. (ii) Since C belongs to group 1 it has less ionization energy than A as atomic size of group 1 elements is more than that of group 17. (iii) Since B is a noble gas element so its electron affinity is zero as its octet is complete.
In simple words: Element A (halogen) attracts electrons more than C (alkali metal) because A is smaller, C loses electrons easier than A because C is bigger, and B (noble gas) doesn't want any electrons because it's already complete.

📝 Teacher's Note: Help students identify elements by their valence electrons — 7 electrons means halogen (group 17), 8 means noble gas (group 18), 1 means alkali metal (group 1). This pattern recognition is crucial.

🎯 Exam Tip: When comparing electron affinity and ionization energy, always consider atomic size first. Smaller atoms have higher electron affinity, larger atoms have lower ionization energy.

Solution 9:
Answer: 1. 18, 7. 2. First. 3. Seventeen. 4. Electron affinity. 5. Decrease, increase 6. Fluorine 7. Zero.
In simple words: These are fill-in-the-blanks covering basic periodic table facts like number of groups, periods, and trends in properties.

📝 Teacher's Note: These are fundamental facts students must memorize. Create flashcards or quick recall exercises to help students remember these numbers and trends.

🎯 Exam Tip: Practice these basic facts until they become automatic. These fill-in-the-blank questions are easy scoring opportunities in exams.

Solution 10:
Answer: 1. False. 2. True. 3. True. 4. True. 5. True. 6. True.
In simple words: Most statements about periodic trends are true, except that fluorine actually has lower electron affinity than chlorine due to its very small size causing electron repulsion.

📝 Teacher's Note: Focus on statement 1 which is false — this is the fluorine exception that students often get wrong. Emphasize that trends have exceptions and students should be aware of them.

🎯 Exam Tip: The fluorine exception (lower electron affinity than chlorine) is a favorite exam question. Remember this exception to score marks.

Solution 11:
Answer: 1. Li < Be < B 2. I < Br < F < Cl 3. \( SiO_2 \) < \( P_2O_5 \) < \( SO_3 \) < \( Cl_2O_7 \) 4. \( I^+ \) < I < \( I^- \)
In simple words: These arrangements show how ionization energy increases across a period, electron affinity has exceptions with halogens, acidic nature of oxides increases across a period, and ion size changes when electrons are added or removed.

📝 Teacher's Note: Teach students to identify the property being arranged first, then apply the correct trend. For ionic sizes, remind them that cations are smaller and anions are larger than their parent atoms.

🎯 Exam Tip: For electron affinity of halogens, remember the order: Cl > F > Br > I (chlorine has the highest, not fluorine). For ionic sizes, remember: cation < atom < anion.

Solution 12:
Answer: The statement that in each period, the atomic size gradually decreases with increase in atomic number means that as move from left to right in a period, nuclear charge increases by one unit in each succeeding element while the number of shells remains the same. Due to this increased nuclear charge, the electrons of all the shells are pulled closer to the nucleus thereby bringing the outer most shell closer to the nucleus. With the result, the atomic size decreases across a period. For example, in the second period from lithium to fluorine, lithium has the largest size while fluorine has the smallest size.
In simple words: As you go across a period, the nucleus gets more positive charge but the number of electron shells stays the same, so the nucleus pulls all electrons tighter, making atoms smaller.

📝 Teacher's Note: Use the analogy of a magnet getting stronger — as nuclear charge increases, it's like the magnet getting stronger and pulling the electrons closer. Draw circles showing decreasing atomic radii across a period.

🎯 Exam Tip: Always mention "nuclear charge increases" and "number of shells remains same" when explaining atomic size trends across a period. Include a specific example for full marks.

Solution 13:
Answer: In the case of noble gases or inert gases there are exceptions and the atomic radius or size of the elements are greater than the other elements of the period to which these elements belong.
In simple words: Noble gases are like balloons that are more puffed up than other atoms in their row because they have complete electron shells which causes more electron repulsion.

📝 Teacher's Note: Explain that noble gases have van der Waals radii (measured differently) rather than covalent radii. This exception often confuses students, so emphasize why noble gases appear larger.

🎯 Exam Tip: Remember this exception — noble gases have larger atomic radii than expected from the general trend due to electron-electron repulsion in complete shells.

Solution 14:
Answer: As we move down a group, the atomic radii increase because a new shell is added at each succeeding element though the number of electrons in the outer most shell remains the same. Thus, the atomic size of elements increases in size downward. Although nuclear charge also increases in going down the group but the effect of nuclear charge on atomic size is much less than the increase due to addition of a new shell. In group 17, the atomic size follows the trend: F < Cl < Br < I
In simple words: Going down a group is like adding more floors to a building — each element gets a new electron shell, making atoms bigger despite the nucleus getting more positive charge.

📝 Teacher's Note: Use the building analogy — each period adds a new "floor" (electron shell). Even though the "foundation" (nucleus) gets stronger, the building gets taller overall.

🎯 Exam Tip: Mention both factors: "new shell added" (main reason for increase) and "nuclear charge increases but has less effect" to show complete understanding.

Solution 15:
Answer: The elements of third period are: Na, Mg, Al, Si, P, S, Cl, Ar. The most metallic element is sodium i.e., Na and the most non-metallic element is chlorine i.e., Cl.
In simple words: In the third row of the periodic table, sodium (far left) is the most metallic and chlorine (far right, excluding noble gas) is the most non-metallic.

📝 Teacher's Note: Point out that we exclude argon when finding the most non-metallic element because noble gases are inert. Students should understand that we're looking for reactive non-metals.

🎯 Exam Tip: When asked for most non-metallic element in a period, choose the halogen (group 17), not the noble gas (group 18), because noble gases are unreactive.

Solution 16:
Answer: Ionization potential or ionization energy is defined as the energy required to remove an electron from an isolated gaseous atom so as to convert it into a gaseous cation. The process may be represented as: M (g) + Ionization energy → \( M^+ \) (g) + \( e^- \). It is known as ionization potential since it is the minimum potential difference (in a discharge tube) required to remove an electron from an isolated gaseous atom to form a gaseous cation. Units: Its units are electron volts (eV) per atom. Its SI units are Kilojoules per mole (\( KJmol^{-1} \)).
In simple words: Ionization energy is like the amount of energy needed to pull an electron completely away from an atom, just like the force needed to pull a ball away from a magnet.

📝 Teacher's Note: Emphasize that this is energy required (input), not released. Use the magnet-ball analogy to help students understand that stronger attraction needs more energy to overcome.

🎯 Exam Tip: Always write the equation M(g) → M⁺(g) + e⁻ and mention both units: eV per atom and KJ/mol for complete marks.

Solution 17:
Answer: The particular with lowest ionization potential in a group will ionize most easily. For example among elements of group 1 value of ionization potential is:

Cesium (Cs) has lowest ionization potential value which is 375 \( KJmol^{-1} \). So it will ionize most easily.
In simple words: The element that needs the least energy to lose an electron will ionize most easily — like how the loosest bolt is easiest to remove.

📝 Teacher's Note: Show students the clear decreasing trend in the data. Relate this to atomic size increase down the group — larger atoms hold outer electrons less tightly.

🎯 Exam Tip: Remember the relationship: lowest ionization energy = easiest to ionize = most reactive metal in a group.

Solution 18:
Answer: Electron affinity is the energy released when an electron is added to an isolated gaseous atom to form the negative ion (anion). Unit: Its units are electron volt (eV). Its SI units are Kilojoules per mole(\( KJmol^{-1} \)).
In simple words: Electron affinity is like how much energy is given out when an atom catches an extra electron, similar to how heat is released when you catch a ball.

📝 Teacher's Note: Contrast this with ionization energy — this is energy released (output), not required (input). Use consistent analogies to help students distinguish between these two concepts.

🎯 Exam Tip: Remember that electron affinity involves energy being released (negative value), while ionization energy involves energy being absorbed (positive value).

Solution 19:
Answer: Out of A and B, A will ionize more easily to form a negative anion because of the high value of electron affinity, energy released during addition of electron will be high hence the resulting anion formed will be more stable than the corresponding atom.
In simple words: Element A will gain electrons more easily because it releases more energy when catching an electron, making the process more favorable, like how a hungry person grabs food more eagerly.

📝 Teacher's Note: Help students understand that higher electron affinity means the element "wants" electrons more strongly, making anion formation more favorable. Connect this to stability concepts.

🎯 Exam Tip: Higher electron affinity = more stable anion formation = easier to gain electrons. Link energy release to stability for full marks.

Solution 20:
Answer: 1. Larger the atomic size, farther is the valence electron from the nucleus and lesser is the pull exerted on it. As a result, electron can be easily removed from the valence shell and hence more metallic is the element. 2. Halogens need only one electron to complete their octet and become stable their atomic size is very less hence the distance between their last shell and nucleus is very less, as a result the force of attraction between the nucleus and the incoming electron is less and hence the electron affinity is high for halogens. 3. When an atom loses or gain electron to form ion, the number of electrons present in the outermost shell also changes. Corresponding to that effective nuclear charge on the changed number of electrons also change which further changes the size of an atom as there is inverse relation between effective nuclear charge and size of atom. 4. K and Li belongs to group 1 i.e., metals and we know that for metals chemical reactivity of elements increases down the group because chemical reactivity increases as electropositive or metallic character increases.
In simple words: Bigger atoms lose electrons easier (more metallic), halogens grab electrons easily because they need just one more, making ions changes atom sizes due to different electron-nucleus balance, and metals get more reactive going down because they get bigger and lose electrons easier.

📝 Teacher's Note: Break this into separate concepts for clarity. Use visual aids to show how atomic size affects electron removal, how halogens are "one short" of stability, and how ion formation affects size.

🎯 Exam Tip: Each point addresses a different concept. For full marks, explain the underlying reason (atomic size, nuclear charge, electron configuration) behind each trend.

Question. The electronegativity of chlorine is higher than sulphur because both of them belong to third group and chlorine follows sulphur. We know that, within a period electronegativity increases as we move from left to right because of decrease in atomic size and increase in nuclear charge.
Answer: This statement is correct. Chlorine has higher electronegativity than sulfur because within a period (third period), electronegativity increases from left to right due to decreasing atomic size and increasing nuclear charge. As we move across the period, atoms become smaller and the nucleus pulls electrons more strongly, making the element more electronegative.

📝 Teacher's Note: Use visual aids showing atomic size trends across periods to help students understand why electronegativity increases. Common mistake: students confuse group trends with period trends.

🎯 Exam Tip: Remember the key phrase "across a period, electronegativity increases" and always mention both factors: decreasing atomic size AND increasing nuclear charge for full marks.

Question. Group 17 elements are non metals because they have 7 electrons in their valence shell and ionize by accepting 1 electron to form an anion. For example group 17 elements F, Cl, Br and I all have 7 electrons in their valence shell and ionize by accepting 1 electron to form F-, Cl-, Br- and I-. Group 1 elements are metals because they have tendency to lose the one electron present in their valence shell and form positive ion. For example, group 1 elements Li, Na, K, Rb, Cs have tendency to lose the one electron present in their valence shell and form positive ions Li+, Na+, K+, Rb+ and Cs+.
Answer: This explanation is correct. Group 17 elements (halogens) are non-metals because they have 7 valence electrons and achieve stability by gaining 1 electron to complete their octet, forming negative ions (anions). Group 1 elements (alkali metals) are metals because they have 1 valence electron and achieve stability by losing this electron, forming positive ions (cations). The tendency to gain or lose electrons determines whether an element is metallic or non-metallic in nature.

📝 Teacher's Note: Emphasize the "rule of 8" - atoms want 8 electrons in their outer shell. Group 17 finds it easier to gain 1, while Group 1 finds it easier to lose 1.

🎯 Exam Tip: Always mention "to achieve stable electronic configuration" when explaining why elements gain or lose electrons - this shows understanding of the underlying principle.

Solution 21:

1. (C) i.e. 2, 8, 2 because it has only 2 electrons in its valence shell which can be lost to form a di positive cation.
2. (C) i.e. 0.72, 0.72
3. (d) i.e. element forms basic oxide because the element is a metal as it has valency 1.
4. (a) i.e. F because it belongs to group 17 whose elements have valency 7 and thus requires only 1 electron to complete their octet.

📝 Teacher's Note: For question 1, help students recognize that 2,8,2 configuration means the element is in Group 2 (alkaline earth metals). For question 4, clarify that "valency 7" means 7 valence electrons, not valency of 7.

🎯 Exam Tip: Electronic configuration questions often test group identification - count valence electrons to determine the group number.

Solution 2000-1:

1. Number of elements in period 1 = 2
In period 2 = 8
In period 3 = 8.
2. Elements in period 1 are Hydrogen (H) and Helium (He).
3. Atomic size of elements decreases on moving from left to right in a period.

📝 Teacher's Note: Students often confuse the number of elements in each period. Use the periodic table to count together - Period 1 has only H and He, while Periods 2 and 3 each have 8 elements.

🎯 Exam Tip: Memorize the pattern: Period 1 has 2 elements, Periods 2 and 3 each have 8 elements. This is a common fill-in-the-blank question.

Solution 2000-2:

1. The elements at the end of period 2 and period 3 both have their outermost shell complete and belong to noble gases.
2. An element in group 7 is likely to be non metallic in character since group 7 element will have 7 electrons in its valence shell.
3. Metallic.

📝 Teacher's Note: Emphasize that noble gases have complete outer shells (8 electrons for Period 2 and 3 noble gases), which makes them very stable and unreactive.

🎯 Exam Tip: Remember that Group 17 (not Group 7 in modern notation) elements are non-metals because they need only 1 electron to complete their octet.

Solution 2001-1:

1. Atomic number.
2. Period, non-metallic.
3. More.
4. Number of outer electrons.

📝 Teacher's Note: These appear to be fill-in-the-blank answers. Help students understand that atomic number determines an element's position and properties in the periodic table.

🎯 Exam Tip: For periodic table questions, atomic number is often the key factor that determines all other properties of elements.

Solution 2002-1:

A group is a vertical column of elements having the same number of valence electrons and same valency in the periodic table. There are 18 groups in the periodic table.

📝 Teacher's Note: Make sure students understand the difference between groups (vertical columns) and periods (horizontal rows). All elements in a group have similar chemical properties.

🎯 Exam Tip: When defining a group, always mention: vertical column, same valence electrons, same valency, and similar properties for complete marks.

Solution 2002-2:

Within a group the element with the greatest metallic character and largest size is expected to be present at the bottom of the group.

📝 Teacher's Note: Use examples like alkali metals (Li to Cs) to show how metallic character increases down the group. Larger atoms lose electrons more easily.

🎯 Exam Tip: Remember the trend: "Down a group, size increases, metallic character increases" - this applies to all groups containing metals.

Solution 2002-3:

Ionization potential decreases down the group because atomic size increases down the group which decreases the effective nuclear charge over the valence electron which further can now be removed easily.

📝 Teacher's Note: Help students visualize this - imagine trying to hold a ball close to you versus far away. The farther the electron, the weaker the attraction to the nucleus.

🎯 Exam Tip: Always connect ionization energy trends to atomic size - larger atoms have lower ionization energy because valence electrons are farther from the nucleus.

Solution 2002-4:

There are 8 elements in period 2.

📝 Teacher's Note: Have students count the elements in Period 2: Li, Be, B, C, N, O, F, Ne. This visual counting helps them remember.

🎯 Exam Tip: Period 2 and Period 3 both have exactly 8 elements each - a fact worth memorizing for quick recall in exams.

Solution 2003-1:

1. Al2(SO4)3.
2. Covalent.
3. The elements of group VIIA all have same number of electrons in their valence shell and same valency.
4. Neon
5. 8 electrons are present in the valence shell of the element with atomic number 18.
6. Electron affinity.
7. Electronic configuration of element in the third period which gains one electron to become an anion is 2, 8, 7.
8. Decreases, number of valence shell electrons/outermost shell electrons, valence shell/ outermost shell.

📝 Teacher's Note: For question 5, help students recognize that atomic number 18 is Argon, a noble gas with 8 valence electrons. For question 7, the configuration 2,8,7 represents Chlorine.

🎯 Exam Tip: When asked about elements that "gain electrons to form anions," think of halogens (Group 17) which have 7 valence electrons.

Solution 2004-1:

1. Na Mg Al Si P S Cl.
2. (a) lower, higher.
3. remains the same.

📝 Teacher's Note: Question 1 shows the elements of Period 3 in order. Use this to teach periodic trends across a period.

🎯 Exam Tip: Learn the order of elements in Periods 2 and 3 by heart - this knowledge helps in many periodic table questions.

Solution 2005-1:

1. b
2. d
3. c
4. a
5. c

📝 Teacher's Note: These are multiple choice answers without the questions. When reviewing, always connect answers back to the underlying periodic table concepts.

🎯 Exam Tip: In MCQ sections on periodic table, elimination method works well - remove obviously wrong options first.

Solution 2006-1:

1. Second period.
2. Nitrogen. It should be placed between oxygen and carbon.
3. Beryllium < nitrogen < fluorine
4. Fluorine (F)

📝 Teacher's Note: For question 3, this appears to be an increasing order of electronegativity. Help students understand that F is the most electronegative element in the periodic table.

🎯 Exam Tip: Fluorine is always the answer when asked for "most electronegative element" - this is a fundamental fact worth remembering.

Solution 2007-1:

1. Thallium
2. Boron
3. 3
4. BCl3
5. The elements in the group to the right of this boron group will be less metallic in character because on moving to the right of the periodic table metallic character decreases as ionization energy deceases and tendency to lose electron also decreases.

📝 Teacher's Note: For question 5, there's a typo - it should say "ionization energy increases" not "deceases." Across periods, metallic character decreases from left to right.

🎯 Exam Tip: Remember the trend: across a period from left to right, metallic character decreases and non-metallic character increases.

Solution 2008-1:

1. False
2. True
3. False
4. True

📝 Teacher's Note: For True/False questions on periodic table, students should always justify their answers with proper reasoning based on periodic trends.

🎯 Exam Tip: In True/False questions, even if you're confident, quickly check your answer against known periodic table facts.

Solution 2008-2:

1. (i) First element in period 2 is Lithium and last element is Neon.
(ii) Atomic size increases on moving from top to bottom of a group.
(iii) Chlorine among halogens has the greatest electron affinity.
(iv) All elements in group 7 have same number of valence shell electrons.
2. (i) metallic
(ii) smallest
3. (i) Ba i.e. Barium will form ion most readily since it is at the bottom of a group its ionization energy is low because its atomic size is more. Due to this effective atomic charge of nucleus over the valence shell electron is least and it can be removed easily.
(ii) Electro negativity of an element measures the capacity of an element to attract the shared pair of electrons in a bond towards itself.

📝 Teacher's Note: For 1(iii), help students understand that electron affinity is highest for elements that most readily accept electrons. Among halogens, chlorine has the highest electron affinity.

🎯 Exam Tip: When explaining why an element forms ions readily, always mention atomic size, ionization energy, and effective nuclear charge for complete marks.

Solution 2009-1:

(d) Fluorine

📝 Teacher's Note: Without seeing the question, this appears to be asking for the most electronegative element, which is always fluorine.

🎯 Exam Tip: Fluorine is the answer to many "most/highest" questions in periodic table - most electronegative, smallest atomic radius in Period 2, etc.

Solution 2009-2:

(i) K is the most electro negative.
(ii) 5 valence electrons are present in G.
(iii) B2H.
(iv) Ionic.
(v)

📝 Teacher's Note: Statement (i) seems incorrect - K (potassium) is not electronegative; it's electropositive. This might be a typo in the source material.

🎯 Exam Tip: Always double-check statements about electronegativity - metals like K are electropositive (give up electrons easily), not electronegative.