NCERT Solutions for Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties

Step 1: Main purpose of the periodic table
🧩 Elements with similar properties are arranged in the same group of the periodic table.

Step 2: Why do elements show similar properties?
👉 The physical and chemical properties of elements mainly depend on the electronic configuration of the valence shell.

Step 3: What is common among elements of the same group?
✅ Same number of valence electrons
✅ Similar valence shell electronic configuration

Step 4: Conclusion
🔥 Because of this similarity in valence shell configuration, elements in the same group show similar chemical behaviour, such as forming similar oxides, having similar valency, and undergoing similar chemical reactions.

👉 (i) Basis of classification
📌 Mendeleev arranged elements in order of increasing atomic mass.

👉 (ii) Main objective
🎯 Elements with similar physical and chemical properties should come in the same group of the periodic table.

👉 (iii) Blank spaces
Mendeleev left blank spaces for undiscovered elements.

👉 (iv) Prediction of elements
He predicted the atomic masses and properties of those elements before they were discovered.

👉 (v) Later discoveries
When those elements were discovered later, their properties were found to be very similar to Mendeleev’s predictions.

👉 (i) Mendeleev’s Periodic Law
According to Mendeleev, the properties of elements are a periodic function of their atomic mass.
This means that when atomic mass increases, the properties of elements repeat periodically.

👉 (ii) Modern Periodic Law
According to the modern periodic law, the properties of elements are a periodic function of their atomic number (Z).
When atomic number increases, the properties repeat in a regular pattern.

👉 (iii) Why Modern Law is more accurate
The atomic number determines the electronic configuration of an element, and electronic configuration controls the chemical properties.

Step 1: A new period starts with a new shell
📌 In the long form of the periodic table, each period begins with a new principal quantum number (n).
✅ The sixth period begins with n = 6.

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Step 2: Possible subshells for n = 6
For n = 6, the azimuthal quantum number (ℓ) can have values:
ℓ = 0, 1, 2, 3

These correspond to the following subshells:
👉 ℓ = 0 → 6s
👉 ℓ = 1 → 6p
👉 ℓ = 2 → 6d
👉 ℓ = 3 → 6f

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Step 3: Subshells actually filled in the sixth period
⚡ According to the Aufbau principle, electrons fill orbitals in order of increasing energy.

👉 The 6d and 6f subshells have higher energy than 7s, so they do not fill immediately.
👉 In the sixth period, 4f and 5d have lower energy than 6p, so they fill first.

✅ Therefore the filling order is:
6s → 4f → 5d → 6p

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Step 4: Maximum electron capacity
👉 s subshell = 2 electrons
👉 p subshell = 6 electrons
👉 d subshell = 10 electrons
👉 f subshell = 14 electrons

Therefore for the sixth period:
👉 6s = 2
👉 4f = 14
👉 5d = 10
👉 6p = 6

Total = 2 + 14 + 10 + 6 = 32
👉 Hence the sixth period contains 32 elements.

Step 1: Electronic configuration of Z = 114
[Rn] 5f¹⁴ 6d¹⁰ 7s² 7p²

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Step 2: Determining the period
👉 The highest principal quantum number (n) present in the electronic configuration determines the period.
Here the highest value is n = 7.
✅ Therefore the element belongs to the 7th period.

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Step 3: Determining the group
Valence shell configuration = 7s² 7p²

👉 Total valence electrons = 4.

For p-block elements: Group number = 10 + valence electrons
10 + 4 = 14
✅ Hence the element belongs to Group 14.

Step 1: Elements in the third period
The third period contains the following elements:

Na (11), Mg (12), Al (13), Si (14), P (15), S (16), Cl (17), Ar (18)

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Step 2: Identify Group 17
👉 Group 17 elements are known as Halogens.
Examples include:
F, Cl, Br, I, At

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Step 3: Element common to both
👉 In the third period, the element belonging to Group 17 is Chlorine (Cl).

✅ Therefore its atomic number is 17.

Step 1: Basis of naming elements
🧪 Many synthetic (man-made) elements are named after scientists, places, or laboratories where they were discovered.

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Step 2: Elements related to Lawrence Berkeley Laboratory
✅ Lawrencium — named in honour of Ernest O. Lawrence.
✅ Berkelium — named after the city Berkeley in California.

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Step 3: Element related to Seaborg’s group
✅ Seaborgium was named in honour of Glenn T. Seaborg, a famous nuclear chemist who contributed greatly to the discovery of many transuranium elements.

Step 1: What do the properties of an element depend on?
🔍 The properties of an element mainly depend on its outermost (valence) electrons.

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Step 2: What is common in the same group?
✅ All elements of the same group have similar valence shell electronic configuration.
👉 This means the number of valence electrons is the same.

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Step 3: Effect of same valence electrons
👉 They usually have the same valency.
👉 They show similar tendency to form ions (for example +1, +2, −1).
👉 They form similar types of compounds.

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Step 4: Conclusion
✅ Therefore elements of the same group show similar chemical properties and many physical properties.

Step 1: What does Atomic Radius indicate?
👉 It represents the size of a neutral atom.
✅ In simple terms: distance from the nucleus to the outermost electron is called Atomic Radius.

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Step 2: What is Ionic Radius?
👉 When an atom loses or gains electrons, it forms an ion and its size changes.
✅ The effective distance from the nucleus to the boundary of the electron cloud in an ion is called Ionic Radius.

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Step 3: Important rule to remember
👉 When a cation (positive ion) is formed, the size decreases because electrons are lost.
👉 When an anion (negative ion) is formed, the size increases because electrons are gained.

(A) Why does atomic radius decrease across a period? (Left → Right)

Step 1: As we move across a period, the atomic number (Z) increases, which means the nuclear charge also increases.

Step 2: Electrons are added to the same outer shell.

Step 3: Shielding effect does not increase much because electrons are added in the same shell.

Step 4: Therefore the effective nuclear charge (Z_eff) increases and the outer electrons are pulled closer to the nucleus.
✅ Result: Atomic radius decreases.
Example: In the 3rd period, Na has a larger radius while Cl has a smaller radius.

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(B) Why does atomic radius increase down a group? (Top → Bottom)

Step 1: As we move down a group, a new electron shell is added for each element.

Step 2: The outer electrons move farther from the nucleus.

Step 3: Inner electrons increase the shielding effect, which reduces the nuclear attraction on outer electrons.
✅ Result: Atomic radius increases.
Example: In Group 1 → Li < Na < K < Rb < Cs (radius increases).

Idea: First find the number of electrons, then choose a species having the same number of electrons. ✅

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(i) F⁻
Atomic number of F = 9
Gains 1 electron → 9 + 1 = 10 e⁻
✅ Therefore Ne (10 electrons) is isoelectronic.

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(ii) Ar
Ar has 18 electrons.
✅ Therefore K⁺ (19 − 1 = 18 electrons) is isoelectronic.

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(iii) Mg²⁺
Atomic number of Mg = 12
Loses 2 electrons → 12 − 2 = 10 e⁻
✅ Therefore O²⁻ (8 + 2 = 10 electrons) is isoelectronic.

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(iv) Rb⁺
Atomic number of Rb = 37
Loses 1 electron → 37 − 1 = 36 e⁻
✅ Therefore Kr (36 electrons) is isoelectronic.

(a) Why are they isoelectronic?
Let us calculate the number of electrons in each species:

👉 N³⁻ : 7 + 3 = 10
👉 O²⁻ : 8 + 2 = 10
👉 F⁻ : 9 + 1 = 10
👉 Na⁺ : 11 − 1 = 10
👉 Mg²⁺ : 12 − 2 = 10
👉 Al³⁺ : 13 − 3 = 10

✅ All species have 10 electrons, therefore they are isoelectronic.

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(b) How to arrange ionic radii?
For isoelectronic species, the rule is:
👉 As the atomic number (Z) increases, the nuclear attraction increases and the ionic radius decreases.

Atomic numbers:
N = 7, O = 8, F = 9, Na = 11, Mg = 12, Al = 13

✅ Largest Z → smallest radius → Al³⁺
✅ Smallest Z → largest radius → N³⁻

Therefore the increasing order of ionic radii is:
Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

(A) Why are cations smaller?

Step 1: A cation is formed when an atom loses one or more electrons.

Step 2: Removal of electrons decreases electron–electron repulsion.

Step 3: The same nuclear charge now acts on fewer electrons, so the effective nuclear charge Z_eff increases.

Step 4: The remaining electrons are pulled closer to the nucleus.
✅ Conclusion: Radius decreases, so the cation becomes smaller.
Example: Na → Na⁺ (Na⁺ is smaller than Na)

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(B) Why are anions larger?

Step 1: An anion is formed when an atom gains one or more electrons.

Step 2: Addition of electrons increases electron–electron repulsion.

Step 3: The nuclear attraction remains the same but has to act on more electrons, so the effective attraction per electron decreases.

Step 4: The electron cloud expands, increasing the atomic radius.
✅ Conclusion: Anions are larger than their parent atoms.
Example: Cl → Cl⁻ (Cl⁻ is larger than Cl)

1) Why “isolated gaseous atom”?

Step 1: The attraction between the nucleus and electrons in an atom can be affected by other atoms or chemical bonds nearby.

Step 2: The value of ionization enthalpy (IE) and electron gain enthalpy (EA) depends on this nuclear attraction.

Step 3: Therefore, the atom is considered isolated in the definition.

Step 4: In the gaseous state, atoms are far apart from each other and can be treated as nearly isolated.
✅ Hence the term isolated gaseous atom is used.

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2) Why “ground state”?

Step 1: The ground state is the lowest energy and most stable state of an atom.

Step 2: If the atom is in an excited state, its energy is already higher.

Step 3: In that case, the energy required to remove an electron (IE) or the energy change during electron gain (EA) would be different.
✅ Therefore, to maintain a standard and comparable definition, the atom is always considered in the ground state.

Step 1: Energy required for ionization (per atom)
Energy of electron in ground state:
E₁ = −2.18 × 10⁻¹⁸ J

After ionization the electron moves to infinity, where energy is:
E_∞ = 0

Ionization energy per atom:
ΔE = E_∞ − E₁
ΔE = 0 − (−2.18 × 10⁻¹⁸)
✅ ΔE = 2.18 × 10⁻¹⁸ J atom⁻¹

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Step 2: Convert to J mol⁻¹ (Mole concept)
Number of particles in 1 mol = Avogadro number N_A = 6.022 × 10²³ mol⁻¹

Ionization enthalpy:
ΔH = (2.18 × 10⁻¹⁸) × (6.022 × 10²³)

= (2.18 × 6.022) × 10⁵
= 13.12796 × 10⁵
✅ ΔH = 1.312796 × 10⁶ J mol⁻¹
≈ 1.313 × 10⁶ J mol⁻¹

(i) Why does Be have higher Δ_iH than B?

Step 1: Electronic configuration
👉 Be (Z = 4): 1s² 2s²
👉 B (Z = 5): 1s² 2s² 2p¹

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Step 2: From which subshell is the electron removed?
👉 In Be, the electron is removed from the 2s orbital.
👉 In B, the electron is removed from the 2p orbital.

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Step 3: Main reason (2s vs 2p)
✅ The 2s electron has greater penetration than a 2p electron.
✅ Therefore it is closer to the nucleus and less shielded.
✅ Also Be has a fully filled stable configuration (2s²).

✅ Hence removing an electron from Be is more difficult → Δ_iH(Be) > Δ_iH(B)

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(ii) Why does O have lower Δ_iH than N and F?

Step 1: Electronic configuration
👉 N (Z = 7): 1s² 2s² 2p³
(half-filled p orbitals, extra stability)

👉 O (Z = 8): 1s² 2s² 2p⁴
(one p orbital contains a pair of electrons)

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Step 2: Effect of pairing in oxygen
👉 In O, two electrons occupy the same 2p orbital.
✅ This increases electron–electron repulsion.
✅ Therefore removing one of the paired electrons becomes easier.

✅ Hence Δ_iH(O) < Δ_iH(N)

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Comparison with F
👉 In F the nuclear charge and effective nuclear charge (Z_eff) are higher.
✅ Therefore removing an electron from F is harder than from O.
✅ Hence Δ_iH(O) < Δ_iH(F)

Step 1: Electronic configurations
👉 Na (Z = 11): 1s² 2s² 2p⁶ 3s¹
👉 Mg (Z = 12): 1s² 2s² 2p⁶ 3s²

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(A) Why is the first ionization enthalpy of Na lower?
1️⃣ Sodium has only one electron in the 3s orbital, which is relatively easy to remove.

2️⃣ Magnesium has higher nuclear charge (+12) compared to sodium (+11), so it holds its electrons more strongly.

✅ Therefore:
Δ_iH₁ (Na) < Δ_iH₁ (Mg)

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(B) Why is the second ionization enthalpy of Na higher?
After removal of the first electron: 👉 Na → Na⁺ + e⁻
Na⁺ configuration:
1s² 2s² 2p⁶ (noble gas configuration like Ne)

👉 Mg → Mg⁺ + e⁻
Mg⁺ configuration:
1s² 2s² 2p⁶ 3s¹

Now the second electron removal: • Removing an electron from Na⁺ means breaking a stable noble gas configuration → requires very high energy.

• Removing an electron from Mg⁺ still involves removing the outer 3s electron → comparatively easier.

✅ Therefore:
Δ_iH₂ (Na) > Δ_iH₂ (Mg)

1. Increase in nuclear charge (Z)
👉 As we move down a group, the number of protons increases, so Z increases.
👉 However, this factor alone does not dominate because other effects become more significant.

2. Increase in atomic size (addition of new shells)
👉 With each successive period, a new electron shell is added.
👉 The outermost electron moves farther from the nucleus.
👉 Greater distance reduces the attractive force between the nucleus and the electron.

3. Increase in shielding effect
👉 Down the group, the number of inner electrons increases.
👉 These inner electrons shield the outer electron from the nuclear attraction.
👉 Thus the nucleus holds the outermost electron less strongly.

4. Final understanding
✅ Although nuclear charge increases, the combined effect of larger atomic size and greater shielding reduces the attraction between the nucleus and the outermost electron.
👉 Therefore the ionization enthalpy decreases down a group.

1) Expected general trend
Down the group:
👉 Atomic size increases
👉 Shielding effect increases
👉 The outer electron is farther from the nucleus.
✅ Therefore the ionization enthalpy should decrease.
This trend is roughly observed from B → Al → In.

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2) Why is IE of Ga slightly higher than Al?
👉 Al: [Ne] 3s² 3p¹
👉 Ga: [Ar] 3d¹⁰ 4s² 4p¹

The presence of 3d¹⁰ electrons in Ga causes poor shielding.
➜ This increases the effective nuclear charge on the outer electron.
✅ Hence Ga holds its outer electron more strongly and its IE becomes slightly higher than Al.

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3) Why is IE of Tl higher than In?
👉 In: [Kr] 4d¹⁰ 5s² 5p¹
👉 Tl: [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹

The presence of 4f¹⁴ and 5d¹⁰ electrons causes very poor shielding.
➜ This increases the effective nuclear charge on the outer electron.
✅ Therefore Tl has a slightly higher ionization enthalpy than In.

(i) Why does F have more negative EA than O?
1. From O to F, the atomic size decreases.
2. The nuclear charge (Z) also increases.
3. Therefore the attraction between the nucleus and the incoming electron becomes stronger.
✅ Result: Fluorine (F) accepts an electron more easily → electron gain enthalpy is more negative.

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(ii) Why does Cl have more negative EA than F?
This is a famous exception.

1. Fluorine has a very small size and its outer 2p orbital is very compact.
2. When an extra electron enters F, strong electron–electron repulsion occurs in this small orbital.
3. Chlorine is larger and its outer 3p orbital has more space, so repulsion is less.

✅ Therefore adding an electron to Cl is easier and more energy is released.
➜ Hence the electron gain enthalpy of Cl is more negative.

Step 1: First electron gain enthalpy
When a neutral oxygen atom gains one electron:
O(g) + e⁻ → O⁻(g)

Energy is released, so the value is negative.
Example:
ΔegH₁ ≈ −141 kJ mol⁻¹

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Step 2: Second electron gain enthalpy
Now oxygen is already a negatively charged ion: O⁻ Adding another electron: O⁻(g) + e⁻ → O²⁻(g)

The incoming electron experiences strong electrostatic repulsion from the already negatively charged ion.

Therefore energy must be absorbed to add the second electron.

✅ Hence the value becomes positive . Example value: ΔegH₂ ≈ +780 kJ mol⁻¹

1) What does Electron Gain Enthalpy represent?
👉 It applies to an isolated gaseous atom .
👉 The process can be written as:
X(g) + e⁻ → X⁻(g)
👉 Energy may be released (negative value) or absorbed (positive value).
✅ Thus it is a measurable energy change expressed in kJ mol⁻¹.

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2) What does Electronegativity represent?
👉 It is a property observed within a chemical bond .
👉 When two atoms share electrons, electronegativity indicates which atom pulls the electrons more strongly.
✅ It is a relative scale (for example the Pauling scale ) rather than a direct enthalpy value.

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3) Simple way to remember
👉 Electron Gain Enthalpy = “accepting an electron” (isolated atom in gaseous state).
👉 Electronegativity = “pulling shared electrons” inside a chemical bond.

1️⃣ On the Pauling scale , the electronegativity of nitrogen is approximately 3.0 . However, this value represents a general or average value .

2️⃣ In different compounds, the environment around nitrogen changes: 👉 The oxidation state of nitrogen may change.
👉 The hybridisation of nitrogen may change.
👉 Therefore the ability of nitrogen to attract electrons also changes slightly.

3️⃣ Hence it is not correct to say that nitrogen always has exactly 3.0 electronegativity in every compound.

Example:
In NO and NO₂, nitrogen has different oxidation states , so its electron-attracting behavior changes slightly.

(a) When an atom gains an electron (Why does radius increase in an anion?)
1. The atom gains one or more electrons and forms an anion.
2. The nuclear charge remains the same, but the number of electrons in the valence shell increases.
3. Increased electrons lead to stronger electron–electron repulsion and shielding.
4. As a result, the effective nuclear attraction (Z_eff) experienced by each electron becomes less effective.
✅ Result: The electron cloud expands → the atomic radius increases.
Example: Cl → Cl⁻ (Cl⁻ is larger)

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(b) When an atom loses an electron (Why does radius decrease in a cation?)
1. The atom loses one or more electrons and forms a cation.
2. The electron–electron repulsion decreases because there are fewer electrons.
3. The nuclear charge remains the same but now attracts fewer electrons, so the effective nuclear charge (Z_eff) increases.
4. Sometimes the entire outer shell may be removed, causing an even greater reduction in size.
✅ Result: Electrons move closer to the nucleus → the atomic radius decreases.
Example: Na → Na⁺ (Na⁺ is smaller)

1️⃣ In isotopes, the atomic number (Z) is the same.
👉 This means the number of protons in the nucleus is identical.

2️⃣ Therefore isotopes have the same:
👉 number of electrons (in neutral atoms)
👉 nuclear charge
👉 electronic configuration

3️⃣ Ionization enthalpy mainly depends on the attraction between the nucleus and the outer electron and the electronic configuration .

✅ Since these factors are the same for isotopes, their first ionization enthalpy is also nearly the same .

1) Tendency to lose or gain electrons
• Metals are electropositive in nature.
👉 They lose one or more valence electrons:
M → M⁺ + e⁻

• Non-metals are electronegative in nature.
👉 They gain electrons:
X + e⁻ → X⁻

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2) Redox nature
• Metals lose electrons and therefore usually act as reducing agents .
• Non-metals gain electrons and therefore usually act as oxidising agents .

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3) Energy and property trends
• Metals usually have:
✅ low ionization enthalpy
✅ low electron gain enthalpy (less negative or sometimes positive)
✅ low electronegativity

• Non-metals usually have:
✅ high ionization enthalpy
✅ more negative electron gain enthalpy
✅ high electronegativity

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4) Nature of oxides
• Metals generally form basic oxides .
Example: Na₂O, CaO

• Non-metals generally form acidic oxides .
Example: CO₂, SO₃

(a) Element with five electrons in outer subshell
• This corresponds to the configuration np⁵ which belongs to the halogen group .
Example:
Fluorine configuration: 1s² 2s² 2p⁵
✅ Therefore the element is Fluorine (F)

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(b) Element that tends to lose two electrons
Elements of Group 2 easily lose two electrons to form M²⁺ ions .
Magnesium configuration:
1s² 2s² 2p⁶ 3s²
Reaction:
Mg → Mg²⁺ + 2e⁻
✅ Therefore the element is Magnesium (Mg)

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(c) Element that tends to gain two electrons
Elements of Group 16 (chalcogens) tend to gain two electrons to complete the octet .
Oxygen configuration:
1s² 2s² 2p⁴
Reaction:
O + 2e⁻ → O²⁻
✅ Therefore the element is Oxygen (O)

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(d) Group containing solid, liquid and gas at room temperature
The halogen group (Group 17) shows different physical states:
• Gas: F₂, Cl₂
• Liquid: Br₂
• Solid: I₂
✅ Therefore the required group is Group 17 (Halogens).

(A) Why does reactivity increase down Group 1?
1. Group 1 elements have one valence electron and easily lose it to form M⁺.
2. Their reactivity depends on how easily this electron can be removed.
3. Down the group:
👉 atomic size increases
👉 shielding effect increases
👉 the outer electron becomes farther from the nucleus and loosely held.
✅ Therefore ionization enthalpy decreases → electron removal becomes easier → reactivity increases.

Hence: Li < Na < K < Rb < Cs

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(B) Why does reactivity decrease down Group 17?
1. Group 17 elements have seven valence electrons and tend to gain one electron to form X⁻.
2. Their reactivity depends on their ability to attract an extra electron (oxidising power).
3. Down the group:
👉 atomic size increases
👉 the incoming electron is added farther from the nucleus.
👉 the nuclear attraction decreases .
✅ Therefore the tendency to gain electrons decreases → reactivity decreases.

Hence: F > Cl > Br > I

1️⃣ The term block in the periodic table refers to the type of subshell being filled by the last electron.

2️⃣ Therefore:
• In the s-block, the s-subshell is filled → ns^1–2
• In the p-block, the p-subshell is filled → ns² np^1–6
• In the d-block, the (n−1)d subshell is filled → (n−1)d^1–10 ns^0–2
• In the f-block, the (n−2)f subshell is filled → (n−2)f^1–14 (n−1)d^0–1 ns²

(i) ns² np⁴, n = 3
1. n = 3 → 3rd period
2. Configuration: 3s² 3p⁴ → filling of p-subshell → p-block
3. For p-block elements:
Group = 10 + valence electrons
Valence electrons = 2 + 4 = 6
Group = 10 + 6 = 16
✅ Position: Period 3, Group 16, p-block
Element: Sulfur (S)

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(ii) (n−1)d² ns², n = 4
1. n = 4 → 4th period
2. (n−1)d = 3d → configuration: 3d² 4s² → filling of d-orbital → d-block
3. For d-block elements:
Group = (d electrons + s electrons)
= 2 + 2 = 4
✅ Position: Period 4, Group 4, d-block
Element: Titanium (Ti)

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(iii) (n−2)f⁷ (n−1)d¹ ns², n = 6
1. n = 6 → 6th period
2. (n−2)f = 4f → 4f⁷ → filling of f-orbital → f-block (Lanthanides)
3. Remaining configuration:
4f⁷ 5d¹ 6s²

✅ Position: Period 6, f-block (Lanthanides), under Group 3
Element: Gadolinium (Gd)

(a) Least reactive element
Element V has a very high ionization enthalpy (2372) and a positive electron gain enthalpy. This means it neither loses nor gains electrons easily. Hence it behaves like a noble gas and is the least reactive element.

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(b) Most reactive metal
Metals are more reactive when their first ionization enthalpy is lowest. Element II has the lowest value 419 kJ mol⁻¹. Hence it easily loses an electron and behaves like an alkali metal.

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(c) Most reactive non-metal
Non-metals are more reactive when their electron gain enthalpy is highly negative. Element III has the most negative value −328 kJ mol⁻¹. Therefore it behaves like a halogen.

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(d) Least reactive non-metal
Element IV has a negative electron gain enthalpy but less negative than III. Hence its tendency to gain electrons is weaker. Therefore IV is the least reactive non-metal.

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(e) Metal forming MX₂
Metals forming M²⁺ ions generally have moderate first and second ionization enthalpies. Element VI has values 738 and 1451, so it can lose two electrons and form a stable MX₂ compound like a Group 2 metal.

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(f) Metal forming covalent MX
Element I shows a very large jump between ΔH₁ and ΔH₂. This means only one electron is easily removed. Therefore it behaves like a Group 1 metal and forms MX type halides.

(a) Li and O → Li₂O
• Li (Group 1) valency = 1 → Li⁺
• O (Group 16) valency = 2 → O²⁻
✅ By cross method → Li₂O

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(b) Mg and N → Mg₃N₂
• Mg (Group 2) valency = 2 → Mg²⁺
• N (Group 15) valency = 3 → N³⁻
✅ Cross method → Mg₃N₂

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(c) Al and I → AlI₃
• Al (Group 13) valency = 3
• I (Group 17) valency = 1
✅ Formula → AlI₃

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(d) Si and O → SiO₂
• Si valency = 4
• O valency = 2
✅ Formula → SiO₂

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(e) P and F → PF₃ or PF₅
• P (Group 15) valency = 3 or 5
• F valency = 1
✅ Therefore stable compounds: PF₃ and PF₅

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(f) Element 71 and F → LuF₃
• Element with atomic number 71 = Lutetium (Lu)
• Lu valency = 3
• F valency = 1
✅ Formula → LuF₃

1️⃣ In the modern periodic table , each period begins with the filling of a new electron shell .

2️⃣ The identity of an electron shell is determined by the principal quantum number (n) .

3️⃣ Therefore, the period number tells us the value of n for the outermost occupied shell.

Example:
• Elements in the 2nd period have outer shell n = 2.
• Elements in the 3rd period have outer shell n = 3.

✅ Hence, the period represents the principal quantum number (n) .

1️⃣ p-subshell Maximum electrons = 6 (3 orbitals × 2 electrons each).
✅ Therefore statement (a) is correct, and the p-block has 6 columns.

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2️⃣ d-subshell Maximum electrons = 10 (5 orbitals × 2 electrons each).
👉 Therefore the d-block has 10 columns, not 8.
✅ Hence statement (b) is incorrect.

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3️⃣ Statement (c) is also correct because the number of columns in a block equals the electron capacity of that subshell:
• s-block → 2
• p-block → 6
• d-block → 10
• f-block → 14

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4️⃣ Statement (d) is also correct because the block (s, p, d, f) corresponds to the value of the azimuthal quantum number (ℓ) of the subshell that receives the last electron.

1️⃣ The valence shell is related to the arrangement of electrons and their attraction toward the nucleus.

2️⃣ The following factors influence the valence shell:
• Principal quantum number (n) determines which shell is the outermost.
• Nuclear charge (Z) controls the attraction between nucleus and electrons (effective nuclear charge Z_eff).
• Core electrons cause the shielding effect , which modifies the attraction felt by valence electrons.

3️⃣ However, the nuclear mass does not change the electronic configuration or the arrangement of valence electrons.

✅ Therefore, nuclear mass does not influence the valence shell.

1️⃣ The species F⁻, Ne and Na⁺ all contain 10 electrons, therefore they are isoelectronic species.

2️⃣ In an isoelectronic series, the number of electrons and the principal quantum number (n) remain almost the same, and electron–electron repulsion is similar.

3️⃣ The main factor that changes is the nuclear charge (Z).
As Z increases, the attraction between the nucleus and electrons becomes stronger and the electron cloud is pulled closer.
✅ Therefore the atomic radius decreases.

Nuclear charges:
• F → Z = 9 → F⁻
• Ne → Z = 10
• Na → Z = 11 → Na⁺

✅ Order of size (radius):
F⁻ > Ne > Na⁺

(a) Correct
Each successive electron is removed from an increasingly positively charged ion , so the attraction between nucleus and electrons increases. Therefore the ionization enthalpy increases.

(b) Correct
When all valence electrons are removed, the atom attains a noble gas configuration . Removing an electron from this stable core requires a very large amount of energy.

(c) Correct
A big jump in ionization enthalpy indicates that all valence electrons have been removed and the next electron must be removed from the inner shell .

(d) Incorrect
Electrons in orbitals with higher n value are farther from the nucleus and experience more shielding . Therefore they are easier to remove . ❌ Hence statement (d) is incorrect.

1️⃣ What does metallic character mean?
👉 Metallic character refers to the tendency of an element to lose electrons easily.
👉 In other words, it shows the electropositive nature of an element.

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2️⃣ Periodic trend of metallic character
• In a group, metallic character increases from top to bottom.
• In a period, metallic character decreases from left to right.

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3️⃣ Comparing the given elements
• K (Potassium) → Group 1, Period 4 → very low ionization enthalpy → most metallic.

• Mg (Magnesium) → Group 2, Period 3 → metallic but less than K.

• Al (Aluminium) → Group 13, Period 3 → more towards the right side of the period → metallic character decreases.

• B (Boron) → Period 2 element → behaves like a metalloid → least metallic.

✅ Therefore the correct order is:
K > Mg > Al > B

1️⃣ Non-metallic character refers to the tendency of an element to attract or gain electrons . It is closely related to electronegativity .

2️⃣ General periodic trends:
• Across a period (left → right), non-metallic character increases .
• Down a group (top → bottom), non-metallic character decreases .

3️⃣ Applying these trends:
• Fluorine (F) is the most electronegative element → highest non-metallic character.
• Nitrogen (N) lies to the right of carbon → N > C.
• Boron (B) lies to the left of carbon → C > B.
• Silicon (Si) lies below carbon in Group 14 → non-metallic character decreases → C > Si.

✅ Therefore the correct order is F > N > C > B > Si .

1️⃣ Meaning of oxidising property
👉 An oxidising agent is a substance that gains electrons or removes electrons from another substance.

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2️⃣ Periodic trend of oxidising power
• Across a period, oxidising power generally increases from left to right.
• Down a group, oxidising power generally decreases.

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3️⃣ Comparing the given elements
• Fluorine (F) → highest electronegativity → strongest oxidising agent.

• Oxygen (O) → very high electron affinity and electronegativity → strong oxidising nature.

• Chlorine (Cl) → strong oxidising halogen but slightly weaker than oxygen in this comparison.

• Nitrogen (N) → relatively lower tendency to gain electrons → least oxidising among these.

✅ Therefore the correct order is:
F > O > Cl > N