Chapter 5: Periodic Classification of Elements

Solution

In Dobereiner's triads, the atomic mass of the middle element is approximately the arithmetic mean (average) of the atomic masses of the first and third elements.

Example: Li (6.9), Na (23.0), K (39.1) → (6.9 + 39.1) / 2 = 23.0 = atomic mass of Na ✓

Solution

The Modern Periodic Table consists of 7 horizontal rows (periods) and 18 vertical columns (groups). Elements are arranged in increasing order of atomic number.

Solution

In N2O5, the valency of N can be deduced from the formula.

Valency of N = 5

Solution

In Al2O3, the valency of Al can be deduced from the formula.

Valency of Al = 3

Solution

In CH4, the valency of C can be deduced from the formula.

Valency of C = 4

Solution

In PCl5, the valency of P can be deduced from the formula.

Valency of P = 5

Solution

In SO3, the valency of S can be deduced from the formula.

Valency of S = 6

Solution

In Na2O, the valency of Na can be deduced from the formula.

Valency of Na = 1

Solution

In FeCl2, the valency of Fe can be deduced from the formula.

Valency of Fe = 2

Solution

In CuO, the valency of Cu can be deduced from the formula.

Valency of Cu = 2

Solution

In FeCl3, the valency of Fe can be deduced from the formula.

Valency of Fe = 3

Solution

In MgCl2, the valency of Mg can be deduced from the formula.

Valency of Mg = 2

Solution

Newlands' Law of Octaves worked well only for lighter elements, up to calcium (Ca, atomic mass 40). After calcium, every 8th element did not have properties similar to the first, and the pattern broke down. This was a major limitation of the law.

Solution

Mendeleev's Periodic Law: 'The properties of elements are a periodic function of their atomic masses.'

The Modern Periodic Law (by Moseley) replaced 'atomic masses' with 'atomic numbers', which resolved many anomalies in Mendeleev's table.

Solution

Electronic configuration: 2, 8, 3

Period = Number of shells = 3 → Period 3

Group: Valence electrons = 3. Since 3 > 2 (p-block), Group = 10 + 3 = 13

This element is Aluminium (Al), atomic number 13.

Solution

Moving left to right across a period, the atomic number increases (more protons), so the nuclear charge increases. Since electrons are added to the same shell, the stronger nuclear charge pulls all electrons closer to the nucleus, resulting in a smaller atomic radius.

Example: Na (186 pm) > Mg (160 pm) > Al (143 pm) > Si (117 pm) ... > Cl (99 pm)

Solution

Moving down a group, new electron shells are added, so atomic size increases. The valence electrons are farther from the nucleus and experience less nuclear attraction. They can be lost more easily, increasing metallic character.

Example in Group 1: Li < Na < K < Rb < Cs (Cs is the most metallic among these)

Solution

Group 17 elements (halogens: F, Cl, Br, I) have 7 valence electrons. They need only 1 more electron to complete their octet. Therefore, their valency is 1.

Valency is the number of electrons gained, lost, or shared — not always equal to the number of valence electrons.

Solution

All elements in the same group have the same number of valence electrons. Since valency depends on the number of valence electrons, the valency remains constant down a group.

Example: Group 1 — Li(1), Na(1), K(1) — all have valency 1.

Solution

Henry Moseley (1913) proposed the Modern Periodic Law: 'The properties of elements are a periodic function of their atomic numbers.' This replaced Mendeleev's law based on atomic masses and resolved anomalies like the positions of cobalt/nickel and tellurium/iodine.

Solution

According to Dobereiner's triad rule:

Atomic mass of middle element = (Atomic mass of 1st + Atomic mass of 3rd) / 2

= (7 + 39) / 2

= 46 / 2

= 23

This corresponds to the triad: Lithium (7), Sodium (23), Potassium (39).

Solution

Elements in the same group have the same number of valence electrons, which is why they show similar chemical properties. For example, Group 1 elements (Li, Na, K, Rb, Cs) all have 1 valence electron.

Elements in the same period have the same number of electron shells.

Solution

For electronegativity:

F, Cl, Br arranged in increasing order: Br < Cl < F

Solution

For ionization energy:

Na, Mg, Al arranged in increasing order: Na < Al < Mg

Solution

For electronegativity:

C, N, O arranged in increasing order: C < N < O

Solution

For metallic character:

Na, Mg, Al arranged in increasing order: Al < Mg < Na

Solution

For atomic radius:

Li, Na, K arranged in increasing order: Li < Na < K

Solution

For metallic character:

Li, Na, K arranged in increasing order: Li < Na < K

Solution

For atomic radius:

C, N, O arranged in increasing order: O < N < C

Solution

For electronegativity:

Na, Mg, Cl arranged in increasing order: Na < Mg < Cl

Solution

For atomic radius:

Na, Mg, Al arranged in increasing order: Al < Mg < Na

Solution

Atomic number 12 means 12 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 2

Solution

Atomic number 6 means 6 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 4

Solution

Atomic number 18 means 18 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 8

Solution

Atomic number 19 means 19 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 8, 1

Solution

Atomic number 1 means 1 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 1

Solution

Atomic number 20 means 20 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 8, 2

Solution

Atomic number 8 means 8 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 6

Solution

Atomic number 11 means 11 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 1

Solution

Atomic number 17 means 17 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 7

Solution

Atomic number 14 means 14 electrons.

Fill shells: K(max 2), L(max 8), M(max 8 for first 20 elements), N...

Electronic configuration: 2, 8, 4

Solution

For atomic radius:

F, Cl, Br arranged in increasing order: F < Cl < Br

Solution

Predicting undiscovered elements was actually an achievement of Mendeleev's table, not a limitation. His predictions of eka-aluminium (Gallium), eka-silicon (Germanium), and eka-boron (Scandium) were remarkably accurate.

The limitations were: uncertain position of hydrogen, no place for isotopes, and incorrect ordering of some elements (like Co before Ni, Te before I).

Solution

Atomic number 12 → Electronic configuration: 2, 8, 2

Valence electrons = 2 → Valency = 2

Since it has 2 valence electrons (s-block element), Group = 2

This element is Magnesium (Mg), an alkaline earth metal.

Solution

Element A (Period 3, Group 1) = Sodium (Na) — an alkali metal (1 valence electron, tendency to lose it).

Element B (Period 3, Group 17) = Chlorine (Cl) — a halogen / non-metal (7 valence electrons, tendency to gain 1).

In any period, elements on the left are metals and elements on the right are non-metals, with metallic character decreasing left to right.

Solution

We need to consider both period and group trends:

• Na vs Mg (same period 3): Mg is to the right, so Mg has smaller atomic size. Mg < Na
• Ca vs K (same period 4): Ca is to the right, so Ca has smaller atomic size. Ca < K
• Na vs K (same group 1): K is below Na, so K has larger atomic size. Na < K
• Mg vs Ca (same group 2): Ca is below Mg, so Ca has larger atomic size. Mg < Ca

Final order: Mg < Na < Ca < K

Solution

(a) Electronic configuration: Atomic number 16 → 2, 8, 6

(b) Period: Number of shells = 3 → Period 3

(c) Group: Valence electrons = 6. Since 6 > 2, Group = 10 + 6 = Group 16

(d) Valency: Valence electrons = 6. It needs 2 more electrons to complete octet. Valency = 8 − 6 = 2

(e) Metal or Non-metal: Group 16 is on the right side of the periodic table. The element is Sulphur (S) — a non-metal.

Solution

Achievements:

1. Prediction of new elements: Mendeleev left gaps for undiscovered elements and predicted their properties (eka-aluminium = Ga, eka-silicon = Ge, eka-boron = Sc). When discovered, the predicted properties matched remarkably.
2. Accommodation of noble gases: Noble gases (He, Ne, Ar, etc.) discovered later could be placed in a new group without disrupting the table.
3. Correction of atomic masses: The position of some elements suggested their accepted atomic masses were incorrect, which was later confirmed.

Limitations:

1. Position of hydrogen: Hydrogen has properties of both alkali metals (Group I) and halogens (Group VII), so no definite position could be assigned.
2. No explanation for isotopes: Isotopes of an element (e.g., Cl-35 and Cl-37) have different atomic masses but should occupy the same position — the table based on atomic mass could not handle this.

Solution

Let's analyse:

• Li (2,1) — Period 2, Group 1
• F (2,7) — Period 2, Group 17
• Na (2,8,1) — Period 3, Group 1
• Cl (2,8,7) — Period 3, Group 17

Fluorine has the smallest radius because it is in Period 2 (fewer shells than Na, Cl) and at the far right of its period (high nuclear charge for its shell count). It is the smallest atom among these four.

Solution

Newlands' Law of Octaves (1866): When elements are arranged in order of increasing atomic masses, the properties of every 8th element are similar to those of the 1st element, like the eighth note in a musical scale.

Reasons for rejection:

1. Not valid beyond calcium: The 'octave' pattern worked only for lighter elements up to calcium (atomic mass 40). For heavier elements, the 8th element did NOT have properties similar to the 1st.
2. No room for new elements: Newlands assumed that all elements had been discovered. When new elements were found later (e.g., noble gases), they could not be fitted into his table without disrupting the entire arrangement.

Additional issues: He sometimes placed two dissimilar elements in the same slot (e.g., cobalt and nickel) and placed chemically different elements in the same group (e.g., iron with sulphur).

Solution

Metallic character decreases as we move from left to right across a period. In Period 3:

Na (Group 1) > Mg (Group 2) > Al (Group 13) > Si (Group 14)

Sodium (Na) is the leftmost and therefore the most metallic. It has the greatest tendency to lose its single valence electron.

Solution

Electronic configurations:

• Nitrogen (Z=7): 2, 5 — Period 2, Group 15
• Phosphorus (Z=15): 2, 8, 5 — Period 3, Group 15

(a) Metallic character:

Phosphorus has greater metallic character than nitrogen. As we move down a group, atomic size increases, so the valence electrons are farther from the nucleus and more loosely held. This makes it easier to lose electrons, increasing metallic character. Nitrogen is a typical non-metal (gas), while phosphorus shows some metallic characteristics (e.g., it exists as a solid and has allotropes with different conductivities).

(b) Atomic size:

Phosphorus has a larger atomic size than nitrogen. Phosphorus has 3 electron shells while nitrogen has only 2. The additional shell increases the distance between the outermost electrons and the nucleus, making the atom bigger.

Solution

Across a period (left → right):

As we move across a period, the atomic number increases by 1 at each step, meaning one more proton is added to the nucleus and one more electron is added to the same outermost shell. The additional proton increases the nuclear charge, which pulls all electrons closer to the nucleus. Since electrons are added to the same shell (shielding effect is roughly constant), the effective nuclear attraction on the outermost electrons increases, reducing the atomic radius.

Down a group (top → bottom):

As we move down a group, a new electron shell is added at each period. Although nuclear charge also increases, the additional inner shells provide significant shielding (screening effect), reducing the effective nuclear charge felt by the outermost electrons. The outermost shell is now farther from the nucleus, so the atomic size increases.

Solution

(a) Identification:

• X (Z=11): 2, 8, 1 → Sodium (Na), Period 3, Group 1
• Y (Z=17): 2, 8, 7 → Chlorine (Cl), Period 3, Group 17
• Z (Z=19): 2, 8, 8, 1 → Potassium (K), Period 4, Group 1

(b) Increasing atomic size:

• Na vs Cl: Same period (3), Cl is to the right → Cl is smaller: Cl < Na
• Na vs K: Same group (1), K is below → K is larger: Na < K

Order: Cl < Na < K

(c) Most metallic: Potassium (K) — Group 1, Period 4. It is the largest atom among the three and loses its valence electron most easily.

(d) Most non-metallic: Chlorine (Cl) — Group 17, it has the highest tendency to gain electrons among these three elements.

Solution

(a) Electronic configurations:

• A (Z=12): 2, 8, 2 → Magnesium (Mg)
• B (Z=17): 2, 8, 7 → Chlorine (Cl)

(b) Position:

• A: 3 shells → Period 3; 2 valence electrons → Group 2
• B: 3 shells → Period 3; 7 valence electrons → Group 10+7 = Group 17

(c) Element A (Mg) is a metal with 2 valence electrons. It will lose 2 electrons to form a cation: Mg²⁺.

(d) A is a metal (Mg) and B is a non-metal (Cl). A will transfer its 2 electrons to 2 atoms of B. This forms an ionic bond.

Compound formula: Mg²⁺ + 2Cl⁻ → MgCl₂ (Magnesium chloride)