The basic theme of organization in the periodic table is to classify all known elements into groups and periods based on their properties. This classification allows us to study elements in families rather than individually, making the study of chemistry systematic and rational. Elements with similar properties are placed together, which helps in predicting the properties of new elements.

Mendeleev used Atomic Mass as the fundamental property to classify elements. He arranged elements in increasing order of their atomic masses.

Did he stick to it? No, he did not strictly adhere to the order of atomic mass. He prioritized similarity in chemical properties. For instance, he placed Iodine (atomic mass 126.90) after Tellurium (atomic mass 127.60) so that Iodine could be grouped with Fluorine, Chlorine, and Bromine, which share similar properties.

Mendeleev’s Periodic Law: The physical and chemical properties of elements are a periodic function of their Atomic Masses.

Modern Periodic Law: The physical and chemical properties of elements are a periodic function of their Atomic Numbers.

The shift from mass to number (proton count) resolved anomalies like the position of isotopes and inverted pairs (Ar-K, Co-Ni, Te-I).

The period number ($n$) indicates the principal quantum number of the valence shell. For the 6th period, $n=6$.
The orbitals filled in this period are determined by the $(n+l)$ rule: $6s, 4f, 5d,$ and $6p$.

Since each element adds one electron, the 6th period can accommodate a maximum of 32 elements.

Electronic Configuration:
Noble gas core for Period 7 starts after Radon ($Z=86$).
$Z=114$: $[Rn] 5f^{14} 6d^{10} 7s^2 7p^2$.

Period: Highest principal quantum number is $n=7$. So, 7th Period.

Group: It belongs to the p-block. Group number = $10 + \text{valence electrons}$.
Valence electrons = $2(s) + 2(p) = 4$. Group = $10 + 4 = 14$.
Group 14 (Carbon Family).

Period 3: $n=3$. Orbitals involved: $3s, 3p$.
Group 17 (Halogens): General configuration $ns^2 np^5$.

Electronic configuration: $[Ne] 3s^2 3p^5$.
Atomic Number ($Z$) = Atomic number of Neon (10) + 2 + 5 = 17.

Element: Chlorine ($Cl, Z=17$).

(i) Lawrence Berkeley Laboratory:
Lawrencium ($Z=103$), Berkelium ($Z=97$). Also Seaborgium ($Z=106$) is associated.

(ii) Seaborg’s Group:
Seaborgium ($Z=106$). Glenn T. Seaborg was instrumental in discovering many transuranium elements (Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No).

Elements in the same group have the same number of valence electrons and identical valence shell electronic configurations. Since chemical properties (like bonding, reactivity, valency) depend primarily on the valence electrons, elements in the same group exhibit similar chemical properties.

Atomic Radius: The distance from the center of the nucleus to the boundary of the outermost shell of electrons in an isolated atom. (Practically measured as covalent, metallic, or van der Waals radius).

Ionic Radius: The effective distance from the nucleus of an ion up to which it has an influence on its electron cloud. It is the radius of an atom after it has gained or lost electrons.

In a Period (Left to Right): Atomic radius decreases.
Reason: Electrons are added to the same valence shell. Nuclear charge increases, pulling the electron cloud closer to the nucleus.

In a Group (Top to Bottom): Atomic radius increases.
Reason: New principal energy shells are added. The shielding effect of inner shells increases, reducing the effective nuclear pull on outer electrons, despite the increase in nuclear charge.

Definition: Atoms and ions that contain the same number of electrons.

(i) $F^-$ ($9+1=10 e^-$): Isoelectronic with $\mathbf{Ne, Na^+, Mg^{2+}}$.
(ii) $Ar$ ($18 e^-$): Isoelectronic with $\mathbf{Cl^-, K^+, Ca^{2+}}$.
(iii) $Mg^{2+}$ ($12-2=10 e^-$): Isoelectronic with $\mathbf{F^-, Na^+, Ne}$.
(iv) $Rb^+$ ($37-1=36 e^-$): Isoelectronic with $\mathbf{Kr, Br^-, Sr^{2+}}$.

Species: $N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}, Al^{3+}$.
(a) Common: All have 10 electrons (Isoelectronic).

(b) Order of Increasing Ionic Radii:
For isoelectronic species, radius decreases as nuclear charge (atomic number) increases.
Nuclear Charge: $N(7) < O(8) < F(9) < Na(11) < Mg(12) < Al(13)$.
Radii Order: $Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$.

Cation: Smaller than parent atom. Electrons are removed, but nuclear charge remains the same. The remaining electrons are pulled more strongly by the nucleus (higher effective nuclear charge). Sometimes a whole shell is removed.

Anion: Larger than parent atom. Electrons are added. Nuclear charge is constant, but inter-electronic repulsion increases, causing the electron cloud to expand. Effective nuclear charge per electron decreases.

Isolated Gaseous Atom: In solids and liquids, atoms are bonded or influenced by neighbors. To measure intrinsic properties like ionization energy, atoms must be free from external influences, which is possible in the gaseous state.

Ground State: This specifies that the atom is in its lowest energy state. If an atom is excited, less energy is needed to remove an electron. For standard comparison, the starting point must be the ground state.

Ground state energy $E_1 = -2.18 \times 10^{-18} \text{ J}$.
Ionization energy is the energy required to remove the electron to infinity ($E=0$).
$\Delta E = E_{\infty} – E_1 = 0 – (-2.18 \times 10^{-18}) = 2.18 \times 10^{-18} \text{ J/atom}$.

For 1 mole ($N_A = 6.022 \times 10^{23}$):
$IE = 2.18 \times 10^{-18} \times 6.022 \times 10^{23} = \mathbf{1.312 \times 10^6 \text{ J mol}^{-1}}$ (or $1312 \text{ kJ mol}^{-1}$).

(i) Be > B:
Be ($1s^2 2s^2$) has a stable, fully filled 2s subshell. B ($1s^2 2s^2 2p^1$) has one electron in 2p. The 2p electron is less penetrating and more shielded by the 2s electrons, making it easier to remove than a 2s electron.

(ii) O < N:
N ($2s^2 2p^3$) has a stable half-filled p-subshell. O ($2s^2 2p^4$) has 4 electrons in p. Removing one electron from Oxygen results in a stable half-filled configuration ($2p^3$), and relieves electron-electron repulsion in the paired orbital. Hence, O has lower IE.

1st IE: Na ($3s^1$) vs Mg ($3s^2$). It is easier to remove the single valence electron from Na to achieve a stable noble gas core (Ne) than to break the stable filled 3s pair of Mg. Thus $IE_1(Na) < IE_1(Mg)$.

2nd IE: After losing one electron, $Na^+$ has the stable configuration of Neon ($2s^2 2p^6$). Removing a second electron disrupts this stability and requires immense energy. $Mg^+$ ($3s^1$) still has one valence electron, which is easily removed to form stable $Mg^{2+}$. Thus $IE_2(Na) > IE_2(Mg)$.

1. Increase in Atomic Size: The distance between the nucleus and valence electrons increases.
2. Screening Effect: The number of inner shells increases, shielding the valence electrons from the nuclear pull.

These two factors dominate over the increase in nuclear charge, resulting in a decrease in Ionization Enthalpy.

Trend: B > Tl > Ga > Al > In.
The deviation (Ga > Al and Tl > In) is due to poor shielding.

• Ga vs Al: Ga follows the 3d transition series. d-electrons shield the nucleus poorly. The effective nuclear charge on valence electrons increases, making IE of Ga slightly higher than Al.
• Tl vs In: Tl follows the 4f inner transition series (Lanthanoids). f-electrons have very poor shielding effect (Lanthanoid Contraction). This significantly increases effective nuclear charge, raising the IE of Tl.

(i) O or F: F. Fluorine needs only 1 electron to complete its octet and has a higher effective nuclear charge, making energy release more favorable.

(ii) F or Cl: Cl. Fluorine is a 2nd period element with a very small size. The incoming electron faces significant inter-electronic repulsion from the existing compact electron cloud. Chlorine (3rd period) is larger, accommodating the extra electron more easily. Hence, Cl has the most negative electron gain enthalpy.

Answer: Positive.

When an electron is added to a neutral oxygen atom ($O$), energy is released (exothermic), forming $O^-$.
However, when a second electron is added to the negatively charged $O^-$ ion to form $O^{2-}$, there is strong electrostatic repulsion between the incoming electron and the existing negative charge of the ion. To overcome this repulsion, energy must be supplied (endothermic). Thus, the second electron gain enthalpy is positive.

Electron Gain Enthalpy ($\Delta_{eg}H$): This is the energy change when an electron is added to an isolated gaseous atom. It provides a measure of the tendency of an isolated atom to accept an electron. Units: kJ/mol.

Electronegativity: This is the tendency of an atom in a chemical compound to attract the shared pair of electrons towards itself. It is a relative property and has no specific units.

The statement is incorrect. Electronegativity is not a constant property of an isolated atom; it depends on the element’s environment, such as its oxidation state and hybridization. For example, nitrogen has a higher electronegativity in $NO_2$ than in $NH_3$. The value of 3.0 on the Pauling scale is an average reference value.

(a) Gains an electron (Anion Formation): The nuclear charge ($Z$) remains constant, but the number of electrons increases. This increases electron-electron repulsion, causing the electron cloud to expand. The effective nuclear charge per electron decreases, resulting in a larger radius ($Anion > Atom$).

(b) Loses an electron (Cation Formation): The nuclear charge remains constant, but there are fewer electrons. The effective nuclear charge per electron increases, pulling the remaining electrons closer. Often, the valence shell is completely removed. This results in a smaller radius ($Cation < Atom$).

Expected to be the Same.

Ionization enthalpy depends primarily on the electronic configuration and the nuclear charge (number of protons). Isotopes of an element have the same number of protons and electrons (same configuration); they only differ in the number of neutrons (mass). Since the electrical forces are identical, their ionization enthalpies are essentially the same.

(a) Five electrons in outer subshell ($np^5$): Group 17 (Halogens). E.g., F, Cl.

(b) Tend to lose two electrons ($ns^2$): Group 2 (Alkaline Earth Metals). E.g., Mg, Ca.

(c) Tend to gain two electrons ($ns^2 np^4$): Group 16 (Chalcogens). E.g., O, S.

(d) Group having metal, non-metal, liquid, gas: Use Group 1 or Group 17. Group 17 contains F, Cl (Gas), Br (Liquid), I (Solid Non-metal), At/Ts (Radioactive/Metal). Group 1 contains H (Gas), others Metal (solid, though Fr is liquid near RT).
Best answer: Group 17 (Halogens) or Group 1.

Group 1 (Alkali Metals): Reactivity depends on the ability to lose an electron (low Ionization Enthalpy). As we go down the group, atomic size increases and IE decreases. Thus, electron loss becomes easier, and reactivity increases ($Li < Na < K \dots$).

Group 17 (Halogens): Reactivity depends on the ability to gain an electron (high negative Electron Gain Enthalpy). As we go down the group, atomic size increases and electron affinity decreases. Thus, reactivity decreases ($F > Cl > Br > I$).

• s-block: $ns^{1-2}$
• p-block: $ns^2 np^{1-6}$
• d-block: $(n-1)d^{1-10} ns^{0-2}$
• f-block: $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$

(i) $ns^2 np^4$ for $n=3$:
Period = 3. p-block. Group = $10 + 2 + 4 = 16$. (Sulphur).

(ii) $(n-1)d^2 ns^2$ for $n=4$:
Period = 4. d-block. Group = $2 + 2 = 4$. (Titanium).

(iii) $(n-2)f^7 (n-1)d^1 ns^2$ for $n=6$:
Period = 6. f-block (Lanthanoids). Group = 3. (Gadolinium).

Analysis of Table:
I (520, 7300): Large jump after 1st IE $\to$ Group 1 (Li).
II (419, 3051): Large jump after 1st IE $\to$ Group 1 (K).
III (1681, 3374, -328): High IE, High Neg EGE $\to$ Group 17 (F).
IV (1008, 1846, -295): Med IE, High Neg EGE $\to$ Group 17 (Cl/I).
V (2372, 5251, +48): Very High IE, Positive EGE $\to$ Group 18 (Noble Gas, He).
VI (738, 1451, -40): Jump after 2nd IE $\to$ Group 2 (Mg).

(a) Least reactive: V (Noble Gas).
(b) Most reactive metal: II (Lowest IE).
(c) Most reactive non-metal: III (Highest negative EGE, Fluorine).
(d) Least reactive non-metal: IV (Iodine) [V is inert, not usually classified in reactivity series].
(e) Metal forming $MX_2$: VI (Group 2, valency 2).
(f) Metal forming $MX$: I (Group 1, valency 1).

Valencies based on groups:
Li (+1), Mg (+2), Al (+3), Si (+4), P (-3/+5), O (-2), N (-3), I (-1), F (-1), Lu (+3).

• (a) Li & O $\to$ $Li_2O$
• (b) Mg & N $\to$ $Mg_3N_2$
• (c) Al & I $\to$ $AlI_3$
• (d) Si & O $\to$ $SiO_2$
• (e) P & F $\to$ $PF_3$ or $PF_5$
• (f) Element 71 (Lutetium) & F $\to$ $LuF_3$

(c) Principal quantum number ($n$).
The period number corresponds to the highest principal quantum number ($n$) of the elements in that period.

(b) is Incorrect.
The d-block has 10 columns (Groups 3 to 12), not 8, because a maximum of 10 electrons can occupy the five d-orbitals.

(c) Nuclear Mass.
Chemical properties and valence shell behavior depend on nuclear charge ($Z$) and electron configuration ($n$, shielding). Nuclear mass (neutrons) affects physical properties like density and isotopes but not the valence shell chemistry.

(a) Nuclear Charge ($Z$).
Since they have the same number of electrons (10), the shielding and electron interaction are similar. The size difference arises purely from the difference in the number of protons (nuclear charge) pulling on those electrons.

(d) is Incorrect.
“Removal of electron from orbitals bearing lower n value is easier…”
Lower $n$ means the electron is closer to the nucleus and more tightly bound. Therefore, removing an electron from a lower $n$ value is harder (requires more energy), not easier.

Trend: Metallic character increases down a group and decreases across a period.
Positions: K (Group 1) > Mg (Group 2) > Al (Group 13) > B (Group 13, Period 2, Non-metal).
Order: K > Mg > Al > B.
Correct Option: (d).

Trend: Non-metallic character increases across a period and decreases down a group.
Period 2: B < C < N < F.
Si is below C (Group 14), so Si < C.
Order: Si < B < C < N < F? Wait, B vs Si. B is semi-metal/non-metal. Si is metalloid. B is generally more non-metallic than Si due to smaller size.
Correct Option: (c) F > N > C > B > Si.

Oxidizing power relates to the tendency to gain electrons (Electronegativity/EGE).
F is the strongest oxidant. O is second (high EN). Cl is third. N is weakest.
Order: F > O > Cl > N.
Correct Option: (b).