NEET Classification of Elements and Periodicity in Properties Previous Year Questions with Solutions

Electron gain enthalpy (Δ_eg H) is the enthalpy change when an isolated gaseous atom in its ground state gains an electron to form a gaseous uninegative ion. For halogens, this process is highly exothermic.

• • Negative Δ_eg H indicates energy release.
• • Trend question compares magnitude (exothermicity).

Although electronegativity decreases uniformly from F to I, the electron gain enthalpy of fluorine is unexpectedly less negative than that of chlorine. In fluorine (n=2), the incoming electron enters the compact 2p subshell where it experiences strong repulsive forces from the other seven valence electrons. In chlorine (n=3), the incoming electron enters the larger 3p subshell where interelectronic repulsions are substantially weaker.

• • Δ_eg H of Cl = -349 kJ mol⁻¹.
• • Δ_eg H of F = -328 kJ mol⁻¹.

Beyond chlorine, as we descend to bromine and iodine, the increase in atomic size and nuclear shielding causes electron affinity to steadily decrease.

• • Trend: Cl > F > Br > I.
• • Matches Option (3).

Across the second period from left to right, first ionization enthalpy generally increases with increasing nuclear charge. However, oxygen has an anomalous dip: - Nitrogen (Z = 7): Δ_i H₁ = 1402 kJ mol⁻¹ - Oxygen (Z = 8): Δ_i H₁ = 1314 kJ mol⁻¹ - Fluorine (Z = 9): Δ_i H₁ = 1681 kJ mol⁻¹ Thus, Δ_i H₁ of O is lower than that of both N and F.

• • Assertion A is factually correct.

Examine the valence electron configurations: - Nitrogen: [He] 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half-filled p subshell, highly stable, high exchange energy). - Oxygen: [He] 2s² 2p_x² 2p_y¹ 2p_z¹ (contains one pair of electrons in the 2p_x orbital). In oxygen, the two electrons in the 2p_x orbital repel each other strongly. Removing one electron relieves this interelectronic repulsion and produces the symmetric, exchange-stabilized half-filled 2p³ configuration of O⁺.

• • Reason R is factually correct.

Because the product ion O⁺ achieves a stable half-filled subshell and interelectronic repulsion in 2p⁴ is relieved, less energy is required to ionize oxygen than nitrogen. Therefore, Reason R directly explains Assertion A.

• • Matches Option (1).

Analyze the canonical NCERT data for Gallium, Caesium, and Pauling electronegativities:

• • Both Ga and Cs have exceptionally LOW melting points and exist as liquids at room temperature during summer. Statement A is FALSE.
• • Pauling scale values: EN(N) = 3.0 and EN(Cl) = 3.0. Their electronegativity values ARE identical. Statement B is FALSE.

Count the total number of electrons in each chemical species:

• • All species have exactly 18 electrons and are therefore isoelectronic. Statement C is TRUE.

Examine electronic configuration effects on ionization enthalpy and group trends in atomic radius:

• • Mg ([Ne]3s²) has a fully filled 3s orbital with greater penetration than the 3p electron of Al ([Ne]3s² 3p¹), so IE₁(Mg) > IE₁(Al). Statement D is FALSE.
• • Down the alkali metal group (Group 1), atomic radius increases monotonically with increasing principal quantum number: r(Li) < r(Na) < r(K) < r(Rb) < r(Cs). Statement E is TRUE.
• • Conclusion: C and E only are true, matching Option (2).

In modern chemistry and the NCERT curriculum, "Main Group Elements" (or "Representative Elements") are defined strictly as elements belonging to the s-block (Groups 1 and 2) and the p-block (Groups 13 to 18):

• • d-block elements are transition elements, and f-block elements are inner transition elements.

Identify the block and identity of configurations A, B, and C:

• • B: [Ar] 3d³ 4s² ⟹ Vanadium (d-block, Group 5, transition metal). Not main group.
• • C: [Kr] 4d¹⁰ 5s² 5p⁵ ⟹ Iodine (p-block, Group 17, halogen). Main Group.

Identify the block and identity of configurations D and E:

• • Neither D nor E is a main group element.
• • Therefore, only configurations A and C belong to main group elements, matching Option (2).

Writing ground-state electronic configurations: - Li (Z = 3): 1s² 2s¹ - Be (Z = 4): 1s² 2s² (completely filled 2s orbital) - B (Z = 5): 1s² 2s² 2p¹ - C (Z = 6): 1s² 2s² 2p² - N (Z = 7): 1s² 2s² 2p³ (half-filled, stable 2p subshell)

Although nuclear charge increases from Be (+4) to B (+5), the 2p electron in boron is shielded from the nucleus by the inner 2s electrons and occupies a higher energy level than 2s electrons. Furthermore, removing an electron from Be disrupts a completely filled 2s² subshell with greater penetration power. As a consequence, it requires less energy to remove the 2p¹ electron of boron than the 2s² electron of beryllium:

Comparing all values: - Li: 520 kJ/mol - B: 801 kJ/mol - Be: 899 kJ/mol - C: 1086 kJ/mol - N: 1402 kJ/mol Arranging in strictly increasing order:

Electronegativity is the qualitative measure of an atom’s ability in a chemical compound to attract shared electrons towards itself. Across a period, increasing effective nuclear charge and decreasing atomic radius attract valence electrons more strongly. Down a group, increased shielding and greater atomic radius reduce electronegativity.

On the Pauling electronegativity scale: - Silicon (Si, Period 3, Group 14): χ = 1.8 - Carbon (C, Period 2, Group 14): χ = 2.5 - Nitrogen (N, Period 2, Group 15): χ = 3.0 - Oxygen (O, Period 2, Group 16): χ = 3.5 - Fluorine (F, Period 2, Group 17): χ = 4.0 (most electronegative element in the periodic table)

Arranging in strictly increasing order yields: Si < C < N < O < F, perfectly matching Option (3).

Each element attains the electron configuration of the nearest noble gas (Neon, 1s² 2s² 2p⁶, total 10 electrons) by gaining or losing electrons:

All four resulting ions possess the identical number of electrons (10), but differing nuclear charges (number of protons Z in the nucleus):

As nuclear charge decreases, the inward electrostatic pull on the 10 electrons weakens and interelectronic repulsions dominate, causing the electron cloud to expand. Consequently, ionic radius increases with decreasing Z:

Nitrogen forms the N³⁻ ion (radius ≈ 171 pm), which is the largest among all ions in this isoelectronic group.

Isoelectronic ions are ions of different chemical elements that possess the identical total number of electrons and identical ground-state electronic configurations.

- Pair 1 ($O^{2-}, F^-$): - Oxygen ($Z=8$): $8 - (-2) = 10\text{ electrons}$ - Fluorine ($Z=9$): $9 - (-1) = 10\text{ electrons}$ (Isoelectronic, Neon configuration $[Ne]$) - Pair 2 ($Na^+, Mg^{2+}$): - Sodium ($Z=11$): $11 - 1 = 10\text{ electrons}$ - Magnesium ($Z=12$): $12 - 2 = 10\text{ electrons}$ (Isoelectronic, Neon configuration $[Ne]$) - Pair 3 ($Mn^{2+}, Fe^{3+}$): - Manganese ($Z=25$): $25 - 2 = 23\text{ electrons}$ ($[Ar]3d^5$) - Iron ($Z=26$): $26 - 3 = 23\text{ electrons}$ ($[Ar]3d^5$) (Isoelectronic, both half-filled $d^5$)

Calculate electrons for iron(II) and manganese(II): - $Fe^{2+}$ ($Z = 26$): $26 - 2 = 24\text{ electrons}$ ($[Ar]3d^6$) - $Mn^{2+}$ ($Z = 25$): $25 - 2 = 23\text{ electrons}$ ($[Ar]3d^5$) Because $24 \neq 23$, this pair does not have the same number of electrons and is NOT an isoelectronic pair.

Roots used: nil = 0, un = 1, bi = 2, tri = 3, quad = 4, pent = 5, hex = 6, sept = 7, oct = 8, enn = 9, ending with "-ium".

Un (1) + nil (0) + un (1) + ium = Atomic number 101 ($$\text{Unu}$$). Official IUPAC name is Mendelevium ($$\text{Md}$$). Correct match.

Un (1) + nil (0) + tri (3) + um = Atomic number 103 ($$\text{Unt}$$). Official IUPAC name is Lawrencium ($$\text{Lr}$$). Correct match.

Un (1) + nil (0) + hex (6) + ium = Atomic number 106 ($$\text{Unh}$$). Official IUPAC name is Seaborgium ($$\text{Sg}$$). Correct match.

Un (1) + un (1) + un (1) + ium = Atomic number 111 ($$\text{Uuu}$$). Official IUPAC name is Roentgenium ($$\text{Rg}$$). Darmstadtium ($$\text{Ds}$$) is element 110 (Ununnilium, $$Z = 110$$). Thus, (d) matching with (iv) is incorrect.

Carbon monoxide ($$\text{CO}$$) is neutral; it shows neither acidic nor basic properties and does not form salts with acids or bases (other neutral oxides are $$\text{NO}$$ and $$\text{N}_2\text{O}$$). Thus, (a) matches (ii).

Barium is a highly electropositive alkaline earth metal (Group 2). Its oxide, $$\text{BaO}$$, dissolves in water to form the strong base barium hydroxide ($$\text{Ba(OH)}_2$$) and reacts vigorously with acids. Thus, (b) matches (i).

$$\text{Al}_2\text{O}_3$$ exhibits both acidic and basic characteristics, dissolving in hydrochloric acid to form aluminium chloride and in sodium hydroxide to form sodium aluminate ($$\text{Na[Al(OH)}_4]$$). Thus, (c) matches (iv) Amphoteric.

Chlorine is a strongly electronegative halogen in its highest oxidation state (+7). $$\text{Cl}_2\text{O}_7$$ dissolves in water to form perchloric acid ($$\text{HClO}_4$$), one of the strongest inorganic acids. Thus, (d) matches (iii) Acidic.