Comprehensive Study Guide to P-Block Elements (Groups 13 to 18)

General Introduction to P-Block Elements

The pp-block of the periodic table encompasses elements belonging to Groups 1313 through 1818. In these elements, the valence electrons enter the outermost pp-orbitals. Their general valence shell electronic configuration is ns2 np1−6ns^2\,np^{1-6} (with the exception of Helium, which possesses a 1s21s^2 configuration).

The maximum oxidation state exhibited by a pp-block element equals the total number of valence electrons (the sum of the ss- and pp-electrons). This is commonly designated as the group oxidation state. Besides the group oxidation state, pp-block elements also exhibit lower oxidation states that typically differ from the group oxidation state by units of two. The relative stability of these lower oxidation states increases down the group for the heavier members due to the inert pair effect (the reluctance of the valence ss-electrons to participate in bonding owing to poor shielding by intervening dd- and ff-electrons).


Group 15 Elements: The Nitrogen Family

Group 1515 elements, historically known as the pnictogens (meaning "suffocation-forming"), include Nitrogen (NN), Phosphorus (PP), Arsenic (AsAs), Antimony (SbSb), Bismuth (BiBi), and the synthetic element Moscovium (McMc).

Electronic Configuration and Atomic Numbers

The atomic numbers and electronic configurations follow the standard Aufbau buildup:

  • Nitrogen (NN): Z=7Z = 7; Electronic Configuration: [He] 2s2 2p3[He]\,2s^2\,2p^3

  • Phosphorus (PP): Z=15Z = 15; Electronic Configuration: [Ne] 3s2 3p3[Ne]\,3s^2\,3p^3

  • Arsenic (AsAs): Z=33Z = 33; Electronic Configuration: [Ar] 3d10 4s2 4p3[Ar]\,3d^{10}\,4s^2\,4p^3

  • Antimony (SbSb): Z=51Z = 51; Electronic Configuration: [Kr] 4d10 5s2 5p3[Kr]\,4d^{10}\,5s^2\,5p^3

  • Bismuth (BiBi): Z=83Z = 83; Electronic Configuration: [Xe] 4f14 5d10 6s2 6p3[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^3

  • Moscovium (McMc): Z=115Z = 115

The general valence shell electronic configuration is ns2 np3ns^2\,np^3. The ss-orbital is completely filled, and the pp-orbitals are half-filled, conferring exceptional electronic stability.

Occurrence

  • Nitrogen: Molecular nitrogen (N2N_2) comprises 78%78\% by volume of the atmosphere. In the earth's crust, it occurs as sodium nitrate (NaNO3NaNO_3, Chile saltpetre) and potassium nitrate (KNO3KNO_3, Indian saltpetre). It is also found in the form of proteins in plants and animals.

  • Phosphorus: Occurs primarily in minerals of the apatite family, Ca9(PO4)6⋅CaX2Ca_9(PO_4)_6 \cdot CaX_2 (where X=F,Cl, or OHX = F, Cl,\text{ or } OH), such as fluorapatite (3Ca3(PO4)2⋅CaF23Ca_3(PO_4)_2 \cdot CaF_2 or Ca9(PO4)6⋅CaF2Ca_9(PO_4)_6 \cdot CaF_2). It is an essential constituent of animal and plant matter, bones, living cells, and phosphoproteins present in milk and eggs.

  • Arsenic, Antimony, and Bismuth: Occur primarily as sulphide minerals in the earth's crust.

Atomic and Physical Properties

Quantitative Data Table (NCERT Table 7.1)

Property

NN

PP

AsAs

SbSb

BiBi

Atomic Number

77

1515

3333

5151

8383

Atomic Mass (g mol−1g\,mol^{-1})

14.0114.01

30.9730.97

74.9274.92

121.75121.75

208.98208.98

Electronic Configuration

[He] 2s2 2p3[He]\,2s^2\,2p^3

[Ne] 3s2 3p3[Ne]\,3s^2\,3p^3

[Ar] 3d10 4s2 4p3[Ar]\,3d^{10}\,4s^2\,4p^3

[Kr] 4d10 5s2 5p3[Kr]\,4d^{10}\,5s^2\,5p^3

[Xe] 4f14 5d10 6s2 6p3[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^3

ΔiH1\Delta_i H_1 (kJ mol−1kJ\,mol^{-1})

14021402

10121012

947947

834834

703703

ΔiH2\Delta_i H_2 (kJ mol−1kJ\,mol^{-1})

28562856

19031903

17981798

15951595

16101610

ΔiH3\Delta_i H_3 (kJ mol−1kJ\,mol^{-1})

45774577

29102910

27362736

24432443

24662466

Electronegativity

3.03.0

2.12.1

2.02.0

1.91.9

1.91.9

Covalent Radius (pmpm)

7070

110110

121121

141141

148148

Ionic Radius (pmpm)

171 (E3−)171\,(E^{3-})

212 (E3−)212\,(E^{3-})

222 (E3−)222\,(E^{3-})

76 (E3+)76\,(E^{3+})

103 (E3+)103\,(E^{3+})

Melting Point (KK)

6363

317317

10891089

904904

544544

Boiling Point (KK)

77.277.2

554554

888888

18601860

18371837

Density (g cm−3g\,cm^{-3} at 298 K298\,K)

0.879 (63 K)0.879\,(63\,K)

1.8231.823

5.7785.778

6.6976.697

9.8089.808

Detailed Periodic Trends
  1. Atomic and Covalent Radii: Covalent radius increases down the group (N<P<As<Sb<BiN < P < As < Sb < Bi). A substantial increase is observed from NN (70 pm70\,pm) to PP (110 pm110\,pm). From AsAs (121 pm121\,pm) to BiBi (148 pm148\,pm), only a small increase is noted due to the poor shielding effect of completely filled dd- and ff-orbitals.

  2. Ionisation Enthalpy: Decreases down the group due to the increase in atomic size (ΔiH1:N>P>As>Sb>Bi\Delta_i H_1: N > P > As > Sb > Bi). Because of their stable half-filled p3p^3 configurations, Group 1515 elements exhibit higher first ionisation enthalpies than the corresponding Group 1414 and Group 1616 elements. Note an anomaly in higher ionisation steps: ΔiH2\Delta_i H_2 and ΔiH3\Delta_i H_3 of BiBi are slightly higher than those of SbSb due to poor shielding by 4f4f-electrons.

  3. Electronegativity: Decreases down the group with increasing atomic size (N(3.0)>P(2.1)>As(2.0)>Sb(1.9)=Bi(1.9)N (3.0) > P (2.1) > As (2.0) > Sb (1.9) = Bi (1.9)).

  4. Boiling and Melting Points:

    • Boiling Point: Increases from Nitrogen to Antimony and then slightly decreases at Bismuth: N<P<As<Bi<SbN < P < As < Bi < Sb (or decreasing: Sb>Bi>As>P>NSb > Bi > As > P > N). The decrease from SbSb to BiBi is attributed to the inert pair effect reducing bonding tendencies.

    • Melting Point: Increases up to Arsenic and then decreases down to Bismuth: N<P<Bi<Sb<AsN < P < Bi < Sb < As (decreasing: As>Sb>Bi>P>NAs > Sb > Bi > P > N). Arsenic possesses a giant layered covalent lattice yielding a peak melting point of 1089 K1089\,K.

  5. Density: Increases monotonically down the group (N<P<As<Sb<BiN < P < As < Sb < Bi).

  6. Physical Nature: Dinitrogen is a diatomic gas (N2N_2), whereas all other elements are polyatomic solids (P4,As4,Sb4P_4, As_4, Sb_4). Nitrogen and phosphorus are non-metals, arsenic and antimony are metalloids, and bismuth is a typical metal. Except for nitrogen, all members exhibit allotropy.

Chemical Properties and Oxidation States

  • Common Oxidation States: −3,+3, and +5-3, +3, \text{ and } +5.

  • Stability of −3-3: The tendency to exhibit a −3-3 oxidation state decreases down the group due to increases in atomic size and metallic character. Bismuth hardly forms any compounds in the −3-3 state.

  • Stability of +5+5 vs +3+3: The stability of the +5+5 oxidation state decreases down the group, while the stability of the +3+3 oxidation state increases down the group. This is caused by the inert pair effect, where the outermost ns2ns^2 electrons remain unshared. For bismuth, Bi(V)Bi(V) compounds are powerful oxidising agents; BiF5BiF_5 is the only stable pentahalide.

  • Disproportionation Reactions:

    • For nitrogen, all oxidation states from +1+1 to +4+4 tend to disproportionate in acid solution:     3HNO2→H+HNO3+2NO+H2O3HNO_2 \xrightarrow{H^+} HNO_3 + 2NO + H_2O

    • For phosphorus, nearly all intermediate oxidation states (+1+1 to +4+4) disproportionate into +5+5 and −3-3 in acidic and alkaline media:     4H3PO3→Δ3H3PO4+PH34H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3

    • For heavier elements (As,Sb,BiAs, Sb, Bi), intermediate oxidation states become increasingly stable toward disproportionation.

  • Maximum Covalency: Nitrogen is restricted to a maximum covalency of 44 (11 ss + 33 pp orbitals) because it lacks vacant dd-orbitals in its valence shell. Heavier elements can expand their covalency to 55 or 66 using vacant dd-orbitals (e.g., PF6−PF_6^-).

Anomalous Properties of Nitrogen

Nitrogen differs from the rest of the group due to:

  1. Small atomic size.

  2. High electronegativity (3.03.0).

  3. High ionisation enthalpy (1402 kJ mol−11402\,kJ\,mol^{-1}).

  4. Non-availability of dd-orbitals in its valence shell.

Specific Consequences of Anomalous Nature
  • Multiple Bonding (pπ−pπp\pi-p\pi): Nitrogen possesses the unique ability to form strong pπ−pπp\pi-p\pi multiple bonds with itself (N≡NN \equiv N) and with other small, highly electronegative atoms (C,OC, O). Heavier members cannot form pπ−pπp\pi-p\pi bonds because their atomic orbitals are too large and diffuse to undergo effective lateral overlap.

  • Bond Enthalpy of Dinitrogen: Dinitrogen exists as a diatomic molecule with a triple bond (1 σ1\,\sigma and 2 π2\,\pi). Its bond dissociation enthalpy is exceptionally high (941.4 kJ mol−1941.4\,kJ\,mol^{-1}), making it chemically inert at room temperature.

  • Single Bond Enthalpies and Catenation: The single N−NN-N bond is significantly weaker than the single P−PP-P bond (N−N:160 kJ mol−1N-N: 160\,kJ\,mol^{-1} vs P−P:201 kJ mol−1P-P: 201\,kJ\,mol^{-1}). This weakness arises from strong interelectronic repulsions among the non-bonding electron pairs situated close together due to the short N−NN-N bond length. Consequently, nitrogen has a much weaker catenation property than phosphorus.

  • Lack of dπ−pπd\pi-p\pi and dπ−dπd\pi-d\pi Bonding: Nitrogen cannot form dπ−pπd\pi-p\pi bonds, unlike phosphorus and heavier elements (e.g., R3P=OR_3P=O or R3P=CH2R_3P=CH_2). Furthermore, phosphorus and arsenic form dπ−dπd\pi-d\pi bonds with transition metals when compounds such as P(C2H5)3P(C_2H_5)_3 and As(C6H5)3As(C_6H_5)_3 act as ligands.

Group 15 Hydrides (EH3EH_3)

All group 15 elements form volatile hydrides of the formula EH3EH_3 (NH3,PH3,AsH3,SbH3,BiH3NH_3, PH_3, AsH_3, SbH_3, BiH_3):

  • Basic Strength: NH3>PH3>AsH3>SbH3>BiH3NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3. Basicity depends on the electron pair availability on the central atom. As size increases, electron density disperses over a larger volume, diminishing basicity.

  • Thermal Stability and Bond Dissociation Enthalpy: NH3>PH3>AsH3>SbH3>BiH3NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3. As the central atom size increases, the E−HE-H bond length increases, bond strength decreases, and thermal stability falls.

  • Reducing Power: NH3<PH3<AsH3<SbH3<BiH3NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3. BiH3BiH_3 is the strongest reducing agent among the hydrides because its Bi−HBi-H bond is the weakest and cleaves most readily. NH3NH_3 is only a mild reducing agent.

  • Bond Angles: NH3 (107.8∘)>PH3 (93.6∘)>AsH3 (91.8∘)>SbH3 (91.3∘)>BiH3 (90∘)NH_3\,(107.8^\circ) > PH_3\,(93.6^\circ) > AsH_3\,(91.8^\circ) > SbH_3\,(91.3^\circ) > BiH_3\,(90^\circ).

  • Boiling Point Anomaly: PH3<AsH3<NH3<SbH3<BiH3PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3. Ammonia (NH3NH_3) has an abnormally high boiling point compared to PH3PH_3 and AsH3AsH_3 due to extensive intermolecular hydrogen bonding.

Group 15 Oxides and Halides

  • Oxides: Elements form trioxides (E2O3E_2O_3) and pentoxides (E2O5E_2O_5).

    • The oxide in the higher oxidation state is more acidic than that in the lower oxidation state: E2O5>E2O3E_2O_5 > E_2O_3.

    • Acidic character decreases down the group:

    • N2O3,P2O3N_2O_3, P_2O_3: Purely acidic.

    • As2O3,Sb2O3As_2O_3, Sb_2O_3: Amphoteric.

    • Bi2O3Bi_2O_3: Predominantly basic.

  • Halides: Elements form trihalides (EX3EX_3) and pentahalides (EX5EX_5).

    • According to Fajans' rules, higher oxidation state halides (EX5EX_5) are more covalent than lower oxidation state halides (EX3EX_3) because cations with a +5+5 state have higher polarising power.

    • Nitrogen cannot form pentahalides due to the absence of vacant dd-orbitals.

    • All trihalides of Group 1515 elements (except BiF3BiF_3, which is predominantly ionic) are covalent. All trihalides of nitrogen are unstable except NF3NF_3.


Group 16 Elements: The Oxygen Family (Chalcogens)

Group 1616 comprises Oxygen (OO), Sulphur (SS), Selenium (SeSe), Tellurium (TeTe), Polonium (PoPo), and Livermorium (LvLv). They are termed chalcogens ("ore-forming") because copper and other metal ores occur predominantly as oxides and sulphides.

Electronic Configuration and Atomic Numbers

  • Oxygen (OO): Z=8Z = 8; Configuration: [He] 2s2 2p4[He]\,2s^2\,2p^4

  • Sulphur (SS): Z=16Z = 16; Configuration: [Ne] 3s2 3p4[Ne]\,3s^2\,3p^4

  • Selenium (SeSe): Z=34Z = 34; Configuration: [Ar] 3d10 4s2 4p4[Ar]\,3d^{10}\,4s^2\,4p^4

  • Tellurium (TeTe): Z=52Z = 52; Configuration: [Kr] 4d10 5s2 5p4[Kr]\,4d^{10}\,5s^2\,5p^4

  • Polonium (PoPo): Z=84Z = 84; Configuration: [Xe] 4f14 5d10 6s2 6p4[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^4

  • Livermorium (LvLv): Z=116Z = 116

The general valence shell electronic configuration is ns2 np4ns^2\,np^4.

Occurrence

  • Oxygen: Most abundant element on Earth (46.6%46.6\% by mass of the earth's crust; dry air contains 20.946%20.946\% oxygen by volume).

  • Sulphur: Abundance in the earth's crust is 0.03−0.1%0.03 - 0.1\%.

    • Combined as sulphates: Gypsum (CaSO4⋅2H2OCaSO_4 \cdot 2H_2O), Epsom salt (MgSO4⋅7H2OMgSO_4 \cdot 7H_2O), Baryte (BaSO4BaSO_4).

    • Combined as sulphides: Galena (PbSPbS), Zinc blende (ZnSZnS), Copper pyrites (CuFeS2CuFeS_2).

    • Traces of H2SH_2S occur in volcanic gases. Organic materials (eggs, proteins, garlic, onion, mustard, hair, wool) contain sulphur.

  • Selenium & Tellurium: Found as metal selenides and tellurides in sulphide ores.

  • Polonium: Occurs in nature as a radioactive decay product of thorium and uranium minerals; half-life of 210Po^{210}Po is 13.8 days13.8\text{ days}.

Periodic Trends and Atomic Properties

  1. Atomic and Covalent Radii: Increase regularly down the group: O<S<Se<Te<PoO < S < Se < Te < Po.

  2. Ionisation Enthalpy: Decreases down the group: O>S>Se>Te>PoO > S > Se > Te > Po. Group 1616 elements have lower first ionisation enthalpies than corresponding Group 1515 elements because removing one electron yields a stable half-filled p3p^3 configuration.

  3. Electronegativity: Decreases down the group (O (3.50)>S (2.58)>Se (2.55)>Te (2.01)>Po (1.76)O\,(3.50) > S\,(2.58) > Se\,(2.55) > Te\,(2.01) > Po\,(1.76)). Oxygen is the second most electronegative element after fluorine.

  4. Electron Gain Enthalpy (ΔegH\Delta_{eg} H):

    • Experimental values: S(−200)>Se(−195)>Te(−190)>Po(−174)>O(−141 kJ mol−1)S (-200) > Se (-195) > Te (-190) > Po (-174) > O (-141\,kJ\,mol^{-1}).

    • Crucial Anomaly: Oxygen has a less negative electron gain enthalpy than sulphur. This is due to the very small size of the oxygen atom, which causes high interelectronic repulsions in the compact 2p2p-subshell upon electron addition.

  5. Melting and Boiling Points: Increase down the group from OO to TeTe. The large difference between the melting and boiling points of oxygen and sulphur is explained on the basis of their atomicity: oxygen exists as a discrete diatomic molecule (O2O_2), whereas sulphur exists as an octatomic polyatomic molecule (S8S_8).

  6. Allotropy: All elements of Group 1616 exhibit allotropy.

    • Sulphur Allotropes: Rhombic (α\alpha-sulphur) and Monoclinic (β\beta-sulphur). The transition temperature is 369 K369\,K, below which α\alpha is stable and above which β\beta is stable. Both contain S8S_8 puckered crown rings.

    • At elevated temperatures (≈1000 K\approx 1000\,K), sulphur exists primarily as S2S_2 vapour, which is paramagnetic like O2O_2 because it possesses two unpaired electrons in antibonding π∗\pi^* orbitals.

Chemical Properties

  • Oxidation States: Common states are −2,+2,+4,+6-2, +2, +4, +6.

    • Oxygen shows −2-2 predominantly, but exhibits +2+2 in OF2OF_2, +1+1 in O2F2O_2F_2, −1-1 in peroxides (H2O2H_2O_2), and −1/2-1/2 in superoxides (KO2KO_2).

    • The stability of the +6+6 oxidation state decreases down the group, while the stability of the +4+4 oxidation state increases down the group due to the inert pair effect.

  • Hydrides (H2EH_2E):

    • Acidic Strength: H2O<H2S<H2Se<H2TeH_2O < H_2S < H_2Se < H_2Te. (Due to a decrease in E−HE-H bond dissociation enthalpy).

    • Thermal Stability: H2O>H2S>H2Se>H2TeH_2O > H_2S > H_2Se > H_2Te.

    • Reducing Power: All hydrides except water are reducing agents. Reducing strength increases down the group: H2S<H2Se<H2TeH_2S < H_2Se < H_2Te.

    • Boiling Point: H2S<H2Se<H2Te<H2OH_2S < H_2Se < H_2Te < H_2O. Water has an abnormally high boiling point due to extensive intermolecular hydrogen bonding.

    • Drago's Rule: For H2S,H2Se,H_2S, H_2Se, and H2TeH_2Te, the central atom has an electronegativity ≤2.5\le 2.5 and bonded to hydrogen; hybridization does not effectively take place. The bond angles remain close to 90∘90^\circ (H2S≈92∘,H2Se≈91∘,H2Te≈90∘H_2S \approx 92^\circ, H_2Se \approx 91^\circ, H_2Te \approx 90^\circ), in contrast to H2OH_2O (104.5∘104.5^\circ, sp3sp^3).

  • Oxides (EO2EO_2 and EO3EO_3):

    • SO2SO_2 is a gas; SeO2SeO_2 and TeO2TeO_2 are crystalline solids.

    • Reducing/Oxidising property: SO2SO_2 is reducing (+4+4 to +6+6), whereas TeO2TeO_2 is an oxidising agent (+4+4 to 00 or +2+2).

    • Acidic strength: EO3>EO2EO_3 > EO_2. Acidic nature decreases down the group (SO2>SeO2>TeO2SO_2 > SeO_2 > TeO_2).

  • Halides:

    • Hexahalides (EX6EX_6): Only hexafluorides are stable; all are gaseous with octahedral geometry (sp3d2sp^3d^2). SF6SF_6 is exceptionally stable due to steric protection around sulphur.

    • Tetrahalides (EX4EX_4): SF4SF_4 is a gas, SeF4SeF_4 is a liquid, TeF4TeF_4 is a solid. They have a see-saw shape (sp3dsp^3d hybridisation with one equatorial lone pair).

    • Monohalides: Exist as dimers (S2F2,S2Cl2,S2Br2,Se2Cl2,Se2Br2S_2F_2, S_2Cl_2, S_2Br_2, Se_2Cl_2, Se_2Br_2). Dimeric monohalides undergo disproportionation:     2Se2Cl2→SeCl4+3Se2Se_2Cl_2 \rightarrow SeCl_4 + 3Se


Group 17 Elements: The Halogens

Group 1717 contains Fluorine (FF), Chlorine (ClCl), Bromine (BrBr), Iodine (II), Astatine (AtAt, radioactive), and Tennessine (TsTs). The name halogens is derived from Greek words meaning "salt producers".

Electronic Configuration and Atomic Numbers

  • Fluorine (FF): Z=9Z = 9; Configuration: [He] 2s2 2p5[He]\,2s^2\,2p^5

  • Chlorine (ClCl): Z=17Z = 17; Configuration: [Ne] 3s2 3p5[Ne]\,3s^2\,3p^5

  • Bromine (BrBr): Z=35Z = 35; Configuration: [Ar] 3d10 4s2 4p5[Ar]\,3d^{10}\,4s^2\,4p^5

  • Iodine (II): Z=53Z = 53; Configuration: [Kr] 4d10 5s2 5p5[Kr]\,4d^{10}\,5s^2\,5p^5

  • Astatine (AtAt): Z=85Z = 85; Configuration: [Xe] 4f14 5d10 6s2 6p5[Xe]\,4f^{14}\,5d^{10}\,6s^2\,6p^5

  • Tennessine (TsTs): Z=117Z = 117

General valence shell electronic configuration: ns2 np5ns^2\,np^5.

Occurrence

  • Fluorine: Occurs mainly as insoluble fluorides: Fluorspar (CaF2CaF_2), Cryolite (Na3AlF6Na_3AlF_6), and Fluoroapatite (3Ca3(PO4)2⋅CaF23Ca_3(PO_4)_2 \cdot CaF_2). Present in small quantities in soil, river plants, bones, and animal teeth.

  • Chlorine, Bromine, Iodine: Found in seawater as chlorides, bromides, and iodides (2.5%2.5\% NaClNaCl by mass). Dry sea beds contain deposits of carnallite (KCl⋅MgCl2⋅6H2OKCl \cdot MgCl_2 \cdot 6H_2O). Marine life accumulates iodine; seaweeds contain up to 0.5%0.5\% iodine, and Chile saltpetre contains up to 0.2%0.2\% sodium iodate (NaIO3NaIO_3).

Atomic and Physical Properties

  1. Atomic and Ionic Radii: Halogens possess the smallest atomic radii in their respective periods. Radii increase down the group: F<Cl<Br<IF < Cl < Br < I.

  2. Ionisation Enthalpy: Very high, decreasing down the group: F>Cl>Br>IF > Cl > Br > I.

  3. Electronegativity: Highest in their respective periods: F (4.0)>Cl (3.2)>Br (3.0)>I (2.7)F\,(4.0) > Cl\,(3.2) > Br\,(3.0) > I\,(2.7). Fluorine is the most electronegative element in the periodic table.

  4. Electron Gain Enthalpy (ΔegH\Delta_{eg} H):

    • Order: Cl>F>Br>ICl > F > Br > I.

    • Chlorine has a more negative electron gain enthalpy than fluorine (Cl:−349 kJ mol−1Cl: -349\,kJ\,mol^{-1} vs F:−333 kJ mol−1F: -333\,kJ\,mol^{-1}) because the small size of the fluorine atom results in strong interelectronic repulsions in the compact 2p2p-subshell.

  5. Bond Dissociation Enthalpy (X2X_2):

    • Order: Cl2>Br2>F2>I2Cl_2 > Br_2 > F_2 > I_2.

    • F2F_2 has an anomalously low bond dissociation enthalpy (158.8 kJ mol−1158.8\,kJ\,mol^{-1} compared to Cl2:242.6 kJ mol−1Cl_2: 242.6\,kJ\,mol^{-1} and Br2:192.8 kJ mol−1Br_2: 192.8\,kJ\,mol^{-1}) due to large lone pair-lone pair repulsions in the compact F2F_2 molecule.

  6. Standard Reduction Potential and Oxidising Power:

    • Standard electrode potentials: E∘(F2/F−)=+2.87 VE^\circ(F_2/F^-) = +2.87\,V, E∘(Cl2/Cl−)=+1.36 VE^\circ(Cl_2/Cl^-) = +1.36\,V, E∘(Br2/Br−)=+1.09 VE^\circ(Br_2/Br^-) = +1.09\,V, E∘(I2/I−)=+0.54 VE^\circ(I_2/I^-) = +0.54\,V.

    • Oxidising power decreases down the group: F2>Cl2>Br2>I2F_2 > Cl_2 > Br_2 > I_2.

    • F2F_2 is the strongest oxidising agent and displaces all other halide ions from aqueous solution:      F2+2X−→2F−+X2(X=Cl,Br,I)F_2 + 2X^- \rightarrow 2F^- + X_2 \quad (X = Cl, Br, I)      Cl2+2X−→2Cl−+X2(X=Br,I)Cl_2 + 2X^- \rightarrow 2Cl^- + X_2 \quad (X = Br, I)      Br2+2I−→2Br−+I2Br_2 + 2I^- \rightarrow 2Br^- + I_2

  7. Physical State and Colour:

    • F2F_2: Yellow gas.

    • Cl2Cl_2: Greenish-yellow gas.

    • Br2Br_2: Red-brown liquid.

    • I2I_2: Violet solid.

    • Colour arises from the absorption of visible light, causing excitation of outer electrons to higher energy levels.

Chemical Properties

  • Oxidation States:

    • Fluorine shows strictly a −1-1 oxidation state (and lacks variable oxidation states or disproportionation reactions due to the absence of dd-orbitals and highest electronegativity).

    • Chlorine, bromine, and iodine show −1,+1,+3,+5, and +7-1, +1, +3, +5, \text{ and } +7 oxidation states. Compounds in intermediate oxidation states undergo disproportionation (e.g., 3Cl2+6OH−→Δ5Cl−+ClO3−+3H2O3Cl_2 + 6OH^- \xrightarrow{\Delta} 5Cl^- + ClO_3^- + 3H_2O).

  • Hydrogen Halides (HXHX):

    • Acidic Strength & Reducing Power: HF<HCl<HBr<HIHF < HCl < HBr < HI.

    • Thermal Stability: HF>HCl>HBr>HIHF > HCl > HBr > HI.

    • Boiling Point: HCl<HBr<HI<HFHCl < HBr < HI < HF. HFHF has an exceptionally high boiling point due to strong intermolecular hydrogen bonding.

    • Melting Point: HCl<HBr<HF<HIHCl < HBr < HF < HI.

  • Oxides of Halogens:

    • Fluorine: Forms OF2OF_2 and O2F2O_2F_2. O2F2O_2F_2 oxidises plutonium to PuF6PuF_6 in spent nuclear fuel processing.

    • Chlorine: Forms Cl2O,ClO2,Cl2O6,Cl2O7Cl_2O, ClO_2, Cl_2O_6, Cl_2O_7. Chlorine oxides tend to explode; ClO2ClO_2 is used as a bleaching agent for paper pulp, textiles, and water purification.

    • Iodine: I2O5I_2O_5 is a very good oxidising agent and is used in the quantitative estimation of carbon monoxide (COCO).

    • Stability Order of Oxides: I>Cl>BrI > Cl > Br.

Interhalogen Compounds

Interhalogens are formed when two different halogens react. They have the general formula XXn′XX'_n (n=1,3,5,7n = 1, 3, 5, 7), where XX is the larger, less electronegative halogen.

Properties of Interhalogen Compounds
Preparation Conditions
  • Equal volume Cl2+F2→437 K2ClF\text{Equal volume } Cl_2 + F_2 \xrightarrow{437\,K} 2ClF

  • Excess F2+Cl2→573 K2ClF3\text{Excess } F_2 + Cl_2 \xrightarrow{573\,K} 2ClF_3

  • Equimolar I2+Cl2→2ICl\text{Equimolar } I_2 + Cl_2 \rightarrow 2ICl

  • Excess Cl2+I2→2ICl3\text{Excess } Cl_2 + I_2 \rightarrow 2ICl_3

  • Br2+3F2 (diluted with water)→2BrF3Br_2 + 3F_2 \,(\text{diluted with water}) \rightarrow 2BrF_3

  • Excess F2+Br2→2BrF5\text{Excess } F_2 + Br_2 \rightarrow 2BrF_5

Hydrolysis and Shapes (NCERT Table 7.11)
  • Hydrolysis Rule: Gives halide of the smaller halogen and an oxoacid of the larger halogen:   XX′+H2O→HX′+HOXXX' + H_2O \rightarrow HX' + HOX   BrF5+3H2O→5HF+HBrO3(bromic acid / halate)BrF_5 + 3H_2O \rightarrow 5HF + HBrO_3 \quad (\text{bromic acid / halate})

  • Geometries:

    • XX1′XX'_1 (ClF,BrF,ICl,IBrClF, BrF, ICl, IBr): Linear.

    • XX3′XX'_3 (ClF3,BrF3,IF3,ICl3ClF_3, BrF_3, IF_3, ICl_3): Bent T-shaped (sp3dsp^3d hybridisation with 22 lone pairs).

    • XX5′XX'_5 (ClF5,BrF5,IF5ClF_5, BrF_5, IF_5): Square pyramidal (sp3d2sp^3d^2 hybridisation with 11 lone pair). Note: ClF5ClF_5 is a colourless liquid at room temperature.

    • XX7′XX'_7 (IF7IF_7): Pentagonal bipyramidal (sp3d3sp^3d^3 hybridisation with 00 lone pairs).


Group 18 Elements: The Noble Gases

Group 1818 comprises Helium (HeHe), Neon (NeNe), Argon (ArAr), Krypton (KrKr), Xenon (XeXe), Radon (RnRn, radioactive), and Oganesson (OgOg).

General Characteristics

  • Electronic Configuration: ns2 np6ns^2\,np^6 (Helium is 1s21s^2). Possess completely filled shells and high chemical inertness.

  • Atomic Radii: Noble gases exhibit the largest atomic radii in their periods because their radii are expressed as van der Waals radii.

  • Ionisation Enthalpy: Extremely high; decreases down the group from HeHe to RnRn.

  • Electron Gain Enthalpy: Have large positive values because added electrons must enter the next higher principal quantum shell.

  • Melting and Boiling Points: Exceptionally low because monoatomic noble gas atoms are held only by weak dispersion forces. Helium has the lowest boiling point of any known substance (4.2 K4.2\,K).

  • Occurrence: All noble gases except radon occur in the atmosphere (≈1%\approx 1\% by volume, of which argon is the major constituent). Radon is obtained as a radioactive decay product of Radium:   88226Ra→86222Rn+24He^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^4_2He

Bartlett's Discovery of Xenon Compounds

Neil Bartlett observed that platinum hexafluoride (PtF6PtF_6) reacts with molecular oxygen (O2O_2) to form an ionic solid, dioxygenyl hexafluoroplatinate(V): O2+PtF6→O2+[PtF6]−O_2 + PtF_6 \rightarrow O_2^+ [PtF_6]^-

Recognizing that the first ionisation enthalpy of molecular oxygen (1175 kJ mol−11175\,kJ\,mol^{-1}) is almost identical to that of xenon (1170 kJ mol−11170\,kJ\,mol^{-1}), Bartlett reacted xenon with PtF6PtF_6, successfully isolating the first noble gas compound: Xe+[PtF6]−Xe^+ [PtF_6]^- (a red solid).

Xenon Fluorides and Oxides

Synthesis Conditions
  • Xe+F2 (excess Xe)→673 K, 1 barXeF2(s)Xe + F_2 \,(\text{excess } Xe) \xrightarrow{673\,K,\,1\,bar} XeF_2(s)

  • Xe+2F2 (1:5 ratio)→873 K, 7 barXeF4(s)Xe + 2F_2 \,(1:5\text{ ratio}) \xrightarrow{873\,K,\,7\,bar} XeF_4(s)

  • Xe+3F2 (1:20 ratio)→573 K, 60−70 barXeF6(s)Xe + 3F_2 \,(1:20\text{ ratio}) \xrightarrow{573\,K,\,60-70\,bar} XeF_6(s)

  • XeF4+O2F2→143 KXeF6+O2XeF_4 + O_2F_2 \xrightarrow{143\,K} XeF_6 + O_2

Hydrolysis Reactions
  • XeF2XeF_2: Undergoes redox reaction with water:   2XeF2(s)+2H2O(l)→2Xe(g)+4HF(aq)+O2(g)2XeF_2(s) + 2H_2O(l) \rightarrow 2Xe(g) + 4HF(aq) + O_2(g)

  • XeF4XeF_4: Disproportionates in water:   6XeF4+12H2O→4Xe+2XeO3+24HF+3O26XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2

  • Complete Hydrolysis of XeF6XeF_6: Forms xenon trioxide (XeO3XeO_3):   XeF6+3H2O→XeO3+6HFXeF_6 + 3H_2O \rightarrow XeO_3 + 6HF

  • Partial Hydrolysis of XeF6XeF_6:   XeF6+H2O→XeOF4+2HFXeF_6 + H_2O \rightarrow XeOF_4 + 2HF   XeF6+2H2O→XeO2F2+4HFXeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF

Fluoride Ion Donor and Acceptor Reactions
  • Xenon fluorides react with fluoride ion acceptors to form cationic species:   XeF2+PF5→[XeF]+[PF6]−XeF_2 + PF_5 \rightarrow [XeF]^+ [PF_6]^-   XeF4+SbF5→[XeF3]+[SbF6]−XeF_4 + SbF_5 \rightarrow [XeF_3]^+ [SbF_6]^-

  • XeF6XeF_6 acts as a fluoride acceptor with alkali metal fluorides to form fluoroxenate anions:   XeF6+MF→M+[XeF7]−(M=Na,K,Rb,Cs)XeF_6 + MF \rightarrow M^+ [XeF_7]^- \quad (M = Na, K, Rb, Cs)

Geometry and Structures of Xenon Compounds

Molecule

Hybridisation

Bond Pairs

Lone Pairs

Shape / Geometry

XeF2XeF_2

sp3dsp^3d

22

33

Linear

XeF4XeF_4

sp3d2sp^3d^2

44

22

Square planar

XeF6XeF_6

sp3d3sp^3d^3

66

11

Distorted octahedral

XeOF4XeOF_4

sp3d2sp^3d^2

55

11

Square pyramidal

XeO3XeO_3

sp3sp^3

33

11

Pyramidal

XeO4XeO_4

sp3sp^3

44

00

Tetrahedral


Group 13 Elements: The Boron Family

Group 1313 includes Boron (BB), Aluminium (AlAl), Gallium (GaGa), Indium (InIn), and Thallium (TlTl). General electronic configuration: ns2 np1ns^2\,np^1.

Atomic and Physical Properties of Group 13 Elements
Quantitative Data Table (NCERT Table 11.2)

Property

BB

AlAl

GaGa

InIn

TlTl

Atomic Number

55

1313

3131

4949

8181

Atomic Mass (g mol−1g\,mol^{-1})

10.8110.81

26.9826.98

69.7269.72

114.82114.82

204.38204.38

Atomic Radius (pmpm)

8888

143143

135135

167167

170170

Ionic Radius M3+M^{3+} (pmpm)

2727

53.553.5

62.062.0

80.080.0

88.588.5

ΔiH1\Delta_i H_1 (kJ mol−1kJ\,mol^{-1})

801801

577577

579579

558558

589589

ΔiH2\Delta_i H_2 (kJ mol−1kJ\,mol^{-1})

24272427

18161816

19791979

18201820

19711971

ΔiH3\Delta_i H_3 (kJ mol−1kJ\,mol^{-1})

36593659

27442744

29622962

27042704

28772877

Electronegativity

2.02.0

1.51.5

1.61.6

1.71.7

1.81.8

Density (g cm−3g\,cm^{-3} at 298 K298\,K)

2.352.35

2.702.70

5.905.90

7.317.31

11.8511.85

Melting Point (KK)

24532453

933933

303303

430430

576576

Boiling Point (KK)

39233923

27402740

26762676

23532353

17301730

E∘(M3+/M) (V)E^\circ(M^{3+}/M)\,(V)

—

−1.66-1.66

−0.56-0.56

−0.34-0.34

+1.26+1.26

E∘(M+/M) (V)E^\circ(M^+/M)\,(V)

—

+0.55+0.55

−0.79-0.79

−0.18-0.18

−0.34-0.34

Anomalies and Trends
  1. Atomic Radius Anomaly: B (88)<Ga (135)<Al (143)<In (167)<Tl (170 pm)B\,(88) < Ga\,(135) < Al\,(143) < In\,(167) < Tl\,(170\,pm). Gallium has a smaller atomic radius than aluminium due to the poor shielding effect of the ten intervening 3d3d-electrons, leading to a higher effective nuclear charge (ZeffZ_{\text{eff}}).

  2. Ionic Radius (M3+M^{3+}): Follows a normal increasing trend: B3+<Al3+<Ga3+<In3+<Tl3+B^{3+} < Al^{3+} < Ga^{3+} < In^{3+} < Tl^{3+}.

  3. First Ionisation Enthalpy (ΔiH1\Delta_i H_1): Discontinuous trend: B (801)>Tl (589)>Ga (579)>Al (577)>In (558 kJ mol−1)B\,(801) > Tl\,(589) > Ga\,(579) > Al\,(577) > In\,(558\,kJ\,mol^{-1}).

    • Ga>AlGa > Al due to poor shielding of 3d3d-electrons.

    • Tl>In,Ga,AlTl > In, Ga, Al due to the presence of 1414 intervening 4f4f-electrons providing very poor shielding (lanthanoid contraction).

  4. Electronegativity Trend: Decreases from BB to AlAl, then increases marginally down to TlTl: Al (1.5)<Ga (1.6)<In (1.7)<Tl (1.8)<B (2.0)Al\,(1.5) < Ga\,(1.6) < In\,(1.7) < Tl\,(1.8) < B\,(2.0).

  5. Melting and Boiling Points:

    • Boron has an unusually high melting point (2453 K2453\,K) because of its extremely strong three-dimensional covalent lattice.

    • Gallium has an unusually low melting point (303 K303\,K) and exists as a liquid in summer. Its high boiling point (2676 K2676\,K) makes it ideal in high-temperature thermometry.

    • Melting point order: B>Al>Tl>In>GaB > Al > Tl > In > Ga.

  6. Oxidation States & Inert Pair Effect:

    • Group oxidation state is +3+3; lower oxidation state is +1+1.

    • Stability of +3+3 state: B>Al>Ga>In>TlB > Al > Ga > In > Tl.

    • Stability of +1+1 state: B<Al<Ga<In<TlB < Al < Ga < In < Tl.

    • Tl+Tl^+ is significantly more stable than Tl3+Tl^{3+}; hence, Tl3+Tl^{3+} acts as a powerful oxidising agent (E∘=+1.26 VE^\circ = +1.26\,V).

  7. Chemical Reactivity:

    • Oxides: B2O3B_2O_3 is acidic; Al2O3Al_2O_3 and Ga2O3Ga_2O_3 are amphoteric; In2O3In_2O_3 and Tl2O3Tl_2O_3 are basic.

    • Reactions with Acid and Alkali: Aluminium dissolves in dilute mineral acids and aqueous bases (amphoteric):      2Al(s)+6HCl(aq)→2Al3+(aq)+6Cl−(aq)+3H2(g)2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^-(aq) + 3H_2(g)      2Al(s)+2NaOH(aq)+6H2O(l)→2Na+[Al(OH)4]−(aq)+3H2(g)2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na^+ [Al(OH)_4]^-(aq) + 3H_2(g)      Note: Concentrated HNO3HNO_3 passivates AlAl by forming a thin protective oxide film.

    • Halides and Dimerisation: Monomeric trihalides are electron-deficient (66 valence electrons) and act as strong Lewis acids. Relative Lewis acid strength of boron trihalides: BF3<BCl3<BBr3<BI3BF_3 < BCl_3 < BBr_3 < BI_3 (back-bonding efficiency decreases as 2pπ−2pπ>2pπ−3pπ>2pπ−4pπ>2pπ−5pπ2p\pi-2p\pi > 2p\pi-3p\pi > 2p\pi-4p\pi > 2p\pi-5p\pi).

    • AlCl3AlCl_3 dimerises to Al2Cl6Al_2Cl_6 via chlorine bridge bonds. Boron trihalides cannot dimerise due to the small size of boron and the unavailability of dd-orbitals.

Structure of Dimeric Aluminium Chloride Al2Cl6

Group 14 Elements: The Carbon Family

Group 1414 contains Carbon (CC), Silicon (SiSi), Germanium (GeGe), Tin (SnSn), and Lead (PbPb). General electronic configuration: ns2 np2ns^2\,np^2.

Atomic and Physical Properties of Group 14 Elements
Quantitative Data Table (NCERT Table 11.3)

Property

CC

SiSi

GeGe

SnSn

PbPb

Atomic Number

66

1414

3232

5050

8282

Atomic Mass (g mol−1g\,mol^{-1})

12.0112.01

28.0928.09

72.6072.60

118.71118.71

207.2207.2

Covalent Radius (pmpm)

7777

118118

122122

140140

146146

ΔiH1\Delta_i H_1 (kJ mol−1kJ\,mol^{-1})

10861086

786786

761761

708708

715715

Electronegativity

2.52.5

1.81.8

1.81.8

1.81.8

1.91.9

Density (g cm−3g\,cm^{-3} at 298 K298\,K)

3.51 (diam)3.51\,(\text{diam})

2.342.34

5.325.32

7.267.26

11.3411.34

Melting Point (KK)

43734373

16931693

12181218

505505

600600

Boiling Point (KK)

35503550

31233123

28962896

28962896

20242024

Key Trends
  1. Atomic Radii: C<Si<Ge<Sn<PbC < Si < Ge < Sn < Pb. A marked increase occurs from CC to SiSi, followed by a small increase from SiSi to PbPb due to inefficient shielding by dd- and ff-electrons.

  2. Ionisation Enthalpy: Decreases from CC to SnSn, then increases slightly at PbPb: C (1086)>Si (786)>Ge (761)>Pb (715)>Sn (708 kJ mol−1)C\,(1086) > Si\,(786) > Ge\,(761) > Pb\,(715) > Sn\,(708\,kJ\,mol^{-1}). The inversion at PbPb is caused by the poor shielding effect of the 1414 4f4f-electrons (lanthanoid contraction).

  3. Melting and Boiling Points: Group 1414 elements have much higher melting and boiling points than corresponding Group 1313 elements due to very strong covalent network structures. Lowest melting point in Group 1414 is Tin (SnSn, Z=50Z=50, 505 K505\,K).

  4. Electronegativity: Decreases sharply from C (2.5)C\,(2.5) to Si (1.8)Si\,(1.8), remains identical from SiSi to Sn (1.8)Sn\,(1.8), and increases slightly to Pb (1.9)Pb\,(1.9).

  5. Oxidation States:

    • Exhibit +4+4 and +2+2 oxidation states.

    • Stability of +4+4 decreases down the group: C>Si>Ge>Sn>PbC > Si > Ge > Sn > Pb.

    • Stability of +2+2 increases down the group: Ge2+<Sn2+<Pb2+Ge^{2+} < Sn^{2+} < Pb^{2+}.

    • Sn2+Sn^{2+} acts as a reducing agent (Sn2+→Sn4++2e−Sn^{2+} \rightarrow Sn^{4+} + 2e^-), whereas Pb4+Pb^{4+} acts as a strong oxidising agent (Pb4++2e−→Pb2+Pb^{4+} + 2e^- \rightarrow Pb^{2+}) due to the inert pair effect.

  6. Oxides:

    • Monoxides (EOEO): COCO (neutral); SiO,GeOSiO, GeO (acidic); SnO,PbOSnO, PbO (amphoteric).

    • Dioxides (EO2EO_2): CO2,SiO2,GeO2CO_2, SiO_2, GeO_2 (acidic); SnO2,PbO2SnO_2, PbO_2 (amphoteric).

    • Acidic character: EO2>EOEO_2 > EO.

  7. Halides and Complexes:

    • Tetrahalides (EX4EX_4) are predominantly covalent (sp3sp^3, tetrahedral), except SnF4SnF_4 and PbF4PbF_4 which are ionic.

    • PbI4PbI_4 does not exist because Pb4+Pb^{4+} readily oxidises I−I^- to I2I_2.

    • Species like [SiF6]2−,[GeCl6]2−,[Sn(OH)6]2−[SiF_6]^{2-}, [GeCl_6]^{2-}, [Sn(OH)_6]^{2-} exist because central atoms have vacant dd-orbitals. However, [SiCl6]2−[SiCl_6]^{2-} does not exist due to the steric crowding of six large chloride ions around Si4+Si^{4+} and weak interaction between lone pairs of Cl−Cl^- and Si4+Si^{4+}.

    • CCl4CCl_4 resists hydrolysis due to the absence of vacant dd-orbitals. Other tetrahalides undergo rapid hydrolysis:      SiCl4+4H2O→Si(OH)4 (silicic acid)+4HClSiCl_4 + 4H_2O \rightarrow Si(OH)_4\,(\text{silicic acid}) + 4HCl

  8. Catenation and Multiple Bonding:

    • Catenation decreases rapidly down the group: C≫Si>Ge≈SnC \gg Si > Ge \approx Sn. Lead does not exhibit catenation.

    • Bond dissociation enthalpies: C−C (348)>Si−Si (297)>Ge−Ge (260)>Sn−Sn (240 kJ mol−1)C-C\,(348) > Si-Si\,(297) > Ge-Ge\,(260) > Sn-Sn\,(240\,kJ\,mol^{-1}).

    • Carbon forms stable pπ−pπp\pi-p\pi multiple bonds (C=C,C≡C,C=O,C=S,C≡NC=C, C\equiv C, C=O, C=S, C\equiv N), enabling diverse allotropic forms (diamond, graphite, fullerenes).


Exam Practice Questions and Previous Years' Solved Problems

Group 15 Questions
Question 1 [08 April, 2024 (Shift-II)]

Identify the incorrect statements about group 15 elements: (A) Dinitrogen is a diatomic gas which acts like an inert gas at room temperature. (B) The common oxidation states of these elements are −3,+3-3, +3 and +5+5. (C) Nitrogen has unique ability to form pπ−pπp\pi-p\pi multiple bonds. (D) The stability of +5+5 oxidation states increases down the group. (E) Nitrogen shows a maximum covalency of 66.

Choose the correct answer from the options given below: (a) (A), (B), (D) only (b) (A), (C), (E) only (c) (B), (D), (E) only (d) (D) and (E) only

Solution: Statements (A), (B), and (C) are correct. Statement (D) is incorrect because the stability of the +5+5 oxidation state decreases down the group due to the inert pair effect. Statement (E) is incorrect because nitrogen lacks vacant dd-orbitals, restricting its maximum covalency to 44. Therefore, the incorrect statements are (D) and (E). Correct Option: (d)

Question 2 [03 April, 2025 (Shift-I)]

Given below are two statements:

  • Statement I: The N−NN-N single bond is weaker and longer than that of P−PP-P single bond.

  • Statement II: Compounds of group 15 elements in +3+3 oxidation states readily undergo disproportionation reactions.

In the light of above statements, choose the correct answer from the options given below: (a) Statement I is true but statement II is false (b) Both statement I and statement II are false (c) Statement I is false but statement II is true (d) Both statement I and statement II are true

Solution:

  • Statement I is false: The N−NN-N single bond is indeed weaker than the P−PP-P single bond due to strong interelectronic repulsions of non-bonding electrons, but it is shorter (not longer) due to nitrogen's smaller atomic radius (70 pm70\,pm vs 110 pm110\,pm).

  • Statement II is false: Compounds of group 15 in intermediate oxidation states of nitrogen (+1+1 to +4+4) and phosphorus disproportionate, but +3+3 states of heavier members (As,Sb,BiAs, Sb, Bi) are stable and do not readily disproportionate. Correct Option: (b)

Question 3 [22 Jan, 2025 (Shift-II)]

The maximum covalency of a non-metallic group 15 element 'E' with weakest E−EE-E bond is: (a) 44 (b) 55 (c) 33 (d) 66

Solution: The non-metallic elements of Group 15 are Nitrogen (NN) and Phosphorus (PP). Between the two, the single N−NN-N bond is weaker than the single P−PP-P bond. The element with the weakest E−EE-E single bond is Nitrogen (NN). The maximum covalency of Nitrogen is 44. Correct Option: (a)

Question 4 [28 Jan, 2025 (Shift-II)]

A group 15 element forms dπ−dπd\pi-d\pi bond with transition metals. It also forms hydride which is a strongest base among the hydrides of other group members that form dπ−dπd\pi-d\pi bond. The atomic number of the element is _____.

Solution: The elements capable of forming dπ−dπd\pi-d\pi bonds with transition metals must possess vacant dd-orbitals, which includes P,As,Sb,BiP, As, Sb, Bi. Among their hydrides (PH3,AsH3,SbH3,BiH3PH_3, AsH_3, SbH_3, BiH_3), basicity decreases down the group: PH3>AsH3>SbH3>BiH3PH_3 > AsH_3 > SbH_3 > BiH_3. The strongest base among these is PH3PH_3, formed by Phosphorus (PP). The atomic number of Phosphorus is 1515. Answer: 15

Question 5 [29 Jan, 2025 (Shift-II)]

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family: (a) 1012 kJ mol−11012\,kJ\,mol^{-1} (b) 834 kJ mol−1834\,kJ\,mol^{-1} (c) 1402 kJ mol−11402\,kJ\,mol^{-1} (d) 947 kJ mol−1947\,kJ\,mol^{-1}

Solution: The main component of the apatite family is Phosphorus (PP). The first ionisation enthalpies of the first four Group 15 elements are: N=1402 kJ mol−1N = 1402\,kJ\,mol^{-1}, P=1012 kJ mol−1P = 1012\,kJ\,mol^{-1}, As=947 kJ mol−1As = 947\,kJ\,mol^{-1}, Sb=834 kJ mol−1Sb = 834\,kJ\,mol^{-1}. The value for phosphorus is 1012 kJ mol−11012\,kJ\,mol^{-1}. Correct Option: (a)

Question 6 [30 Jan, 2024 (Shift-II)]

Choose the correct statements about the hydrides of group 15 elements: A. The stability of the hydrides decreases in the order NH3>PH3>AsH3>SbH3>BiH3NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3. B. The reducing ability of the hydrides increases in the order NH3<PH3<AsH3<SbH3<BiH3NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3. C. Among the hydrides, NH3NH_3 is strong reducing agent while BiH3BiH_3 is mild reducing agent. D. The basicity of the hydrides increases in the order NH3<PH3<AsH3<SbH3<BiH3NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3.

Choose the most appropriate from the option given below: (a) B and C only (b) C and D only (c) A and B only (d) A and D only

Solution: Statements A and B are correct. Statement C is incorrect because BiH3BiH_3 is the strongest reducing agent and NH3NH_3 is mild. Statement D is incorrect because basicity decreases down the group (NH3>PH3>AsH3>SbH3>BiH3NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3). Correct Option: (c)

Question 7 [01 Feb, 2024 (Shift-I)]

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

  • Assertion (A): PH3PH_3 has lower boiling point than NH3NH_3.

  • Reason (R): In liquid state NH3NH_3 molecules are associated through vander waal's forces, but PH3PH_3 molecules are associated through hydrogen bonding.

Choose the most appropriate answer: (a) Both (A) and (R) are correct and (R) is not the correct explanation of (A) (b) (A) is not correct but (R) is correct (c) Both (A) and (R) are correct but (R) is the correct explanation of (A) (d) (A) is correct but (R) is not correct

Solution: Assertion (A) is correct (NH3NH_3 has a higher boiling point than PH3PH_3). Reason (R) is false because in the liquid state, NH3NH_3 molecules associate via hydrogen bonding, while PH3PH_3 molecules associate through weak van der Waals forces. Correct Option: (d)

Question 8 [01 Feb, 2024 (Shift-II)]

The strongest reducing agent among the following is: (a) NH3NH_3 (b) SbH3SbH_3 (c) BiH3BiH_3 (d) PH3PH_3

Solution: Due to the lowest bond dissociation enthalpy and longest bond length, BiH3BiH_3 cleaves most readily to release hydrogen, making it the strongest reducing agent. Correct Option: (c)

Question 9

Given below are two statements:

  • Statement-I: The pentavalent oxide of group-15 element, E2O5E_2O_5, is less acidic than trivalent oxide, E2O3E_2O_3, of the same element.

  • Statement-II: The acidic character of trivalent oxide of group 15 elements, E2O3E_2O_3, decreases down the group.

Solution:

  • Statement-I is false: The oxide in the higher oxidation state (E2O5E_2O_5) has higher positive oxidation state and polarising power, making it more acidic than E2O3E_2O_3.

  • Statement-II is true: As metallic character increases down the group, the acidic character of E2O3E_2O_3 decreases (N2O3,P2O3N_2O_3, P_2O_3 acidic; As2O3,Sb2O3As_2O_3, Sb_2O_3 amphoteric; Bi2O3Bi_2O_3 basic).


Group 16 Questions
Question 10 [08 April, 2025 (Shift-II)]

Given below are two statements:

  • Statement I: H2SeH_2Se is more acidic than H2TeH_2Te.

  • Statement II: H2SeH_2Se has higher bond enthalpy for dissociation than H2TeH_2Te.

Choose the correct answer: (a) Both statement I and Statement II are false. (b) Both statement I and Statement II are true. (c) Statement I is true but Statement II is false. (d) Statement I is false but Statement II is true.

Solution:

  • Statement I is false: Acidic character increases down the group (H2O<H2S<H2Se<H2TeH_2O < H_2S < H_2Se < H_2Te); hence, H2TeH_2Te is more acidic than H2SeH_2Se.

  • Statement II is true: The Se−HSe-H bond is shorter and stronger than the Te−HTe-H bond, so H2SeH_2Se has a higher bond dissociation enthalpy than H2TeH_2Te. Correct Option: (d)

Question 11 [02 April, 2025 (Shift-II)]

The nature of oxide (TeO2TeO_2) and hydride (TeH2TeH_2) formed by TeTe, respectively are: (a) Oxidising and acidic (b) Reducing and basic (c) Reducing and acidic (d) Oxidising and basic

Solution: In Group 16 dioxides, SO2SO_2 is reducing while TeO2TeO_2 is an oxidising agent. The hydrides of Group 16 are acidic, with TeH2TeH_2 (H2TeH_2Te) being acidic. Correct Option: (a)

Question 12 [24 Jan, 2025 (Shift-I)]

The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of: (a) Atomic size (b) Atomicity (c) Electron gain enthalpy (d) Electronegativity

Solution: Oxygen exists as discrete diatomic molecules (O2O_2), whereas sulphur exists as octatomic polyatomic molecules (S8S_8). This difference in atomicity gives rise to much stronger van der Waals forces in sulphur. Correct Option: (b)

Question 13 [29 Jan, 2024 (Shift-II)]

Anomalous behaviour of oxygen is due to its: (a) large size and high electronegativity (b) small size and low electronegativity (c) small size and high electronegativity (d) large size and low electronegativity

Solution: Anomalous behavior of oxygen arises directly from its small size, high electronegativity, and lack of vacant dd-orbitals. Correct Option: (c)

Question 14 [27 Jan, 2024 (Shift-I)]

Given below are two statements:

  • Statement I: Oxygen being the first member of group 16 exhibits only −2-2 oxidation state.

  • Statement II: Down the group 16 stability of +4+4 oxidation state decreases and +6+6 oxidation state increases.

Choose the most appropriate answer: (a) Statement I is correct but Statement II is incorrect. (b) Both Statement I and Statement II are incorrect. (c) Statement I is incorrect but Statement II is correct. (d) Both Statement I and Statement II are correct.

Solution:

  • Statement I is incorrect because oxygen also exhibits +2+2 in OF2OF_2, +1+1 in O2F2O_2F_2, −1-1 in H2O2H_2O_2, and −1/2-1/2 in KO2KO_2.

  • Statement II is incorrect because down the group, the stability of +4+4 increases and +6+6 decreases due to the inert pair effect. Correct Option: (b)


Group 17 Questions
Question 15 [02 April, 2025 (Shift-I)]

The property/properties that show irregularity in first four elements of group-17 is/are: (A) Covalent radius (B) Electron affinity (C) Ionic radius (D) First ionization energy

Choose the correct answer: (a) B and D only (b) A and C only (c) B only (d) A, B, C and D

Solution: Covalent radius, ionic radius, and first ionization energy show regular periodic trends (F<Cl<Br<IF < Cl < Br < I for radii; F>Cl>Br>IF > Cl > Br > I for ionization energy). Only electron affinity (electron gain enthalpy) shows an irregularity, where Chlorine has higher electron affinity than Fluorine (Cl>F>Br>ICl > F > Br > I). Correct Option: (c)

Question 16 [08 April, 2024 (Shift-I)]

Among the following halogens F2,Cl2,Br2F_2, Cl_2, Br_2 and I2I_2, which can undergo disproportionation reaction? (a) Only I2I_2 (b) Cl2,Br2Cl_2, Br_2 and I2I_2 (c) F2,Cl2F_2, Cl_2 and Br2Br_2 (d) F2F_2 and Cl2Cl_2

Solution: Fluorine (F2F_2) is the most electronegative element and exhibits only a −1-1 oxidation state in its compounds; thus, it cannot be oxidised to positive oxidation states and cannot undergo disproportionation. Chlorine, bromine, and iodine display multiple oxidation states and readily disproportionate in alkaline solutions. Correct Option: (b)

Question 17 [09 April, 2024 (Shift-I)]

The F−F^- ions make the enamel on teeth much harder by converting hydroxyapatite into much harder fluoroapatite having the formula: (a) [3(Ca3(PO4)2)⋅CaF2][3(Ca_3(PO_4)_2) \cdot CaF_2] (b) [3(Ca2(PO4)2)⋅Ca(OH)2][3(Ca_2(PO_4)_2) \cdot Ca(OH)_2] (c) [3(Ca3(PO4)3)⋅CaF2][3(Ca_3(PO_4)_3) \cdot CaF_2] (d) [3(Ca3(PO4)2)⋅Ca(OH)2][3(Ca_3(PO_4)_2) \cdot Ca(OH)_2]

Solution: Hydroxyapatite is 3Ca3(PO4)2⋅Ca(OH)23Ca_3(PO_4)_2 \cdot Ca(OH)_2. Reaction with F−F^- replaces the hydroxide group to form fluoroapatite, 3Ca3(PO4)2⋅CaF23Ca_3(PO_4)_2 \cdot CaF_2. Correct Option: (a)

Question 18 [27 Jan, 2024 (Shift-I)]

Element not showing variable oxidation state is: (a) Bromine (b) Iodine (c) Chlorine (d) Fluorine

Solution: Fluorine lacks dd-orbitals and is the most electronegative element, so it exhibits only a −1-1 oxidation state in its compounds. Correct Option: (d)

Question 19 [29 Jan, 2024 (Shift-I)]

Identify the incorrect pair from the following: (a) Fluorspar — BF3BF_3 (b) Cryolite — Na3AlF6Na_3AlF_6 (c) Fluoroapatite — 3Ca3(PO4)2⋅CaF23Ca_3(PO_4)_2 \cdot CaF_2 (d) Carnallite — KCl⋅MgCl2⋅6H2OKCl \cdot MgCl_2 \cdot 6H_2O

Solution: Fluorspar is calcium fluoride (CaF2CaF_2), not BF3BF_3. Correct Option: (a)

Question 20 [30 Jan, 2024 (Shift-II)]

Reduction potential of ions are given below:

  • ClO4−:E∘=1.19 VClO_4^-: E^\circ = 1.19\,V

  • IO4−:E∘=1.65 VIO_4^-: E^\circ = 1.65\,V

  • BrO4−:E∘=1.74 VBrO_4^-: E^\circ = 1.74\,V

The correct order of their oxidising power is: (a) ClO4−>IO4−>BrO4−ClO_4^- > IO_4^- > BrO_4^- (b) BrO4−>IO4−>ClO4−BrO_4^- > IO_4^- > ClO_4^- (c) BrO4−>ClO4−>IO4−BrO_4^- > ClO_4^- > IO_4^- (d) IO4−>BrO4−>ClO4−IO_4^- > BrO_4^- > ClO_4^-

Solution: Higher reduction potential corresponds directly to greater oxidising power. Since E∘(BrO4−)=1.74 V>E∘(IO4−)=1.65 V>E∘(ClO4−)=1.19 VE^\circ(BrO_4^-) = 1.74\,V > E^\circ(IO_4^-) = 1.65\,V > E^\circ(ClO_4^-) = 1.19\,V, the order is BrO4−>IO4−>ClO4−BrO_4^- > IO_4^- > ClO_4^-. Correct Option: (b)

Question 21 [29 Jan, 2023 (Shift-I)]

The bond dissociation energy is highest for: (a) Cl2Cl_2 (b) I2I_2 (c) Br2Br_2 (d) F2F_2

Solution: Bond dissociation energies of the halogens follow the order: Cl2 (242.6 kJ mol−1)>Br2 (192.8 kJ mol−1)>F2 (158.8 kJ mol−1)>I2 (151.1 kJ mol−1)Cl_2\,(242.6\,kJ\,mol^{-1}) > Br_2\,(192.8\,kJ\,mol^{-1}) > F_2\,(158.8\,kJ\,mol^{-1}) > I_2\,(151.1\,kJ\,mol^{-1}). Chlorine has the highest value. Correct Option: (a)

Question 22 [01 Feb, 2023 (Shift-I)]

Given below are two statements:

  • Statement-I: Chlorine can easily combine with oxygen to form oxides, and the product has a tendency to explode.

  • Statement-II: Chemical reactivity of an element can be determined by its reaction with oxygen and halogens.

Choose the correct answer: (a) Both the statements-I and Statement-II are true (b) Statement-I is true but Statement-II is false (c) Statement-I is false but Statement-II is true (d) Both the Statements-I and II are false

Solution: Both statements are true. Chlorine oxides (Cl2O,ClO2,Cl2O6,Cl2O7Cl_2O, ClO_2, Cl_2O_6, Cl_2O_7) are highly reactive and tend to explode violently. Standard chemical reactivity of elements is generally assessed by reaction with electronegative elements like oxygen and halogens. Correct Option: (a)

Question 23 [13 April, 2023 (Shift-I)]

ClF5ClF_5 at room temperature is a: (a) Colourless gas with trigonal bipyramidal geometry. (b) Colourless gas with square pyramidal geometry. (c) Colourless liquid with square pyramidal geometry. (d) Colourless liquid with trigonal bipyramidal geometry.

Solution: According to NCERT Table 7.11, ClF5ClF_5 is a colourless liquid with a square pyramidal shape (sp3d2sp^3d^2 hybridisation with 11 lone pair). Correct Option: (c)

Question 24 [13 April, 2023 (Shift-II)]

The correct group of halide ions which can be oxidised by oxygen in acidic medium is: (a) Br−Br^- only (b) Cl−,Br− and I−Cl^-, Br^-\text{ and } I^- only (c) Br− and I−Br^-\text{ and } I^- only (d) I−I^- only

Solution: In acidic medium, oxygen (E∘=+1.23 VE^\circ = +1.23\,V) can only oxidise iodide ions (E∘(I2/I−)=+0.54 VE^\circ(I_2/I^-) = +0.54\,V) spontaneously: 4I−(aq)+4H+(aq)+O2(g)→2I2(s)+2H2O(l)4I^-(aq) + 4H^+(aq) + O_2(g) \rightarrow 2I_2(s) + 2H_2O(l) Oxygen cannot oxidise bromide (E∘=+1.09 VE^\circ = +1.09\,V under standard conditions kinetic barriers prevent it) or chloride (E∘=+1.36 VE^\circ = +1.36\,V). Correct Option: (d)

Question 25 [27 June, 2022 (Shift-II)]

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

  • Assertion A: Fluorine forms one oxoacid.

  • Reason R: Fluorine has smallest size amongst all halogens and is highly electronegative.

Choose the most appropriate answer: (a) Both A and R are correct and R is the correct explanation of A. (b) Both A and R are correct but R is NOT the correct explanation of A. (c) A is correct but R is not correct. (d) A is not correct but R is correct.

Solution: Assertion A is correct (HOFHOF, hypofluorous acid, is the only oxoacid of fluorine). Reason R is also correct: because of its small size, high electronegativity, and lack of dd-orbitals, fluorine cannot act as a central atom surrounded by oxygen or form higher oxoacids. Correct Option: (a)

Question 26 [25 July, 2022 (Shift-I)]

The interhalogen compound formed from the reaction of bromine with excess of fluorine is: (a) Hypohalite (b) Halate (c) Perhalate (d) Halite

Solution: Reaction of bromine with excess fluorine produces bromine pentafluoride (BrF5BrF_5): Br2+5F2 (excess)→2BrF5Br_2 + 5F_2 \,(\text{excess}) \rightarrow 2BrF_5 Hydrolysis of BrF5BrF_5 produces bromic acid (HBrO3HBrO_3), whose salts are halates (bromates): BrF5+3H2O→5HF+HBrO3BrF_5 + 3H_2O \rightarrow 5HF + HBrO_3 Correct Option: (b)


Group 13 Questions
Question 27 [07 April, 2025 (Shift-II)]

The correct statements from the following are: (A) Tl3+Tl^{3+} is a powerful oxidising agent (B) Al3+Al^{3+} does not get reduced easily (C) Both Al3+Al^{3+} and Tl3+Tl^{3+} are very stable in solution (D) Tl+Tl^+ is more stable than Tl3+Tl^{3+} (E) Al3+Al^{3+} and Tl+Tl^+ are highly stable

Choose the correct answer: (a) (A), (B), (C), (D) and (E) (b) (A), (B), (D) and (E) only (c) (B), (D) and (E) only (d) (A), (C) and (D) only

Solution: Due to the inert pair effect, Tl+Tl^+ is much more stable than Tl3+Tl^{3+}, making Tl3+Tl^{3+} an unstable ion that acts as a powerful oxidising agent (E∘=+1.26 VE^\circ = +1.26\,V). Therefore, statement (C) is false because Tl3+Tl^{3+} is not stable in solution. Statements (A), (B), (D), and (E) are all correct. Correct Option: (b)

Question 28 [04 April, 2025 (Shift-II)]

The elements of Group 13 with highest and lowest first ionisation enthalpies are respectively: (a) B & GaB\,\&\,Ga (b) B & TlB\,\&\,Tl (c) Tl & BTl\,\&\,B (d) B & InB\,\&\,In

Solution: The first ionisation enthalpies are: B=801 kJ mol−1B = 801\,kJ\,mol^{-1}, Al=577 kJ mol−1Al = 577\,kJ\,mol^{-1}, Ga=579 kJ mol−1Ga = 579\,kJ\,mol^{-1}, In=558 kJ mol−1In = 558\,kJ\,mol^{-1}, Tl=589 kJ mol−1Tl = 589\,kJ\,mol^{-1}. The highest is Boron (BB) and the lowest is Indium (InIn). Correct Option: (d)

Question 29 [04 April, 2025 (Shift-I)]

Given below are the pairs of group 13 elements showing their relation in terms of atomic radius: (B<Al),(Al<Ga),(Ga<In) and (In<Tl)(B < Al), (Al < Ga), (Ga < In) \text{ and } (In < Tl) Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius (M3+M^{3+}) than the other one. The atomic number of the element (X) is: (a) 3131 (b) 4949 (c) 1313 (d) 8181

Solution: The incorrect pair is (Al<Ga)(Al < Ga) because the atomic radius of AlAl (143 pm143\,pm) is greater than that of GaGa (135 pm135\,pm). Comparing their ionic radii (M3+M^{3+}): Al3+=53.5 pmAl^{3+} = 53.5\,pm and Ga3+=62.0 pmGa^{3+} = 62.0\,pm. The element with higher ionic radius is Gallium (GaGa), which has an atomic number of 3131. Correct Option: (a)

Question 30 [03 April, 2025 (Shift-II)]

The correct orders among the following are:

  • Atomic radius: B<Al<Ga<In<TlB < Al < Ga < In < Tl

  • Electronegativity: Al<Ga<In<Tl<BAl < Ga < In < Tl < B

  • Density: Tl<In<Ga<Al<BTl < In < Ga < Al < B

  • 1st Ionisation Energy: In<Al<Ga<Tl<BIn < Al < Ga < Tl < B

Choose the correct answer: (a) B and D Only (b) A and C Only (c) C and D Only (d) A and B Only

Solution:

  • Atomic radius is incorrect (Ga<AlGa < Al).

  • Electronegativity order is correct (Al (1.5)<Ga (1.6)<In (1.7)<Tl (1.8)<B (2.0)Al\,(1.5) < Ga\,(1.6) < In\,(1.7) < Tl\,(1.8) < B\,(2.0)).

  • Density is incorrect (density increases down the group: B<Al<Ga<In<TlB < Al < Ga < In < Tl).

  • 1st Ionisation Energy is correct (In (558)<Al (577)<Ga (579)<Tl (589)<B (801 kJ mol−1)In\,(558) < Al\,(577) < Ga\,(579) < Tl\,(589) < B\,(801\,kJ\,mol^{-1})). Therefore, statements B and D are correct. Correct Option: (a)

Question 31 [02 April, 2025 (Shift-I)]

Given below are two statements:

  • Statement (I): The metallic radius of AlAl is less than that of GaGa.

  • Statement (II): The ionic radius of Al3+Al^{3+} is less than that of Ga3+Ga^{3+}.

Choose the most appropriate answer: (a) Both Statement I and Statement II are incorrect (b) Statement I is incorrect but Statement II is correct (c) Statement I is correct but Statement II is incorrect (d) Both Statement I and Statement II are correct

Solution:

  • Statement (I) is incorrect because the metallic radius of AlAl (143 pm143\,pm) is greater than that of GaGa (135 pm135\,pm).

  • Statement (II) is correct because the ionic radius of Al3+Al^{3+} (53.5 pm53.5\,pm) is less than that of Ga3+Ga^{3+} (62.0 pm62.0\,pm). Correct Option: (b)

Question 32 [27 Jan, 2024 (Shift-I)]

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

  • Assertion (A): Melting point of Boron (2453 K2453\,K) is unusually high in group 13 elements.

  • Reason (R): Solid Boron has very strong crystalline lattice.

Choose the most appropriate answer: (a) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (b) Both (A) and (R) are correct and (R) is the correct explanation of (A) (c) (A) is true but (R) is false (d) (A) is false but (R) is true

Solution: Boron forms an extremely hard non-metallic solid composed of B12B_{12} icosahedral units in a three-dimensional covalent lattice, resulting in an unusually high melting point (2453 K2453\,K). Correct Option: (b)

Question 33 [04 April, 2024 (Shift-II)]

The correct order of the first ionisation enthalpy is: (a) B>Al>GaB > Al > Ga (b) Ga>Al>BGa > Al > B (c) Al>Ga>TlAl > Ga > Tl (d) Tl>Ga>AlTl > Ga > Al

Solution: The values are: Tl=589 kJ mol−1Tl = 589\,kJ\,mol^{-1}, Ga=579 kJ mol−1Ga = 579\,kJ\,mol^{-1}, and Al=577 kJ mol−1Al = 577\,kJ\,mol^{-1}. Thus, the order is Tl>Ga>AlTl > Ga > Al. Correct Option: (d)

Question 34 [08 April, 2024 (Shift-I)]

Given below are two statements: One is labelled as Assertion (A) and other is labelled as Reason (R).

  • Assertion (A): The stability order of +1+1 oxidation state of Ga,InGa, In and TlTl is Ga<In<TlGa < In < Tl.

  • Reason (R): The inert pair effect stabilises the lower oxidation state down the group.

Choose the correct answer: (a) A is false, but R is true. (b) Both A and R are true but R is not the correct explanation of A. (c) Both A and R are true and R is the correct explanation of A. (d) A is true, but R is false.

Solution: The inert pair effect increases down the group, which increasingly stabilises the +1+1 oxidation state for the heavier elements (Ga+<In+<Tl+Ga^+ < In^+ < Tl^+). Correct Option: (c)

Question 35 [06 April, 2024 (Shift-I)]

Given below are two statements:

  • Statement I: Gallium is used in the manufacturing of thermometers.

  • Statement II: A thermometer containing gallium is useful for measuring the freezing point (256 K256\,K) of brine solution.

Choose the correct answer: (a) Statement I is true but Statement II is false (b) Statement I is false but Statement II is true (c) Both Statement I and Statement II are true (d) Both Statement I and Statement II are false

Solution: Gallium has a melting point of 303 K303\,K (30∘C30^\circ C); below this temperature, gallium freezes solid. Therefore, it cannot be used to measure low temperatures like the freezing point of brine solution (256 K256\,K). It is used for measuring high temperatures up to its boiling point (2676 K2676\,K). Correct Option: (a)

Question 36 [05 April, 2024 (Shift-I)]

Given below are two statements:

  • Statement I: In group 13, the stability of +1+1 oxidation state increases down the group.

  • Statement II: The atomic size of gallium is greater than that of aluminium.

Choose the most appropriate answer: (a) Both Statement I and Statement II are correct. (b) Statement I is incorrect but Statement II is correct. (c) Statement I is correct but Statement II is incorrect. (d) Both Statement I and Statement II are incorrect.

Solution: Statement I is correct (inert pair effect). Statement II is incorrect because the atomic radius of GaGa (135 pm135\,pm) is smaller than that of AlAl (143 pm143\,pm). Correct Option: (c)

Question 37 [31 Jan, 2024 (Shift-II)]

Given below are two statements:

  • Statement I: Group 13 trivalent halides get easily hydrolysed by water due to their covalent nature.

  • Statement II: AlCl3AlCl_3 upon hydrolysis in acidified aqueous solution forms octahedral [Al(H2O)6]3+[Al(H_2O)_6]^{3+} ion.

Choose the correct answer: (a) Both statement I and statement II are false. (b) Both statement I and statement II are true. (c) Statement I is false but statement II is true. (d) Statement I is true but statement II is false.

Solution: Both statements are true. Group 13 halides hydrolyse in water; in acidified aqueous solution, AlCl3AlCl_3 forms the octahedral complex ion [Al(H2O)6]3+[Al(H_2O)_6]^{3+} because aluminium expands its coordination number using vacant 3d3d-orbitals. Correct Option: (b)


Group 14 Questions
Question 38 [28 Jan, 2025 (Shift-I)]

Consider the following elements: In,Tl,Al,Pb,Sn and GeIn, Tl, Al, Pb, Sn\text{ and } Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are: (a) +2 and +3+2\text{ and }+3 (b) +4 and +1+4\text{ and }+1 (c) +1 and +4+1\text{ and }+4 (d) +4 and +3+4\text{ and }+3

Solution: Comparing first ionisation enthalpies among the given elements:

  • Ge:761 kJ mol−1Ge: 761\,kJ\,mol^{-1}

  • Pb:715 kJ mol−1Pb: 715\,kJ\,mol^{-1}

  • Sn:708 kJ mol−1Sn: 708\,kJ\,mol^{-1}

  • Tl:589 kJ mol−1Tl: 589\,kJ\,mol^{-1}

  • Ga:579 kJ mol−1Ga: 579\,kJ\,mol^{-1}

  • Al:577 kJ mol−1Al: 577\,kJ\,mol^{-1}

  • In:558 kJ mol−1In: 558\,kJ\,mol^{-1}

The element with the highest first ionisation enthalpy is Germanium (GeGe, 761 kJ mol−1761\,kJ\,mol^{-1}), whose most stable oxidation state is +4+4. The element with the lowest first ionisation enthalpy is Indium (InIn, 558 kJ mol−1558\,kJ\,mol^{-1}), whose most stable oxidation state is +3+3. Correct Option: (d)

Question 39 [23 Jan, 2025 (Shift-II)]

Given below are the atomic numbers of some group 14 elements. The atomic number of the element with lowest melting point is: (a) 5050 (b) 66 (c) 1414 (d) 8282

Solution: Melting points of Group 14 elements: C (4373 K)>Si (1693 K)>Ge (1218 K)>Pb (600 K)>Sn (505 K)C\,(4373\,K) > Si\,(1693\,K) > Ge\,(1218\,K) > Pb\,(600\,K) > Sn\,(505\,K). The lowest melting point belongs to Tin (SnSn), which has an atomic number of 5050. Correct Option: (a)

Question 40 [04 April, 2025 (Shift-II)]

Given below are two statements:

  • Statement (I): The first ionisation enthalpy of group 14 elements is higher than the corresponding elements of group 13.

  • Statement (II): Melting points and boiling points of group 13 elements are in general much higher than those the corresponding elements of group 14.

Choose the most appropriate answer: (a) Statement I is correct but Statement II is incorrect (b) Statement I is incorrect but Statement II is correct (c) Both Statement I and Statement II are incorrect (d) Both Statement I and Statement II are correct

Solution:

  • Statement (I) is correct because Group 14 elements have a higher effective nuclear charge (ZeffZ_{\text{eff}}) and smaller atomic radii than Group 13 elements.

  • Statement (II) is incorrect because Group 14 elements have much higher melting and boiling points than the corresponding elements of Group 13 due to strong four-coordinate network covalent bonding. Correct Option: (a)

Question 41 [07 April, 2025 (Shift-I)]

The group 14 elements AA and BB have the first ionisation enthalpy values of 708708 and 715 kJ mol−1715\,kJ\,mol^{-1} respectively. The above values are lowest among their group members. The nature of their ions A2+A^{2+} and B4+B^{4+} respectively is: (a) both reducing (b) both oxidising (c) reducing and oxidising (d) oxidising and reducing

Solution: From the NCERT data table for Group 14, element AA (ΔiH1=708 kJ mol−1\Delta_i H_1 = 708\,kJ\,mol^{-1}) is Tin (SnSn), and element BB (ΔiH1=715 kJ mol−1\Delta_i H_1 = 715\,kJ\,mol^{-1}) is Lead (PbPb). For tin, the +4+4 state is more stable than +2+2, so Sn2+Sn^{2+} (A2+A^{2+}) acts as a reducing agent (Sn2+→Sn4++2e−Sn^{2+} \rightarrow Sn^{4+} + 2e^-). For lead, due to the inert pair effect, the +2+2 state is more stable than +4+4, so Pb4+Pb^{4+} (B4+B^{4+}) acts as an oxidising agent (Pb4++2e−→Pb2+Pb^{4+} + 2e^- \rightarrow Pb^{2+}). Correct Option: (c)

Question 42 [06 April, 2024 (Shift-I)]

Which of the following material is not a semiconductor? (a) Silicon (b) Graphite (c) Germanium (d) Copper oxide

Solution: Silicon, germanium, and copper oxide are well-known semiconductors. Graphite has delocalised π\pi-electrons free to move across hexagonal layers, making it a good metallic conductor of electricity. Correct Option: (b)

Question 43 [06 April, 2024 (Shift-II)]

The number of ions from the following that are expected to behave as oxidising agent is: Sn4+,Sn2+,Pb2+,Tl3+,Pb4+,Tl+Sn^{4+}, Sn^{2+}, Pb^{2+}, Tl^{3+}, Pb^{4+}, Tl^+ (a) 11 (b) 33 (c) 44 (d) 22

Solution: An ion behaves as an oxidising agent if it easily gains electrons to attain a more stable lower oxidation state. Down Groups 13 and 14, lower oxidation states are stabilized by the inert pair effect:

  • Tl3+Tl^{3+} is readily reduced to Tl+Tl^+ (E∘=+1.26 VE^\circ = +1.26\,V), behaving as a strong oxidising agent.

  • Pb4+Pb^{4+} is readily reduced to Pb2+Pb^{2+}, behaving as a strong oxidising agent.

  • Sn4+,Sn2+,Pb2+,Sn^{4+}, Sn^{2+}, Pb^{2+}, and Tl+Tl^+ are stable in their states and do not act as oxidising agents under normal conditions (Sn2+Sn^{2+} is reducing). Thus, exactly 22 ions (Tl3+Tl^{3+} and Pb4+Pb^{4+}) behave as oxidising agents. Correct Option: (d)

Question 44 [01 Feb, 2024 (Shift-II)]

Given below are two statements:

  • Statement I: SiO2SiO_2 and GeO2GeO_2 are acidic while SnOSnO and PbOPbO are amphoteric in nature.

  • Statement II: Allotropic forms of carbon are due to property of catenation and pπ−dπp\pi-d\pi bond formation.

Choose the most appropriate answer: (a) Both Statement I and Statement II are false (b) Statement I is true but Statement II is false (c) Statement I is false but Statement II is true (d) Both Statement I and Statement II are true

Solution:

  • Statement I is true: SiO2SiO_2 and GeO2GeO_2 are acidic dioxides, whereas SnOSnO and PbOPbO are amphoteric monoxides.

  • Statement II is false: Carbon forms allotropes through catenation and pπ−pπp\pi-p\pi multiple bond formation. Carbon lacks dd-orbitals entirely and cannot form pπ−dπp\pi-d\pi bonds. Correct Option: (b)

Question 45 [31 Jan, 2024 (Shift-I)]

Consider the oxides of group 14 elements: SiO2,GeO2,SnO2,PbO2,CO and GeOSiO_2, GeO_2, SnO_2, PbO_2, CO \text{ and } GeO. The amphoteric oxides are: (a) SnO2,COSnO_2, CO (b) SnO2,PbO2SnO_2, PbO_2 (c) SiO2,GeO2SiO_2, GeO_2 (d) GeO,GeO2GeO, GeO_2

Solution: COCO is neutral; SiO2,GeO,SiO_2, GeO, and GeO2GeO_2 are acidic. Both SnO2SnO_2 and PbO2PbO_2 are amphoteric. Correct Option: (b)

Question 46 [29 Jan, 2024 (Shift-I)]

Given below are two statements:

  • Statement I: The electronegativity of group 14 elements from SiSi to PbPb, gradually decreases.

  • Statement II: Group 14 contains non-metallic, metallic, as well as metalloid elements.

Choose the most appropriate answer: (a) Statement I is false but Statement II is true. (b) Both Statement I and Statement II are false. (c) Statement I is true but Statement II is false. (d) Both Statement I and Statement II are true.

Solution:

  • Statement I is false: The electronegativity values are: C=2.5C = 2.5, Si=1.8Si = 1.8, Ge=1.8Ge = 1.8, Sn=1.8Sn = 1.8, and Pb=1.9Pb = 1.9. It remains essentially constant from SiSi to SnSn and slightly increases at PbPb, rather than gradually decreasing.

  • Statement II is true: Carbon and silicon are non-metals, germanium is a metalloid, and tin and lead are metals. Correct Option: (a)