CBSE Class 12 Chemistry The p Block Elements Notes and Questions

7. p-Block Elements

Points to remember:
The general valence shell electronic configuration of p-block elements is \(ns^2 np^{1-6}\).

GROUP 15 ELEMENTS:
Group 15 elements: N, P, As, Sb & Bi
General electronic configuration: \(ns^2 np^3\)

Physical Properties:
- Dinitrogen is a diatomic gas while all others are solids.
- N & P are non-metals, As & Sb are metalloids, and Bi is a metal. This is due to decrease in ionization enthalpy and increase in atomic size down the group.
- Electronegativity decreases down the group.

Chemical Properties:
- Common oxidation states: -3, +3 & +5.
- Due to the inert pair effect, the stability of the +5 oxidation state decreases down the group, whereas the stability of the +3 oxidation state increases.
- In the case of nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution, e.g.: \(3HNO_3 \rightarrow H_2O + 2NO\)
- Anomalous behavior of nitrogen: Due to its small size, high electronegativity, high ionization enthalpy, and absence of d-orbitals.
- \(N_2\) has a unique ability to form \(p\pi - p\pi\) multiple bonds, whereas the heavier elements of this group do not form \(p\pi - p\pi\) bonds because their atomic orbitals are so large and diffuse that they cannot undergo effective overlapping.
- Nitrogen exists as a diatomic molecule with a triple bond between the two atoms, whereas other elements form single bonds in their elemental state.
- Nitrogen cannot form \(d\pi - p\pi\) bonds due to the non-availability of d-orbitals, whereas other elements can.

Trends In Properties:
- Stability: \(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3\)
- Bond Dissociation Enthalpy: \(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3\)
- Reducing Character: \(BiH_3 > SbH_3 > AsH_3 > PH_3 > NH_3\)
- Basic Character: \(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3\)
- Acidic Character: \(N_2O_3 > P_2O_3 > As_2O_3 > Sb_2O_3 > Bi_2O_3\)

Dinitrogen:
Preparation:
- Commercial preparation: By the liquefaction and fractional distillation of air.
- Laboratory preparation: By treating an aqueous solution of \(NH_4Cl\) with sodium nitrite.
\(NH_4Cl + NaNO_2 \rightarrow N_2 + 2H_2O + NaCl\)
- Thermal decomposition of ammonium dichromate also gives \(N_2\):
\((NH_4)_2Cr_2O_7 \rightarrow N_2 + 4H_2O + Cr_2O_3\)
- Thermal decomposition of barium or sodium azide gives very pure \(N_2\).

Properties:
At high temperatures, nitrogen combines with metals to form ionic nitrides (e.g., \(Mg_3N_2\)) and with non-metals to form covalent nitrides.

AMMONIA PREPARATION:
- In the laboratory, it is prepared by heating an ammonium salt with \(NaOH\) or lime:
\(2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + 2H_2O + CaCl_2\)
- On a large scale, it is manufactured by Haber's process:
\(N_2 + 3H_2 \rightleftharpoons 2NH_3\)
\(\Delta H^\circ = -46.1 \text{ kJ/mol}\)
According to Le Chatelier's principle, the favorable conditions for the manufacture of \(NH_3\) are:
- Optimum temperature: 700 K
- High pressure: 200 atm
- Catalyst: Iron oxides
- Promoter: \(K_2O\) & \(Al_2O_3\)

PROPERTIES:
- Ammonia is a colorless gas with a pungent odor.
- Highly soluble in water.
- In solid and liquid states, it exists as an associated molecule due to hydrogen bonding, which accounts for the high melting and boiling points of \(NH_3\).
- Trigonal pyramidal shape of \(NH_3\) molecule.
- Aqueous solution of ammonia is weakly basic due to the formation of \(OH^-\) ions:
\(ZnSO_4 + 2NH_4OH \rightarrow Zn(OH)_2 + (NH_4)_2SO_4\)
- Ammonia can form coordinate bonds by donating its lone pair of electrons on nitrogen, thus forming complexes:
\(CuSO_4 + 4NH_3 \rightarrow [Cu(NH_3)_4]SO_4\)

NITRIC ACID
PREPARATION: Ostwald's process - It is based upon catalytic oxidation of ammonia by atmospheric oxygen. The main steps are:
1) \(4NH_3 + 5O_2 \xrightarrow[500\text{ K, 9 bar}]{Pt} 4NO + 6H_2O\)
2) \(2NO + O_2 \rightarrow 2NO_2\)
\(3NO_2 + H_2O \rightarrow 2HNO_3 + NO\)

PROPERTIES:
(i) Concentrated \(HNO_3\) is a strong oxidizing agent and attacks most metals except gold and Pt.
(ii) Cr and Al do not dissolve in \(HNO_3\) because of the formation of a passive film of oxide on their surface.
(iii) It oxidizes non-metals, e.g., \(I_2\) to \(HIO_3\), \(C\) to \(CO_2\), \(S\) to \(H_2SO_4\).
(iv) Brown ring test is used to detect nitrate ions (\(NO_3^-\)).

PHOSPHORUS:
ALLOTROPIC FORMS: White, red, \(\alpha\)-black, and \(\beta\)-black.
White phosphorus is more reactive than red phosphorus because white phosphorus exists as discrete \(P_4\) molecules with angular strain, whereas in red phosphorus, several \(P_4\) molecules are linked to form a polymeric chain.

PHOSPHINE
Preparation: It is prepared in the laboratory by heating white phosphorus with concentrated \(NaOH\) solution in an inert atmosphere of \(CO_2\):
\(P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2\)

Phosphorus Halides:
Phosphorus forms two types of halides: \(PX_3\) and \(PX_5\) (\(X = F, Cl, Br, I\)).
Trihalides have a pyramidal shape, and pentahalides have a trigonal bipyramidal structure.

OXOACIDS OF PHOSPHORUS:
- The acids in +3 oxidation state disproportionate to higher and lower oxidation states:
\(4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3\)
- Acids which contain P-H bonds have strong reducing properties, e.g., \(H_3PO_2\).
- Hydrogen atoms attached to oxygen in P-OH form are ionizable and cause basicity.

GROUP 16 ELEMENTS (CHALCOGENS)
Group 16 elements: O, S, Se, Te, Po
General electronic configuration: \(ns^2 np^4\)

ATOMIC & PHYSICAL PROPERTIES:
- Ionisation enthalpy decreases from oxygen to polonium.
- Oxygen atom has less negative electron gain enthalpy than S because of the compact nature of the oxygen atom. However, from S onwards the value again becomes less negative up to polonium.
- Electronegativity gradually decreases from oxygen to polonium; metallic character increases from oxygen to polonium.
- Oxygen & S are non-metals, selenium and tellurium are metalloids, and Po is a radioactive metal.
- Oxygen is a diatomic gas while S, Se, & Te are octa-atomic (\(S_8, Se_8, Te_8\)) molecules which have a puckered 'ring' structure.

CHEMICAL PROPERTIES:
- Common oxidation states: -2, +2, +4 & +6.
- Due to the inert pair effect, the stability of the +6 state decreases down the group and the stability of the +4 state increases.
- Oxygen exhibits +1 state in \(O_2F_2\) and +2 in \(OF_2\).
- Anomalous behavior of oxygen: due to its small size, high electronegativity, and absence of d-orbitals.

TREND IN PROPERTIES:
- Acidic character: \(H_2O < H_2S < H_2Se < H_2Te\)
- Thermal stability: \(H_2O > H_2S > H_2Se > H_2Te\)
- Reducing character: \(H_2S < H_2Se < H_2Te\)
- Boiling point: \(H_2S < H_2Se < H_2Te < H_2O\)
- Reducing property of dioxides: \(SO_2 > SeO_2 > TeO_2\)
- Stability of halides: \(F > Cl > Br > I\)

HALIDES:
- DIHALIDES: \(sp^3\) hybridisation but angular structure.
- TETRAHALIDES: \(sp^3\) hybridisation - see-saw geometry.
- HEXAHALIDES: \(sp^3d^2\), octahedral \(SF_6\).

DIOXYGEN:
Prepared by heating oxygen-containing salts like chlorates and nitrates:
\(2KClO_3 \xrightarrow{\text{heat}} 2KCl + 3O_2\)
\(2Fe^{3+} + SO_2 + 2H_2O \rightarrow 2Fe^{2+} + SO_4^{2-} + 4H^+\)
\(5SO_2 + 2MnO_4^- + 2H_2O \rightarrow 5SO_4^{2-} + 4H^+ + 2Mn^{2+}\)
\(SO_2\) molecule is angular.

OXIDES:
A binary compound of oxygen with another element is called an oxide. Oxides can be classified on the basis of nature:
- Acidic Oxides: Non-metallic oxides. Aqueous solutions are acids. Neutralize bases to form salts. Ex: \(SO_2, CO_2, N_2O_5\), etc.
- Basic Oxides: Metallic oxides. Aqueous solutions are alkalis. Neutralize acids to form salts. Ex: \(Na_2O, K_2O\), etc.
- Amphoteric Oxides: Some metallic oxides exhibit dual behavior. Neutralize both acids & bases to form salts. Ex: \(Al_2O_3, Sb_2O_3, SnO\), etc.

OZONE
PREPARATION:
Prepared by subjecting cold, dry oxygen to silent electrical discharge:
\(3O_2 \rightarrow 2O_3\)

PROPERTIES:
Due to the ease with which it liberates atoms of nascent oxygen, it acts as a powerful oxidizing agent. For e.g., it oxidizes lead sulphide to lead sulphate and iodide ions to iodine:
\(PbS + 4O_3 \rightarrow PbSO_4 + 4O_2\)

SULPHUR DIOXIDE
PREPARATION:
- Burning of S in air:
\(S + O_2 \rightarrow SO_2\)
- Roasting of sulphide minerals (Iron pyrites):
\(4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2\)
(Zinc blend):
\(2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2\)

PROPERTIES:
- Highly soluble in water to form a solution of sulphurous acid:
\(SO_2 + H_2O \rightarrow H_2SO_3\)
- \(SO_2\) reacts with \(Cl_2\) to form sulphuryl chloride:
\(SO_2 + Cl_2 \rightarrow SO_2Cl_2\)
- It reacts with oxygen to form \(SO_3\) in the presence of \(V_2O_5\) catalyst:
\(2SO_2 + O_2 \rightarrow 2SO_3\)
- Moist \(SO_2\) behaves as a reducing agent. It converts Fe(III) ions to Fe(II) ions & decolorizes acidified potassium permanganate(VII) solution (this serves as a test for the gas).

SULPHURIC ACID
PREPARATION:
It is manufactured by the contact process which involves 3 steps:
1. Burning of S or sulphide ores in air to generate \(SO_2\).
2. Conversion of \(SO_2\) to \(SO_3\) in the presence of \(V_2O_5\) catalyst.
3. Absorption of \(SO_3\) in \(H_2SO_4\) to give oleum.

PROPERTIES:
1. In aqueous solution it ionizes in 2 steps:
\(H_2SO_4 + H_2O \rightarrow H_3O^+ + HSO_4^-\)
\(HSO_4^- + H_2O \rightarrow H_3O^+ + SO_4^{2-}\)
2. It is a strong dehydrating agent, e.g., charring action of sugar:
\(C_{12}H_{22}O_{11} \xrightarrow{H_2SO_4} 12C + 11H_2O\)
3. It is a moderately strong oxidizing agent:
\(Cu + 2H_2SO_4(\text{conc.}) \rightarrow CuSO_4 + SO_2 + 2H_2O\)
\(C + 2H_2SO_4(\text{conc.}) \rightarrow CO_2 + 2SO_2 + 2H_2O\)

GROUP 17 ELEMENTS (HALOGENS)
Group 17 elements: F, Cl, Br, I, At
General electronic configuration: \(ns^2 np^5\)

ATOMIC & PHYSICAL PROPERTIES:
i. Atomic & ionic radii increase from fluorine to iodine.
ii. Ionization enthalpy gradually decreases from fluorine to iodine due to increase in atomic size.
iii. Electron gain enthalpy of fluorine is less negative than that of chlorine. It is due to the small size of fluorine & repulsion between the newly added electron & electrons already present in its small 2p orbital.
iv. Electronegativity decreases from fluorine to iodine. Fluorine is the most electronegative element in the periodic table.
v. The color of halogens is due to absorption of radiations in the visible region which results in the excitation of outer electrons to higher energy levels.
vi. Bond dissociation enthalpy of fluorine is smaller than that of chlorine due to high electron-electron repulsion among the lone pairs in fluorine molecules where they are much closer to each other than in the case of chlorine. The trend: \(Cl-Cl > Br-Br > F-F > I-I\).

CHEMICAL PROPERTIES:
OXIDATION STATES: -1. However, chlorine, bromine & iodine exhibit +1, +3, +5, +7 oxidation states also.
Fluorine forms two oxides: \(OF_2\) and \(O_2F_2\). These are essentially oxygen fluorides because of the higher electronegativity of fluorine than oxygen.
Anomalous behavior of fluorine: due to its small size, highest electronegativity, low F-F bond dissociation enthalpy, and absence of d-orbitals.

TRENDS IN PROPERTIES:
- Oxidizing property: \(F_2 > Cl_2 > Br_2 > I_2\)
- Acidic strength: \(HF < HCl < HBr < HI\)
- Stability & bond dissociation enthalpy: \(HF > HCl > HBr > HI\)
- Stability of oxides of halogens: \(I > Cl > Br\)
- Ionic character of halides: \(MF > MCl > MBr > MI\)

CHLORINE
PREPARATION:
1. \(MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O\)
2. \(4NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2\)
3. \(2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2\)
4. DEACON'S PROCESS:
\(4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O\)
5. By electrolysis of brine solution. \(Cl_2\) is obtained at the anode.

PROPERTIES:
i. With cold and dilute \(NaOH\), \(Cl_2\) produces a mixture of chloride and hypochlorite, but with hot and concentrated alkalis, it gives chloride and chlorate:
\(2NaOH + Cl_2 \rightarrow NaCl + NaOCl + H_2O\)
\(6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O\)
ii. With dry slaked lime it gives bleaching powder:
\(2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OH)_2 \cdot CaCl_2 \cdot 2H_2O\)
iii. It is a powerful bleaching agent; bleaching action is due to oxidation:
\(Cl_2 + H_2O \rightarrow 2HCl + [O]\)
Colored substance \(+ [O] \rightarrow\) colorless substance
iv. Action of concentrated \(H_2SO_4\) on NaCl gives HCl gas:
\(NaCl + H_2SO_4 \xrightarrow{420\text{ K}} NaHSO_4 + HCl\)
3:1 ratio of conc. \(HCl\) & \(HNO_3\) is known as aqua regia & it is used for dissolving noble metals like Au and Pt.

OXOACIDS OF HALOGENS:
Interhalogen compounds are prepared by direct combination of halogens.
Ex: \(ClF, ClF_3, BrF_5, IF_7\)
They are more reactive than halogens because X-X' bonds are weaker than X-X bonds in halogens (except F-F).

GROUP 18 ELEMENTS
Group 18 elements: He, Ne, Ar, Kr, Xe & Rn
General electronic configuration: \(ns^2 np^6\)
Atomic radii: Large as compared to other elements in the period since it corresponds to van der Waals radii.
Inert: Due to complete octet of outermost shell, very high ionization enthalpy, and electron gain enthalpies are almost zero.

The first noble gas compound prepared by Neil Bartlett was \(XePtF_6\). Xenon reacted with \(O_2^+PtF_6^-\) leading to the discovery of \(XePtF_6\) since the first ionization enthalpy of molecular oxygen (\(1175 \text{ kJ mol}^{-1}\)) was almost identical with that of xenon (\(1170 \text{ kJ mol}^{-1}\)).

PROPERTIES:
\(Xe + F_2 \xrightarrow{673\text{ K, 1 bar}} XeF_2\)
\(Xe(g) + 2F_2(g) \xrightarrow{873\text{ K, 7 bar}} XeF_4(s)\)
\(Xe(g) + 3F_2(g) \xrightarrow{573\text{ K, 60-70 bar}} XeF_6(s)\)
\(XeF_6 + MF \rightarrow M^+[XeF_7]^-\)
\(XeF_2 + PF_5 \rightarrow [XeF]^+[PF_6]^-\)
\(XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF\) (partial hydrolysis)

SOLVED QUESTIONS

1 MARK QUESTIONS

Question 1. Ammonia has higher boiling point than phosphine. Why?
Answer: Ammonia forms intermolecular hydrogen bonds due to the high electronegativity and small size of nitrogen, whereas phosphine does not form hydrogen bonds.
In simple words: Ammonia molecules stick together strongly using hydrogen bonds, needing more heat to boil.

Exam Tip: Mention intermolecular hydrogen bonding explicitly to get full marks.

Question 2. Why is \(BiH_3\) the strongest reducing agent amongst all the hydrides of group 15 elements?
Answer: Due to the large size of the bismuth atom, the Bi-H bond length is longest and weakest, making it dissociate easily to release hydrogen.
In simple words: Bismuthene has a weak bond that breaks easily to release hydrogen.

Exam Tip: Relate reducing strength directly to bond dissociation enthalpy.

Question 3. Why does \(PCl_3\) fume in moisture?
Answer: In the presence of atmospheric moisture (\(H_2O\)), \(PCl_3\) undergoes hydrolysis, giving fumes of hydrochloric acid (\(HCl\)).
\(PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl\)
In simple words: It reacts with water vapor in the air to produce sharp HCl gas fumes.

Question 4. What happens when \(H_3PO_3\) is heated?
Answer: It disproportionates to give orthophosphoric acid and phosphine.
\(4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3\)
In simple words: Heating phosphorous acid breaks it into phosphoric acid and phosphine gas.

Exam Tip: Identify this reaction as a classic example of disproportionation.

Question 5. Why is \(H_2S\) acidic and \(H_2O\) is neutral?
Answer: The S-H bond is weaker than the O-H bond because the size of the sulphur atom is bigger than that of the oxygen atom. Hence, \(H_2S\) can dissociate to give \(H^+\) ions in aqueous solution.
In simple words: The sulphur-hydrogen bond is weaker than the oxygen-hydrogen bond, so it releases protons easily.

Exam Tip: Contrast bond lengths and bond dissociation energies of O-H and S-H.

Question 6. Name two poisonous gases which can be prepared from chlorine gas?
Answer: Phosgene (\(COCl_2\)) and tear gas (\(CCl_3NO_2\)).
In simple words: Phosgene and tear gas are toxic chemicals made using chlorine.

Question 7. Name the halogen which does not exhibit positive oxidation state.
Answer: Fluorine, being the most electronegative element, does not show positive oxidation states.
In simple words: Fluorine is too greedy for electrons and never shows a positive charge.

Exam Tip: Emphasize that fluorine is the most electronegative element in the periodic table.

Question 8. Iodine forms \(I_3^-\) but \(F_2\) does not form \(F_3^-\) ions. Why?
Answer: Due to the presence of vacant d-orbitals, \(I_2\) accepts electrons from \(I^-\) ions to form \(I_3^-\) ions, but because of the absence of d-orbitals and small size, \(F_2\) does not accept electrons from \(F^-\) ions to form \(F_3^-\) ions.
In simple words: Iodine has empty d-orbitals to hold extra electrons, while fluorine does not.

Exam Tip: Mention vacant d-orbitals as the primary reason for expanded octets in heavier halogens.

Question 9. Draw the structure of peroxosulphuric acid.
Answer: Structure of peroxosulphuric acid (\(H_2SO_5\) or Caro's acid) consists of a central sulphur atom bonded to two -OH groups, one oxo group (\(=O\)), and one peroxy-OH group (\(-O-O-H\)).
In simple words: It has a central sulphur atom linked to a peroxy oxygen-oxygen bond.

Exam Tip: Clearly show the peroxy linkage (\(-O-O-\)) in structures when asked.

Question 10. Phosphorus forms \(PCl_5\) but nitrogen cannot form \(NCl_5\). Why?
Answer: Due to the availability of vacant d-orbitals in phosphorus, it can expand its octet, whereas nitrogen lacks d-orbitals in its valence shell.
In simple words: Phosphorus has empty space in its shell to accommodate five chlorine atoms, but nitrogen does not.

Exam Tip: Relate maximum covalency to the presence or absence of vacant d-orbitals.

2 MARK QUESTIONS (SHORT ANSWER TYPE QUESTIONS)

Question 1. Why is HF acid stored in wax-coated glass bottles?
Answer: This is because HF does not attack wax, but it reacts with glass. It dissolves \(SiO_2\) present in glass, forming hydrofluorosilicic acid:
\(SiO_2 + 6HF \rightarrow H_2SiF_6 + 2H_2O\)
In simple words: HF eats away glass by reacting with silica, so it must be stored in wax-coated containers.

Exam Tip: Include the reaction with silica for complete marks.

Question 2. What is laughing gas? Why is it so called? How is it prepared?
Answer: Nitrous oxide (\(N_2O\)) is called laughing gas because when inhaled, it produces hysterical laughter. It is prepared by gently heating ammonium nitrate:
\(NH_4NO_3 \xrightarrow{\text{heat}} N_2O + 2H_2O\)
In simple words: Nitrous oxide causes uncontrollable laughter and is made by heating ammonium nitrate.

Exam Tip: Ensure the decomposition temperature is controlled to avoid violent explosions.

Question 3. Give reasons for the following:
(i) Conc. \(HNO_3\) turns yellow on exposure to sunlight.
(ii) \(PCl_5\) behaves as an ionic species in solid state.
Answer:
(i) Concentrated \(HNO_3\) decomposes in sunlight to form \(NO_2\), which is brown in colour and dissolves in \(HNO_3\), turning it yellow.
(ii) In the solid state, \(PCl_5\) exists as an ionic compound, specifically as \([PCl_4]^+[PCl_6]^-\).
In simple words: Sunlight breaks nitric acid into brown nitrogen dioxide gas, and solid phosphorus pentachloride splits into ions.

Exam Tip: Note both the ionic formulation and decomposition pathway.

Question 4. What happens when white P is heated with conc. \(NaOH\) solution in an atmosphere of \(CO_2\)? Give equation.
Answer: Phosphine gas is formed along with sodium hypophosphite:
\(P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2\)
In simple words: White phosphorus reacts with caustic soda to release phosphine gas.

Exam Tip: Mention the inert atmosphere of \(CO_2\) to prevent catching fire.

Question 5. How is ozone estimated quantitatively?
Answer: When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating \(O_3\) gas.
In simple words: Ozone liberates iodine from potassium iodide, which is measured by titrating with sodium thiosulphate.

Question 6. Are all the five bonds in \(PCl_5\) molecule equivalent? Justify your answer.
Answer: No. \(PCl_5\) has a trigonal bipyramidal structure. The three equatorial P-Cl bonds are equivalent, while the two axial bonds are different and longer than the equatorial bonds due to greater repulsion.
In simple words: Axial bonds are pushed apart and are longer than equatorial bonds, making them unequal.

Exam Tip: Mention trigonal bipyramidal geometry and bond length differences.

Question 7. \(NO_2\) is colored and readily dimerises. Why?
Answer: \(NO_2\) contains an odd number of valence electrons (17 valence electrons). It behaves as a typical odd-electron molecule. On dimerization, it is converted into a stable \(N_2O_4\) molecule with an even number of electrons, eliminating the odd electron.
In simple words: It has an odd electron making it reactive and colored, so it pairs up into \(N_2O_4\) to become stable.

Question 8. Write the balanced chemical equation for the reaction of \(Cl_2\) with hot and concentrated \(NaOH\). Is this reaction a disproportionation reaction? Justify.
Answer:
Equation:
\(3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O\)
Justification: Yes, chlorine is simultaneously oxidized from 0 to +5 (in \(NaClO_3\)) and reduced from 0 to -1 (in \(NaCl\)).
In simple words: Chlorine is both oxidized and reduced at the same time, which makes it a disproportionation reaction.

Question 9. Account for the following:
(i) \(SF_6\) is less reactive than \(SF_4\).
(ii) Of the noble gases, only xenon forms chemical compounds.
Answer:
(i) In \(SF_6\), there is less steric crowding and repulsion between F atoms than in \(SF_4\), and sulphur is sterically saturated by six fluorine atoms.
(ii) Xenon has a relatively low ionization enthalpy and high polarising power due to its larger atomic size compared to lighter noble gases.
In simple words: Sulfur hexafluoride is tightly packed with fluorines making it unreactive, while xenon is large enough to have its electrons pulled away to form compounds.

Question 10. With what neutral molecule is \(ClO^-\) isoelectronic? Is that molecule a Lewis base?
Answer: \(ClO^-\) has \(17 + 8 + 1 = 26\) electrons. It is isoelectronic with \(ClF\) (chlorine monofluoride, \(17 + 9 = 26\) electrons). Yes, it is a Lewis base due to the presence of lone pairs of electrons on the chlorine and fluorine atoms.
In simple words: It has the same number of electrons as ClF and acts as a Lewis base because of lone pairs.

Question 1.
(i) Why is He used in diving apparatus?
(ii) Noble gases have very low boiling points. Why?
(iii) Why is ICl more reactive than \(I_2\)?
Answer:
(i) Helium has very low solubility in blood even under high pressure, preventing 'bends'.
(ii) Being monoatomic, they have only weak dispersion forces between their atoms.
(iii) The I-Cl bond is weaker than the I-I bond, so it cleaves more easily during reactions.
In simple words: Helium doesn't dissolve in blood under pressure, noble gases have weak forces between atoms, and the iodine-chlorine bond breaks easily.

Exam Tip: Answer each sub-part separately with scientific terms like dispersion forces and bond energy.

Question 2. Complete the following equations:
(i) \(XeF_4 + H_2O \rightarrow\)
(ii) \(Ca_3P_2 + H_2O \rightarrow\)
(iii) \(AgCl_{(s)} + NH_{3(aq)} \rightarrow\)
Answer:
(i) \(6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2\)
(ii) \(Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3\)
(iii) \(AgCl_{(s)} + 2NH_{3(aq)} \rightarrow [Ag(NH_3)_2]Cl_{(aq)}\)
In simple words: These reactions show hydrolysis of xenon fluoride and calcium phosphide, and the complexation of silver chloride.

Question 3.
(i) How is \(XeOF_4\) prepared? Draw its structure.
(ii) When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?
Answer:
(i) Prepared by partial hydrolysis of \(XeF_6\):
\(XeF_6 + H_2O \rightarrow XeOF_4 + 2HF\)
Structure: Square pyramidal.
(ii) Reaction with iron produces hydrogen:
\(Fe + 2HCl \rightarrow FeCl_2 + H_2\)
The liberation of hydrogen prevents the formation of ferric chloride by maintaining a reducing environment.
In simple words: Partial hydrolysis gives \(XeOF_4\), and the hydrogen gas released keeps iron in the ferrous state.

Question 1. Account for the following:
(i) Noble gases form compounds with \(F_2\) and \(O_2\) only.
(ii) Sulphur shows paramagnetic behavior.
(iii) HF is much less volatile than HCl.
(iv) White phosphorous is kept under water.
(v) Ammonia is a stronger base than phosphine.
Answer:
(i) \(F_2\) and \(O_2\) are best oxidizing agents with high electronegativity.
(ii) In the vapour state, sulphur partly exists as \(S_2\) molecules which have two unpaired electrons in the antibonding \(\pi^*\) orbitals, similar to \(O_2\), hence exhibiting paramagnetism.
(iii) HF is associated with strong intermolecular hydrogen bonding.
(iv) The ignition temperature of white phosphorus is very low (303 K). Therefore, on exposure to air, it spontaneously catches fire forming \(P_4O_{10}\). To protect it from air, it is kept under water.
(v) Due to the smaller size of N, the lone pair of electrons on nitrogen is concentrated and readily available for donation compared to larger phosphorus in \(PH_3\).
In simple words: Fluorine and oxygen are strong oxidizers, sulfur vapor has unpaired electrons, HF molecules stick together via hydrogen bonds, white phosphorus catches fire in air, and ammonia's lone pair is more concentrated.

Exam Tip: Address each point using core concepts like hydrogen bonding, electron availability, and molecular orbitals.

Question 2. When conc. \(H_2SO_4\) was added to an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were added into the test tube. On cooling, gas (A) changed into a colourless gas (B).
(a) Identify the gases 'A' and 'B'.
(b) Write the equations for the reactions involved.
Answer:
(a) Gas 'A' is \(NO_2\) (nitrogen dioxide) and gas 'B' is \(N_2O_4\) (dinitrogen tetroxide).
(b) Equations:
\(XNO_3 + H_2SO_4 \rightarrow XHSO_4 + HNO_3\)
\(Cu + 4HNO_3(\text{conc.}) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O\)
\(2NO_2 (\text{on cooling}) \rightarrow N_2O_4\)
In simple words: The brown gas is nitrogen dioxide, which turns into colorless dinitrogen tetroxide when cooled.

Question 3. Arrange the following in the increasing order of the property mentioned:
(i) \(HOCl, HClO_2, HClO_3, HClO_4\) (Acidic strength)
(ii) \(As_2O_3, ClO_2, GeO_3, Ga_2O_3\) (Acidity)
(iii) \(NH_3, PH_3, AsH_3, SbH_3\) (HEH bond angle)
(iv) \(HF, HCl, HBr, HI\) (Acidic strength)
(v) \(MF, MCl, MBr, MI\) (Ionic character)
Answer:
(i) Acidic strength: \(HOCl < HClO_2 < HClO_3 < HClO_4\)
(ii) Acidity: \(Ga_2O_3 < GeO_2 < As_2O_3 < ClO_2\)
(iii) Bond angle: \(SbH_3 < AsH_3 < PH_3 < NH_3\)
(iv) Acidic strength: \(HF < HCl < HBr < HI\)
(v) Ionic character: \(MI < MBr < MCl < MF\)
In simple words: Ordered correctly based on oxidation state, electronegativity differences, and bond sizes.

Exam Tip: Higher oxidation states of central atoms in oxoacids lead to higher acidity.

ASSIGNMENTS

Very short answer type questions:

Question 1. \(PH_3\) has lower boiling point than \(NH_3\). Explain.
Answer: Ammonia forms intermolecular hydrogen bonds due to high nitrogen electronegativity, whereas phosphine molecules are held together only by weak van der Waals forces.
In simple words: Ammonia molecules stick together with hydrogen bonds, while phosphine molecules do not.

Exam Tip: Mention absence of hydrogen bonding in phosphine.

Question 2. Why are halogens coloured?
Answer: Halogens absorb radiations in the visible region, which results in the excitation of outer electrons to higher energy levels.
In simple words: They absorb visible light, making their electrons jump to higher energy levels and display color.

Question 3. What are chalcogens?
Answer: Group 16 elements (oxygen, sulphur, selenium, tellurium, and polonium) are called chalcogens, meaning ore-forming elements.
In simple words: Group 16 elements are called chalcogens because they form ores.

Question 4. Which noble gas is radioactive?
Answer: Radon (\(Rn\)) is a radioactive noble gas.
In simple words: Radon is the radioactive element in Group 18.

Question 5. Explain why fluorine always exhibits an oxidation state of -1 only.
Answer: Fluorine is the most electronegative element in the periodic table and lacks d-orbitals in its valence shell, preventing it from expanding its octet to show positive oxidation states.
In simple words: Fluorine is extremely electronegative and has no empty d-orbitals to show positive states.

Question 6. Which compound led to the discovery of compounds of noble gas?
Answer: Dioxygenyl hexafluoroplatinate (\(O_2^+[PtF_6]^-\)), which led Neil Bartlett to prepare \(XePtF_6\).
In simple words: A platinum-oxygen compound inspired scientists to make the first noble gas compound.

Question 7. Name the most electronegative element.
Answer: Fluorine.
In simple words: Fluorine is the most electronegative element.

Question 8. Why is \(OF_6\) compound not known?
Answer: Due to the small size of oxygen and the absence of vacant d-orbitals in its valence shell, oxygen cannot expand its coordination number to six.
In simple words: Oxygen is too small and lacks empty d-orbitals to hold six fluorine atoms around it.

Question 9. Why is \(N_2\) not particularly reactive?
Answer: Due to the high bond dissociation enthalpy of the triple bond (\(N \equiv N\)) between the two nitrogen atoms.
In simple words: The triple bond between nitrogen atoms is extremely strong and hard to break.

Question 10. Ammonia acts as a ligand. Explain.
Answer: Ammonia has a lone pair of electrons on the nitrogen atom which it can donate to transition metal ions to form coordinate bonds and coordinate complexes.
In simple words: Ammonia donates its lone pair of electrons to metal ions, acting as a ligand.

Question 1. White phosphorous is more reactive than red phosphorous. Explain.
Answer: White phosphorus consists of discrete tetrahedral \(P_4\) molecules with high angular strain (bond angle \(60^\circ\)), making it unstable and reactive, whereas red phosphorus has a polymeric chain structure.
In simple words: White phosphorus has tightly squeezed bonds that break easily, making it reactive.

Question 2. Why do noble gases have comparatively large atomic sizes?
Answer: Because their atomic radii correspond to van der Waals radii, which are measured as half of the interatomic distance in non-bonded closest approach, unlike covalent radii of other elements.
In simple words: Their sizes are measured using van der Waals forces, which span a larger distance because they do not form covalent bonds.

Question 3. Arrange in decreasing order of ionic character: M - F, M - Cl, M - Br, M - I
Answer: \(M - F > M - Cl > M - Br > M - I\)
In simple words: Larger electronegativity difference gives higher ionic character.

Question 4. Phosphinic acid behaves as a monoprotic acid.
Answer: Phosphinic acid (\(H_3PO_2\)) contains only one P-OH bond, meaning only one hydrogen atom is ionizable per molecule.
In simple words: It only has one ionizable hydrogen attached to oxygen.

Question 5. Arrange the following in the order of property indicated:
a) \(As_2O_3, ClO_2, GeO_2, Ga_2O_3\) - Increasing acidity
b) \(H_2O, H_2S, H_2Se, H_2Te\) - Increasing acid strength
Answer:
a) \(Ga_2O_3 < GeO_2 < As_2O_3 < ClO_2\)
b) \(H_2O < H_2S < H_2Se < H_2Te\)
In simple words: Acidity increases across the period for oxides and down the group for hydrides due to weaker bonds.

Question 6. Arrange in decreasing order of bond energy: \(F_2, Cl_2, Br_2, I_2\)
Answer: \(Cl_2 > Br_2 > F_2 > I_2\)
In simple words: Chlorine has the highest bond energy because fluorine's small size creates lone pair repulsion that weakens its bond.

Question 7. Complete the following:
i) \(HNO_3 + P_4O_{10} \rightarrow\)
ii) \(IO_3^- + I^- + H^+ \rightarrow\)
Answer:
i) \(4HNO_3 + P_4O_{10} \rightarrow 4HPO_3 + 2N_2O_5\)
ii) \(IO_3^- + 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O\)
In simple words: Phosphorus pentoxide dehydrates nitric acid to dinitrogen pentoxide, and iodate reacts with iodide in acid to yield iodine.

Question 8. Give the chemical reactions in support of following observations:
a) The +5 oxidation state of Bi is less stable than +3 oxidation state.
b) Sulphur exhibits greater tendency for catenation than selenium.
Answer:
a) \(Bi_2O_5 \rightarrow 2Bi_2O_3 + O_2\) (shows bismuth in +5 decomposes to +3)
b) Sulphur forms stable chains up to \(S_8\) due to strong S-S single bonds compared to weaker Se-Se bonds.
In simple words: Bismuth +5 oxide breaks down to +3 oxide, and sulphur forms longer bonds due to stronger atom-to-atom linkages.

Question 9. How would you account for following?
i) Enthalpy of dissociation of \(F_2\) is much less than that of \(Cl_2\).
ii) Sulphur in vapour state exhibits paramagnetism.
Answer:
i) Due to small size of fluorine, large lone pair - lone pair repulsions weaken the F-F bond.
ii) In vapour state, sulphur exists as \(S_2\) molecules which contain two unpaired electrons in antibonding orbitals.
In simple words: Repulsion between lone pairs weakens fluorine's bond, and sulfur vapor has unpaired electrons like oxygen.

Question 10. Draw structures of following:
a) Peroxomonsulphuric acid \(H_2SO_5\)
b) \(XeF_4\)
Answer:
a) Peroxomonosulphuric acid has a central S atom linked to two -OH groups (one normal, one peroxy), and two oxo groups (\(=O\)).
b) \(XeF_4\) has a square planar geometry with two lone pairs of electrons located axially.
In simple words: Caro's acid has a peroxy linkage, and xenon tetrafluoride has a flat square shape with lone pairs on top and bottom.

Exam Tip: Specify VSEPR geometry for xenon compounds clearly.

Level - III

Question 1. Complete and balance:
i) \(F_2 + H_2O(\text{Cold}) \rightarrow\)
ii) \(BrO_3^- + F_2 + OH^- \rightarrow\)
iii) \(Li + N_2(\text{cold}) \rightarrow\)
iv) \(NH_3 + NaOCl \rightarrow\)
Answer:
i) \(2F_2 + 2H_2O \rightarrow 4HF + O_2\)
ii) \(BrO_3^- + F_2 + 2OH^- \rightarrow BrO_4^- + 2F^- + H_2O\)
iii) \(6Li + N_2 \rightarrow 2Li_3N\)
iv) \(2NH_3 + NaOCl \rightarrow N_2H_4 + NaCl + H_2O\)
In simple words: Fluorine oxidizes water, reacts with bromate, combines with lithium, and converts into hydrazine with sodium hypochlorite.

Question 2. Despite lower electron affinity of \(F_2\), is stronger oxidising agent than \(Cl_2\). Explain.
Answer: The oxidizing power depends on overall energetics which include low bond dissociation enthalpy of \(F-F\) and high hydration enthalpy of \(F^-\) ions.
In simple words: Fluorine easily breaks its bond and forms stable hydrated ions, making it a stronger oxidizer overall.

Question 3. Give reasons:
a) Nitric oxide becomes brown when released in air.
b) \(P_5\) is ionic in nature in the solid state.
Answer:
a) Nitric oxide (\(NO\)) reacts instantly with atmospheric oxygen to form brown nitrogen dioxide (\(NO_2\)): \(2NO + O_2 \rightarrow 2NO_2\).
b) Solid \(PCl_5\) exists as an ionic crystal lattice composed of \([PCl_4]^+[PCl_6]^-\).
In simple words: Colorless NO gas oxidizes in air to form brown \(NO_2\), and solid phosphorus pentachloride consists of ions.

Question 4. Which of the two is more covalent: \(SbCl_3\) or \(SbCl_5\)?
Answer: \(SbCl_5\) is more covalent because the central atom has a higher oxidation state (+5), causing greater polarizing power according to Fajan's rules.
In simple words: Higher charge on the central atom in pentachloride pulls electron clouds tighter, making it more covalent.

Question 5. Addition of \(Cl_2\) to KI solution gives brown colour, but excess of it turns it colourless. Explain.
Answer: Chlorine oxidizes potassium iodide to liberate iodine (\(I_2\)), which imparts a brown colour. In the presence of excess chlorine, iodine is further oxidized to colourless iodic acid (\(HIO_3\)):
\(5Cl_2 + I_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl\)
In simple words: First, brown iodine is released, and then excess chlorine bleaches it into a colorless iodate solution.