CBSE Class 12 Chemistry –Chapter 7 The p-Block Elements- Study Materials

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CBSE Class 12th Chemistry Notes: The p-Block Elements

The p-Block Elements, is an important chapter of CBSE Class 12th Chemistry. This is chapter number 7th of NCERT Class 12th Chemistry textbook. Questions based on this chapter are frequently asked in CBSE Class 12th board examinations.

The main topics covered in these quick notes are:

• Definition of p-block elements

• Nitrogen family

• General properties of nitrogen family

• Anomalous properties of nitrogen

• Reactivity of group 15 elements towards:

o Hydrogen

o Oxygen

o Halogens

o Metals

• Preparation, properties and uses of dinitrogen (N₂)

• Preparation, properties and uses of ammonia (NH₃)

• Oxides of nitrogen

• Preparation, properties and uses of nitric acid (HNO₃)

• Brown ring test

• Allotropes of phosphorus (P)

• Uses of phosphorus

• Preparation and properties phosphine (PH₃)

• Phosphorus halides

The notes of the chapter as follow:

p-block elements

The group number 13 to 18, in which the last electrons or the valence electrons enter in the p-orbital are called the p-block elements. The general electronic configuration of p-block elements is ns² np^{1 ‒ 6}

Nitrogen family

The elements of group 15: nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi) having general electronic configuration

ns² np³, are known as the nitrogen family. The s-orbital in these elements is completely filled and p-orbitals are half-filled, making their electronic configuration extra stable.

General properties of nitrogen family:

Atomic and ionic radii: Covalent and ionic radii increase down the group. There is appreciable increase in covalent radii from N to P. However, from As to Bi only a small increase in covalent radius is observed.due to presence of completely filled d or f-orbitals in heavy elements.

Ionisation enthalpy: Ionisation enthalpy goes on decreasing down the group due to the increase in atomic size. Due to the stable electronic configuration with half filled p-orbital, group 15 elements have higher ionisation energy than group 16 elements. Also due to the smaller size of the elements, the group 15 elements have higher ionisation energy than group 14 elements.

Oxidation states: The common oxidation states are +3, +5, –3. The tendency to show –3 oxidation state decreases down the group due to increased size and hence decreased electronegativity. The stability of +5 oxidation state decreases whereas stability of +3 oxidation state increases due to inert pair effect. Nitrogen and phosphorus with oxidation states from +1 to +4 undergo oxidation as well as reduction in acidic medium. This process is called disproportionation.

For example: 3HNO₂ → HNO₃ + H₂O + 2NO

Anomalous properties of nitrogen

Nitrogen has unique ability to form pπ ‒ pπ multiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O). Thus nitrogen exists as diatomc molecule, N₂ with a triple bond between the two atoms. However, P cannot pπ ‒ pπ from bond, therefore P exists as P₄ ‒ P₄ is highly strained and is chemically reactive while N₂ is chemically inert.

The behaviour of nitrogen differs from rest of the elements due to the following reasons:

(i) It has a small size

(ii) It does not have d – orbitals

(iii) It has high electronegativity

(iv) It has high ionization enthalpy

All the elements of Group 15 form trihydrides, MH₃ having sp³ type hybridization.

Stability: The stability of hydrides decreases down the group due to decrease in bond dissociation energy down the group.

NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃

Boiling point: Boiling point of hydrides (except NH₃) increases down the group as with increase in size due to the Van der Waals forces also increase down the group. Boiling point of NH₃ is highest due to the presence of hydrogen bonding.

PH₃ < AsH₃ < NH₃< SbH₃ < BiH₃

Bond angle: Electronegativity of N is highest. Therefore, the lone pairs will be towards nitrogen and hence more repulsion between bond pairs. Therefore bond angle is the highest. After nitrogen, the electronegativity decreases down the group.

NH₃ (107.8°) > PH3 (93.6°) > AsH₃ (91.8°) ≈ SbH₃ (91.3°) > BiH₃ (90°)

Basicity: Basicity decreases down the group.

NH₃ > PH₃> AsH₃ > SbH₃ > BiH₃.

Reactivity towards oxygen: All the group 15 elements form two types of oxides: trioxides, M₂O₃ and pentaoxides, M₂O₅. The oxide in the higher oxidation state of the element is more acidic than that of lower oxidation state. Their acidic character decreases down the group.

Reactivity towards halogens: Group 15 elements react to form two series of halides: trihalides, MX₃ and pentahalides, MX₅. Trihalides are sp³ hybridised with pyramidal shape whereas pentahalides are sp³d hybridized with trigonal bipyramidal shape. Nitrogen does not form pentahalide due to non-availability of the d– orbitals in its valence shell. All the trihalides of these elements except those of nitrogen are stable. In case of nitrogen, only NF₃ is known to be stable.

Reactivity towards metals: All the group 15 elements react with metals to form binary compounds in –3 oxidation state.

Dinitrogen, N₂

Preparation: N₂is produced commercially by the liquifaction of air followed by fractional distillation.

Liquid N₂ distills out first leaving behind liquid oxygen. In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

NH₄Cl (aq) + NaNO₂ (aq) → N₂(g) + 2H₂O (l) + NaCl (aq)

Small amounts of NO and HNO₃ formed as impurities can be removed by passing the gas through aqueous H₂SO₄containing K₂Cr₂O₇.

(NH₄)₂Cr₂O₇ → N₂ + 4 H₂O + Cr₂O₃

Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide

Ba (N₃)₂ → Ba + 3N₂

Properties: Dinitrogen is a colourless, odourless gas, tasteless and non-toxic gas. It is chemically inert at room temperature due to the presence of triple bond with high bond dissociation energy.

Uses: Some important uses of dinitrogen are as follows:

(i) Liquid N₂ is used as a refrigerant

(ii) It is used to provide inert atmosphere in iron and steel industry

(iii) It is used in the manufacture of HNO₃ and NH₃.

Ammonia, NH₃

Ammonia molecule is trigonal pyramidal with nitrogen atom at the apex. It has 3 bond pairs and 1 lone pair. N is sp³ hybridised.

Preparation: Ammonia (NH₃) is manufactured on the commercial scale by Haber’s process.

Pressure ‒ 200 × 10⁵ Pa

Temperature ‒ 773 K

Catalyst ‒ Iron oxide with small amounts of K₂O and Al₂O₃

In laboratory, ammonia is prepared by reacting NH₄Cl with NaOH.

NH₄Cl + NaOH → NaCl + H₂O + NH₃

Properties: Due to the presence of the lone pair of electrons on the nitrogen atoms, NH₃ is a Lewis base. It can form coordinate covalent bond with the transition metal ion and form complexes, e.g.

Uses: Some important uses of ammonia are:
(i) It is used as a refrigerant
(ii) It is used in the manufacture of nitric acid,
(iii) It is used in the production of nitrogenous fertilisers.

Oxides of nitrogen

Nitrogen forms a total of five oxides from +1 oxidation state to +5 oxidation state. The five oxides of nitrogen are: N₂O, NO, N₂O₃, NO₂ or N₂O₄, N₂O₅.

The following table gives the brief information of various oxides of nitrogen:

Image Source: NCERT Books

Nitric acid, HNO₃

Nitric acid is the most important oxoacid formed by nitrogen. It is a colourless liquid. In the gaseous state, HNO₃ exists as a planar molecule with the structure as shown below:

Preparation: Nitric acid is manufactured by the catalytic oxidation of ammonia in Ostwald process.

Properties: Concentrated HNO₃ is a strong oxidising agent and attacks most metals except noble metals. Cr, Fe and AI do not dissolve in conc. HNO₃ due to the formation of a passive film of oxide on the surface.

The oxidising action of HNO₃ is depends on its concentration and the nature of the reducing agent. The principal product of reduction of HNO₃ is NO when it is dilute but NO₂ when it is concentrated.

For example:

3Cu + 8HNO₃ (dil) → 3Cu(NO₃)₂ + 2NO + 4H₂O

Cu + 4HNO₃ (conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Brown ring test for the detection of nitrates:

Uses: Some important uses of nitric acid are:

(i) HNO₃ is used in the manufacture of fertilisers.

(ii) It is used in the formation of explosives, dynamites, TNT, etc.

(iii) It is also used in the etching of metals.

Phosphorus

Phosphorus is an essential constituent of elements and plants. It has many allotropic forms, the important ones are:

(i) White phosphorus

(ii) Red phosphorus

(iii) Black phosphorus

Properties of white phosphorus:

• Highly toxic and can ignite when exposed to air.

• Waxy, insoluble in H₂O

• More reactive than the other solid phases due to the angular strain in the P₄molecule

• Glows in dark

• Discrete tetrahedral P₄ molecules.

Properties of red phosphorus:

• Prepared by heating white phosphorus at 573K in an inert atmosphere

• Odourless, nonpoisonous and insoluble in H₂O

• Less reactive than white phosphorus

• Polymeric structure consisting of chains of P₄ units linked together

• Higher meltng point and density

Properties of black phosphorus:

• Prepared by heating white phosphorus at 473 K under high pressure.

• Exists in two forms – α black P and β black P

• Thermodynamically most stable, i.e., least reactive

• Has an opaque monoclinic or rhombohedral crystals

Uses: Some important uses of phosphorus are:

(i) Used in the manufacture of fertilisers and food grade phosphates.

(ii) Elemental P is used the manufacture of organo-phosphorus compounds used as pesticides.

Phosphine, PH₃

Phosphine is a highly poisonous, colourless gas and has a smell of rotten fish.

Preparation: Phosphine is prepared by the reaction of calcium phosphide with water or dilute HCl.

Ca₃P₂ + 6H₂O → 2PH₃ + 3Ca(OH)₂

Ca₃P₂ + 6HCl → 2PH₃ + 3CaCl₂

Properties: It is insoluble in water and is a weaker base than ammonia. Like ammonia, it gives phosphonium compounds with acids.

For example: PH₃ + HBr → PH₄Br

PH₃ is non-inflammable when pure but becomes inflammable owing to the presence of P₂H₄ or P₄ vapours.

In water, PH₃ decomposes in the presence of light to give red phosphorus and H₂.

Phosphorus halides

Phosphorus forms two types of halides, PX₃ (X = F, Cl, Br, I) and PX₅ (X = F, Cl, Br).

Phosphorus trichloride, PCl₃

It is a colourless oily liquid.

Preparation:
It is obtained by passing dry chlorine over heated white phosphorus or by the action of thionyl chloride with white phosphorus.

Properties: It has a pyramidal shape, in which phosphorus is sp³ hybridised.

It gets hydrolysed in the presence of moisture.

PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Phosphorus pentachloride, PCl

Preparation:

Phosphorus pentachloride is prepared by the reaction of white phosphorus with excess of dry chlorine or can be prepared by the action of SO₂Cl₂ on phosphorus.

P₄ + 10 Cl₂ → 4 PCl₅

P₄ + 10SO₂Cl₂ → 4 PCl₅ + 10SO₂

Properties:

Oxides of nitrogen

Nitrogen forms a total of five oxides from +1 oxidation state to +5 oxidation state. The five oxides of nitrogen are: N₂O, NO, N₂O₃, NO₂ or N₂O₄, N₂O₅.

The following table gives the brief information of various oxides of nitrogen:

In Part-I we have studied about the Group 15 elements (Nitrogen family), their important compounds like Dinitrogen, Ammonia, Oxides of Nitrogen, Nitric acid, Phosphine, Phosphorus Halides, etc.

In Part-II, we will get acquainted with Group 16 elements. All the these quick notes are based on the latest CBSE syllabus for Class 12th Chemistry.

The main topics covered in these quick notes are:

• Oxygen family or Chalcogens

• General properties of oxygen family

• Anomalous properties of oxygen

• Reactivity of group 16 elements towards:

o Hydrogen

o Oxygen

o Halogens

• Preparation, properties and uses of dioxygen (O₂)

• Simple Oxides

• Preparation, properties and uses of ozone (O₃)

• Depletion of ozone layer

• Preparation, properties and uses of sulphur dioxide (SO₂)

• Oxoacids of sulphur

• Preparation, properties and uses of sulphuric acid (H₂SO₄)

The notes of the chapter are as follow:

Oxygen family

The elements of group 15: oxygen (O), sulphur (S), selenium(Se), tellurium (Te) and polonium (Po) having general electronic configuration ns²np⁴, are known as the oxygen family. All these elements collectively are also known as chalcogens. Polonium is a radioactive element.

General properties of oxygen family

Atomic and ionic radii: Due to increase in the number of shells, atomic and ionic radii increase from top to bottom in the group.

Ionisation enthalpy: Due to the increase in size of the atoms the ionisation enthalpy decreases down the group. IE1of group 16 elements is less than the IE₁ of group 15. This is because group 15 elements have extrastability due to half-filled p-orbitals.

Electron gain enthalpy: Due the compact nature of oxygen, it has less electron gain enthalpy than sulphur. After sulphur, the electron gain enthalpy decreases down the group.

Electronegativity: The electronegativity decreases down the group. This implies that the metallic character increases down the group from oxygen to polonium.

Melting and boiling point: The melting and boiling point increases with increase in atomic number down the group.

Oxidation states: Group 16 elements show ‒2, +2, +4, +6 oxidation states. The stability of ‒2 oxidation state decreases down the group due to increase in atomic size and decrease in electronegativity. O shows only ‒2 oxidation state except when it combines with the most electronegative F with which it shows positive oxidation states. S shows + 6 only with O and F.

Anomalous behaviour of oxygen

Oxygen forms strong hydrogen bonding in H₂O which is not found in H₂S. Also, the maximum covalency of oxygen is four, whereas in a case of other elements of the group, the valence shells can be expanded and covalency exceeds four.

Reasons for the anomalous behaviour of oxygen are:

• Small size and high electronegativity

• Absence of d-orbitals

Reactivity towards hydrogen: All the elements of Group 16 form hydrides of the type H₂E (E = S, Se, Te, Po).

Thermal stability: Thermal stability of group 16 elements decreases down the group.

H₂O > H₂S > H₂Se> H₂Te > H₂Po

This is because the H-E bond length increases down the group, hence the bond dissociation enthalpy decreases down the group.

Acidic nature: Due to the decreasing bond dissociation enthalpy, acidic character of group 16 elements increases down the group.

H₂O < H₂S < H₂Se < H₂Te

Reducing character: The reducing character also decreases down the group due to the decreasing bond dissociation enthalpy.

H₂O < H₂S < H₂Se < H₂Te < H₂Po

Reactivity towards oxygen: All group 16 elements form oxides of the type EO₂and EO₃ Reducing character of dioxides decreases down the group. Acidity also decreases down the group. Besides EO₂ type, sulphur, selenium and tellurium also form EO₃ type oxides. Both types of oxides are acidic in nature.

Reactivity with halogens: Elements of Group 16 form a large number of halides of the type, EX₂ EX₄ and EX₆, where X is a halogen. The stability of halides decreases in the order F⁻ > Cl⁻> Br⁻> I⁻ . This is because E-X bond length increases with increase in size. Among hexa halides, hexafluorides are the most stable because of steric reasons. Dihalides are sp³ hybridised and have tetrahedral geometry. H₂O is a liquid while H₂S is a gas. Because in water due to the small size and high electronegativity of O, strong hydrogen bonding is present there.

Oxygen (O)

Oxygen is the first element of Group 16 with the electronic configuration of 1s² 2s²2p⁴in the ground state. Oxygen has two allotropes: dioxygen (O₂) and trioxygen or ozone (O₃).

Dioxygen (O₂)

Oxygen usually exists in the form of dioxygen.

Preparation:

Dioxygen is prepared in the laboratory by thermal decompositions of oxygen rich compounds such as KClO₃,

Properties:

(i) Oxygen is a colourless, odourless and is a highly reactive tasteless gas.

(ii) Due to the presence of pπ‒ pπ bonding, O₂ is a discrete molecule and intermolecular forces are weak van der Waals forces, hence, O₂ is a gas.

(iii) Dioxygen combines with metals and non-metals to form binary compounds called oxides.

Examples are:

2Ca + O₂ → 2CaO

P₄ + 5O₂ → P₄O₁₀

Uses:

(i) Dioxygen is used in making steel.

(ii) It is used in the production of oxygen containing organic chemicals.

(iii) Dioxygen is also used for sewage treatment, river revival and paper pulp bleaching.

(iv) It is used as an oxidiser in underwater diving and in space shuttles.

Simple Oxides

Oxygen combines with majority of the elements of the periodic table to forms oxides (O^2‒). There are three types of oxides:

(i) Acidic oxides: Oxides of non- metals are usually acidic in nature. For example, SO₂ combines with water to give H₂SO₃, an acid.

SO₂ + H₂O → H₂SO₃

(ii) Basic oxides: Metallic oxides are mostly basic in nature. Basic oxides dissolve in water to give basic solution. For example, CaO combines with water to give Ca(OH)₂, a base.

CaO + H₂O → Ca(OH)₂

(iii) Amphoteric oxides: Some metallic oxides show characteristics of both acidic as well as basic oxides. Such oxides are known as amphoteric oxides. For example, Al₂O₃ reacts with acids as well as alkalies

Al₂O₃ + 6HCl + 9H₂O → 2 [Al(H₂O)₆]³⁺ + 6 Cl^‒

Base Acid

Al₂O₃ + 6NaOH + 3H₂O → 2Na₃[Al(OH)₆]

Acid Base

Ozone (O₃)

Ozone is an allotropic form of oxygen. Ozone has angular structure with a bond angle of about 117^o. Both O = O bonds are of equal bond length due to resonance.

Preparation:

It is formed when dioxygen is irradiated with UV light or silent electric discharge.

3O₂ (g) → 2O₃ (g) ΔH^o = +142 KJ/mol

Properties:

(i) Ozone is a pale blue gas with a characteristic pungent odour.

(ii) Ozone is diamagnetic in nature.

(iii) It is the second most powerful oxidising agent after fluorine. It liberates oxygen gas when acting as an oxidising agent.

2Fe²⁺ + O₃ + 2H⁺ → 2Fe³⁺ + H₂O + O₂

PbS + 4O₂ → PbSO₄ + 4O₂

Depletion of ozone layer:

Thinning of the ozone layer is termed as depletion of ozone layer. The depletion of ozone layer in the stratosphere is caused by the presence of chlorofluoro carbons. CFCs decomposed by UV radiation to produce chlorine which reacts with ozone and this causes a de-crease in the concentration of ozone at a rate faster than its formation from dioxygen. Another cause of depletion of ozone layer is the release of nitrogen oxides into the stratosphere by supersonic jet aeroplanes.

NO + O₃ → NO₂ + O₂

Sulphur (S)

Sulphur exhibits allotropy, two important allotropes of which are:

(a) Yellow Rhombic (α – sulphur)

(b) Monoclinic (β- sulphur)

At 369 K both forms are stable. S₈ puckered shape in both forms and has crown shape.

Sulphur Dioxide (SO₂)

The molecule of SO₂ is angular. It is a resonance hybrid of the two canonical forms:

Preparation:

Sulphur dioxide is formed together with a little (6-8%) sulphur trioxide when sulphur is burnt in air or oxygen:

S (g) + O₂(g) → SO₂(g)

In the laboratory it is readily generated by treating a sulphite with dilute sulphuric acid.

Na₂SO₃ + H₂SO₄ → Na₂SO₄ + H₂O + SO₂

Properties:

Uses:

(i) SO₂ is used in refining petroleum and sugar

(ii) It is used in bleaching wool and silk

(iii) It is also used as a disinfectant and preservative.

(iv) It is used in the manufacture of sulphuric acid, sodium hydrogen sulphite and calcium hydrogen sulphite.

Oxoacids of sulphur

Sulphur forms variety of oxoacids. All oxoacids of sulphur are dibasic.

Sulphur forms a number of oxoacids such as H₂SO₃, H₂S₂O₃, H₂S₂O₄, H₂S₂O₅, H₂S_xO₆ (x = 2 to 5), H₂SO₄, H₂S₂O₇, H₂SO₅, H₂S₂O₈ . Some of these acids are unstable and cannot be isolated. They are known in aqueous solution or in the form of their salts. Structures of some important oxoacids are shown in figure given below

Image Source: NCERT Books

Sulphuric acid (H₂SO₄)

Preparation:

Sulphuric acid is manufactured by the Contact Process as follows:

Properties:

(i) Sulphuric acid is a colourless, dense oily liquid.

(ii) It is dibasic acid or diprotic acid.

(iii) It is a strong dehydrating agent.

(iv) It is a moderately strong oxidizing agent.

Uses:

(i) H₂SO₄ is used in manufacture of fertilisers

(ii) It is also used in petroleum refining, manufacture of pigments, detergents, metallurgical application and storage batteries.

In Part-I and Part-II we have studied about the Group 15 elements (Nitrogen family) and Group 16 elements respectively. In Part-III, you will get acquainted with Group 17 and Group 18 elements. These notes are based on the latest CBSE syllabus for Class 12th Chemistry.

The main topics covered in these quick notes are:

• Halogen family

• General properties of halogen family

• Anomalous properties of fluorine

• Reactivity of halogens towards:

o Hydrogen

o Metals

o Other halogens (Interhalogens)

• Preparation, properties and uses of chlorine (Cl₂)

• Preparation, properties and uses of hydrogen chloride (HCl)

• Oxoacids of halogens

• Noble gases

• General properties of noble gases

• Preparation and properties of xenon fluorine compounds

• Preparation and properties of xenon oxygen compounds

• Uses of noble gases

The notes are as follow:

Group 17 elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I) and astatine (At) are collectively known as halogens and are having the general electronic configuration of ns², np⁵.

Atomic and ionic radii

The halogens have the smallest atomic radii in their respective periods due to maximum effective nuclear charge. Atomic and ionic radii increases down the group due to the addition of a new shell at each step.

Ionisation enthalpy

Due to their small size halogens have little tendency to lose electron. Thus they have very high ionisation enthalpy. Ionisation enthalpy decreases down the group due to the increase in atomic size.

Electron gain enthalpy

Electron gain enthalpy of halogens is very high as they are short of only one electron to attain noble gas configuration. Electron gain enthalpy becomes less negative as we move down the group. However F has less electron gain enthalpy than Cl due to its small size and high electron density.

Electronegativity

Electronegativity decreases down the group. F is the most electronegative element in the periodic table.

Melting and boiling point

The melting and boiling points increases down the group.

Bond dissociation enthalpy

Bond dissociation enthalpy decreases as we move down the group. F₂ has less ΔH_diss. Then Cl₂ due to small size and strong lone pair-lone pair repulsion.

Colour

All halogens exhibit colour due to the absorption of radiations in visible region of light due to which the electrons get excited to higher energy levels.

For example, F₂ has yellow, Cl₂ has greenish yellow, Br₂ has red and I₂ has violet colour.

Oxidation state

The most common oxidation state of halogens is −1. Cl, Br, I also shows positive oxidation states of +1, +3, + 5, + 7. F does not show positive oxidation states due to non-availability of d-orbitals.

Reactivity

All halogens are highly reactive and the reactivity decreases down the group.

Anomalous behaviour of fluorine

Fluorine is anomalous in many properties like, ionisation enthalpy, electronegativity, enthalpy of bond dissociation that are higher than expected from the regular trends among the halogens. Its ionic and covalent radii, melting and boiling points, and electron gain enthalpy is quite lower than expected.
Reasons for the anomalous behaviour of oxygen are:

• Small size and highest electronegativity

• Low F-F bond dissociation enthalpy

• Absence of d-orbitals

Reactivity towards hydrogen

All halogens reacts with H₂ to form hydrogen halides (HX) and the reactivity towards H₂ decreases down the group

Acidic strength

As the size of X increases and the strength of H─X bond decreases down the group, acidic strength dectreses down the group

HF < HCl < HBr < HI

Stability

As the bond dissociation enthalpy decreases down the group so the stability of hydrogen halides also decreases from HF to HI.

HF > HCl > HBr > HI

Boiling point

Due to the increase in size of halogens the van der Waals forces increases down the group resulting in the increase in boiling point from HCl to HI. HF has the highest boiling point due to the presence of strong intermolecular H bonding

HCl < HBr < HI < HF.

Ionic character

Due to the decrease in electronegativity down the group the ionic character of hydrogen halides also decreases down the group.

HF > HCl > HBr > HI

Dipole moment

Due to the decrease in electronegativity down the group the ionic character of hydrogen halides also decreases down the group.

HF > HCl > HBr > HI

Reducing power

As the bond dissociation enthalpy decreases, so it becomes easier to give out the hydrogen atom causing the reducing power to increase from HF to HI.

HF < HCl < HBr < HI

Reactivity towards metals:

Halogens react with metals to form metal halides of the form MX, where M is a monovalent metal.

Ionic character

Due to the decrease in electronegativity down the group the ionic character of metal halides also decreases down the group

MF > MCl > MBr > MI

Reactivity of halogens towards other halogens (Interhalogens):

Binary compounds of two different halogen atoms of general formula XX’_n are called interhalogen compounds where n = 1, 3, 5, or 7. All the interhalogen compounds are covalent in nature.

Some properties of interhalogen compounds are given in the following table:

Image Source: NCERT Textbook

Chlorine (Cl)

Preparation:

Chlorine can be prepared by any of the following processes:

Properties:

• It is a greenish yellow gas with pungent and suffocating odour.

• It is soluble in H₂O

• Reaction with metals and non-metals: Chlorine reacts with a number of metals and non-metals to form chlorides.

For example:

2 Al + 3Cl₂ → 2AlCl₃

S₈ + 4Cl₂ → 4S₂Cl₂

• Reaction with ammonia: When treated with excess ammonia, chlorine gives nitrogen and ammonium chloride whereas when excess chlorine reacts with ammonia, nitrogen trichloride is formed.

8NH₃ + 3Cl₂ → 6NH₄Cl + N₂

Excess

NH₃ + Cl₂ → NCl₃ + 3 HCl

Excess

• Reaction with NaOH: Chlorine reacts differently with cold dilute NaOH and hot concentrated NaOH.

Cl₂ + 2NaOH → NaCl + NaOCl + H₂O

Cold dil.

Cl₂ + 6NaOH → NaCl + NaOCl + H₂O

Hot conc.

Reaction with slaked lime: Cl₂ when treated With dry slaked lime it gives bleaching powder:

2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O

Cl₂ acts as a powerful bleaching agent and its bleaching action is due to its oxidizing nature.

Cl₂ + H₂O → 2HCl + O

Uses:

(i) Chlorine is used for bleaching woodpulp.

(ii) It is used in the extraction of gold and platinum

(iii) It is used in in sterilising drinking water.

(iv) It is used in the manufacture of dyes, drugs and organic compounds like CCl_4, DDT, refrigerants, etc.

Hydrogen chloride (HCl)

Preparation:

It is prepared by heating sodium chloride with concentrated sulphuric acid.

2 NaCl + H₂SO₄ + Heat → Na₂SO₄ + HCl

Properties:

HCl is a colourless gas with pungent odour.

It is extremely soluble in water, HCl + H₂O → H₃O⁺ + Cl^‒

It decomposes salts of weaker acids, Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

When treated with NH₃, it gives white fumes of NH₄Cl, NH₃ + HCl → NH₄Cl

3HCl : 1HNO₃ is called aqua regia, which is used for dissolving noble metals.

Au + 4 H⁺ + NO₃^‒ + 4Cl^‒ → AuCl₄^‒ + NO + 2 H₂O

Uses:

(i) Hydrogen chloride is used in medicine and as a laboratory reagent.

(ii) It is used in the manufacture of chlorine, NH₄Cl and glucose.

Oxoacids of halogens

Fluorine due to its small size and high electronegativity forms only one oxoacid HOF (Hypofluorous acid).

Other halogen form several oxoacids as given in the following table:

Image Source: NCERT Textbook

Group 18, Noble gases

Group 18 elements: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn) having the electronic configuration ns² np⁶_, are named as noble gases. All these are gases and chemically unreactive.

General properties of noble gases

Atomic radii:

Atomic radii of noble gases increases down the group due to the addition of a new shell at each step.

He < Ne < Ar < Kr < Xe < Rn

Ionisation enthalpy:

They have very high ionization enthalpy because of completely filled orbitals. Ionisation enthalpy decreases down the group because of increase in size.

He < Ne < Ar < Kr < Xe < Rn

Electron gain enthalpy:

Because of stable electronic configuration, noble gases have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy.

Melting and boiling point:

Due to the weak dispersion forces they have low melting and boiling point.

Xenon-fluorine compounds

Xenon forms three binary fluorides, XeF₂, XeF₄ and XeF₆ .

Preparation:

Properties:

XeF₂ is linear, XeF₄ is square planar and XeF₆ is distorted octahedral.

XeF₂, XeF₄ and XeF₆ are colourless crystalline solids

They are readily hydrolysed

2XeF₂(s) + 2H₂O(l) → 2Xe(g) + HF(aq) + O₂(g)

They react with fluoride ion acceptors to form cationic species and fluoride ion donors to form fluoroanions.

XeF₂ + PF₅ → [XeF] + [PF₆]^─

XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^─

XeF₆ + MF → M⁺ [XeF₇]^─

[Where, M = Na, K, Rb or Cs]

Xenon-oxygen compounds

Xenon forms some important compounds with oxygen like XeO₃, XeOF₄ XeO₂F₂.

Preparation:

Various xenon-oxygen compounds are prepared as follows:

6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂

XeF₆ + 3H₂O → XeO₃ + 6HF

Partial Hydrolysis XeOF₄

XeF₆ + H₂O → XeOF₄ + 2 HF

Partial Hydrolysis also gives XeO₂F₄

XeF₆ + 2H₂O → XeO₂F₄ + 4HF

Properties:

XeO₃ is a colourless explosive solid having a trigonal pyramidal structure.

XeOF₄ is a colourless volatile liquid with a square pyramidal

Uses of inert gases

Helium is used:

• Gas cooled Nuclear reactors

• In filling balloons for meteorological observations.

• In the oxygen mixture of deep sea divers

• In inflating aeroplane tyres

• Used to provide an inert atmosphere in melting and welding of easily oxidizable metals.

Neon is used:

• In discharge tubes and fluorescent bulbs used for advertising purposes

• In beacon lights for the safety of air navigators as the light can easily pass through the fog for a clear view.

Argon is used:

• To provide an inert atmosphere in high-temperature metallurgical processes (arc welding of metals or alloys)

• For filling electric bulbs.

• In the laboratory for handling substances that are air-sensitive.

• Xenon and Krypton are also used in light bulbs.

P Block Elements MCQ Chapter 7

The elements that can be found in the periodic table from group 13 to group 18 are called the P Block elements.

Below are some of the very important NCERT P Block Elements MCQ Class 12 Chemistry Chapter 7 with Answers. These P Block Elements MCQs have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of CBSE Term 1 examination.

We have given these P Block Elements Class 12 Chemistry MCQs Questions with Answers to help students understand the concept.

The p-Block Elements Class 12 Chemistry MCQs

1. Among the following, which one is a wrong statement.
(a) PH₅ and BiCl₅ do not exist.
(b) pπ-dπ bonds are present in SO₂
(c) SeF₄ and CH₄ have same shape.
(d) I₃ has bent geometry.

Answer/Explanation

Answer: c
Explaination:
(c) SeF₄ has see-saw shape where as CH₄ is tetrahedral.

2. In which of the pair of ions, both species contain S—S bond?

Answer/Explanation

Answer: a
Explaination:

3. Which one of the following order is correct for the bond dissociation enthalpy of halogen molecule?
(a) Br₂ > I₂ > F₂ > Cl₂
(b) F₂ > Cl₂ > Br₂ > I₂
(c) I₂ > Br₂ > Cl₂ > F₂
(d) Cl₂ > Br₂ > F₂ > I₂

Answer/Explanation

Answer: d
Explaination:
(d) In F2, therefore, inter electronic repulsion, therefore, bond dissociation enthalpy is less.

4. Which is strongest acid in the following:
(a) HClO₄
(b) H₂SO₃
(c) H₂SO₄
(d) HClO₃

Answer/Explanation

Answer: a
Explaination:
(a) HClO₄ is strongest because ‘Cl’ has +7 oxidation state.

5. In which of the following pairs, the two species are isostructural

Answer/Explanation

Answer: c
Explaination:
(c) Both are pyramidal.

6. The correct order of oxidising power is
(a) HClO₄ > HClO₃ > HClO₂ > HCIO
(b) HOCl > HClO₂ > HClO₃ > HClO₄
(c) HClO₃ > HClO₄ > HClO₂ > HClO
(d) HCIO₂ > HOCl > HClO₃ > HClO₄

Answer/Explanation

Answer: b
Explaination:
(b) HOCl → HCl + [O]
It is strongest oxidising agent whereas HClO₄ is weakest.

7. The correct order of acid strength is
(a) HClO₄ < HClO₃ < HClO₂ < HClO
(b) HCIO < HClO₂ < HClO₃ < HClO₄
(c) HClO₄ < HClO < HClO₂ < HClO₃
(d) HClO₂ < HClO₃ < HClO₄ < HClO

Answer/Explanation

Answer: b
Explaination:
(b) As oxidation state increases, acid strength increases.

8. Among the following which is strongest oxidising agent.
(a) Br₂
(b) I₂
(c) Cl₂
(d) F₂

Answer/Explanation

Answer: d
Explaination:
(d) F₂ is best oxidising agent.

9. The correct order of bond angles in the following species is

Answer/Explanation

Answer: b
Explaination:
(b) ClO₂^– < Cl₂O < ClO₂ is increasing order of bond angle.

10. Sulphur trioxide can be obtained by which of the following:

Answer/Explanation

Answer:
Explaination:

11. When Cl₂ reacts with hot and cone. NaOH, the oxidation number of chlorine changes from
(a) zero to +1 and zero to +5
(b) 0 to -1 and 0 to +5
(c) zero to -1 and zero to +3
(d) 0 to +1 and 0 to -3

Answer/Explanation

Answer: b
Explaination:
(b) Cl2 has oxidation number 0, in Cl (-1) and in ClO3(+5).
3Cl₂ + 6NaOH(hot and cone.) → 5NaCl + NaClO₃ + 3H₂O

12. Acidity of diprotic acid in aqueous solution increases in the order.
(a) H₂S < H₂Se < H₂Te
(b) H₂Se < H₂S < H₂Te
(c) H₂Te < H₂S < H₂Se
(d) H₂Se < H₂Te < H₂S

Answer/Explanation

Answer: a
Explaination:
(a) Because bond dissociation enthalpy decreases as atomic size increases.

13. Which of the following statement is incorrect?
(a) ONF is isoelectronic with NO^–₂
(h) OF₂ is an oxide of fluoride
(c) Cl₂O₇ is an anhydride of perchloric acid
(d) O₃ molecule is bent

Answer/Explanation

Answer: b
Explaination:
(b) It is fluoride of oxygen.
∵ ‘F’ is more electronegative than O.

14. Chlorine reacts with excess of NH₃ to form
(a) NH₄Cl
(b) N₂ + HCl
(c) N₂ + NH₄Cl
(d) NCl₃ + HCl

Answer/Explanation

Answer: c
Explaination:
(c) 8NH₃ + 3Cl₂ → 6NH₄Cl + N₂

15. Which of the following reactions is an example of redox reaction?
(a) XeF₄ + O₂F₂ → XeF₆ + O₂
(b) XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–
(c) XeF₆ + H₂O → XeOF₄ + 2HF
(d) XeF₆ + 2H₂O → Xeo₂F₂ + 2HF

Answer/Explanation

Answer: a
Explaination:
(a) Redox reaction becaueXe(+4) is getting oxidised to Xe(+6) and 0(+l) is reduced to zero.

16. On addition of cone. H₂SO₄ to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet flames come out. This is because [NCERT Exemplar]
(a) H₂SO₄ reduces HI to I₂
(b) HI is of violet colour
(c) HI gets oxidised to I₂
(d) HI changes to HIO₃

Answer/Explanation

Answer: c
Explaination:
(c) HI gets oxidised to I₂

17. Which of the following pairs of ions are isoelectronic and isostructural? [NCERT Exemplar]

Answer/Explanation

Answer: a
Explaination:
(a) CO^–₃ and NO^–₃ are isoelectronic (32 electrons) and Planar.

18. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy? [NCERT Exemplar]
(a) HF
(b) HCl
(c) HBr
(d) HI

Answer/Explanation

Answer: a
Explaination:
(a) HF has highest bond dissociation enthalpy due to smaller bond length.

19. Bond dissociation enthalpy of E—H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent? [NCERT Exemplar]

(a) NH₃
(b) PH₃
(c) AsH₃
(d) SbH₃

Answer/Explanation

Answer: d
Explaination:
(d) SbH₃ due to lowest bond dissociation enthalpy.

20. Hot cone. H₂SO₄ acts as moderately strong oxidising agent. It oxidises both metals and non-metals. Which of the following element is oxidised by cone. H₂SO₄ into two gaseous products? [NCERT Exemplar]
(a) Cu
(b) S
(c) C
(d) Zn

Answer/Explanation

Answer: c
Explaination:
(c) C + 2H₂SO₄(conc.) → CO₂ + SO₂ + 2H₂O

21. H₂S is more acidic than H₂O because
(a) oxygen is more electronegative than sulphur.
(b) atomic number of sulphur is higher than oxygen.
(c) H — S bond dissociation energy is less as compared to H — O bond.
(d) H — O bond dissociation energy is less also compared to H — S bond.

Answer

Answer: b

22. The boiling points of hydrides of group 16 are in the order
(a) H₂O > H₂Te > H₂S > H₂Se
(b) H₂O > H₂S > H₂Se > H₂Te
(c) H₂O > H₂Te > H₂Se > H₂S
(d) None of these

Answer

Answer: b

23. In the manufacture of sulphuric acid by contact process Tyndall box is used to
(a) convert SO₂ and SO₃
(b) test the presence of dust particles
(c) filter dust particles
(d) remove impurities

Answer

Answer: b

24. Fluorine differs from rest of the halogens in some of its properties. This is due to
(a) its smaller size and high electronegativity.
(b) lack of d-orbitals.
(c) low bond dissociation energy.
(d) All of the these.

Answer

Answer: b

25. The set with correct order of acidity is
(a) HClO < HClO₂ < HClO₃ < HClO₄
(b) HClO₄ < HClO₃ < HClO₂ < HClO
(c) HClO < HClO₄ < HClO₃ < HClO₂
(d) HClO₄ < HClO₂ < HClO₃ < HClO

Answer

Answer: b

26. When chlorine reacts with cold and dilute solution of sodium hydroxide, it forms
(a) Cl^– and ClO^–
(b) Cl^– and ClO₂^–
(c) Cl^– and ClO₃^–
(d) Cl^– and ClO₄^–

Answer

Answer: a

27. The formation of O₂⁺ [PtF₆]^– is the basis for the formation of first xenon compound. This is because
(a) O₂ and Xe have different sizes.
(b) both O₂ and Xe are gases.
(c) O₂ and Xe have comparable electro-negativities.
(d) O₂ and Xe have comparable ionisation enthalpies.

Answer

Answer: d

28. Partial hydrolysis of XeF₄ gives
(a) XeO₃
(b) XeOF₂
(c) XeOF₄
(d) XeF₂

Answer

Answer: b

29. Helium is preferred to be used in balloons instead of hydrogen because it is
(a) incombustible
(b) lighter than hydrogen
(c) more abundant than hydrogen
(d) non polarizable

Answer

Answer: a

30. The increasing order of reducing power of the halogen acids is
(a) HF < HCl < HBr < HI
(b) HI < HBr < HCl < HF
(c) HBr < HCl < HF < HI
(d) HCl < HBr < HF < HI

Answer

Answer: a

Note: In the following questions two or more options may be correct. (Q 21. to Q 24)
31. Which of the following options are not in accordance with the property mentioned against them? [NCERT Exemplar]
(a) F₂ > Cl₂ > Br₂ > I₂ Oxidising power.
(b) MI > MBr > MCI > MF Ionic character of metal halide.
(c) F₂ > Cl₂ >Br₂ > I₂ Bond dissociation enthalpy.
(d) HI < HBr < HCl < HF Hydrogen-halogen bond strength.

Answer/Explanation

Answer:
Explaination:
(b) MF > MCl > HBr > MI Ionic character
(c) Cl₂ > Br₂ > F₂ > I₂

32. Which of the following statements are correct? [NCERT Exemplar]
(a) Among halogens, radius ratio between iodine and fluorine is maximum.
(b) Leaving F—F bond, all halogens have weaker X—X bond than X—X’ bond in interhalogens.
(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(d) Interhalogen compounds are more reactive than halogen compounds.

Answer/Explanation

Answer:
Explaination:
(a), (c) and (d) are correct.
F₂ is more reactive than interhalogen compounds.
(b) is not correct, other halogens are less reactive than interhalogen compounds

33. Which of the following statements are correct for SO₂ gas? [NCERT Exemplar]
(a) It acts as bleaching agent in moist conditions.
(b) It’s molecule has linear geometry.
(c) It’s dilute solution is used as disinfectant.
(d) It can be prepared by the reaction of dilute H2S04 with metal sulphide.

Answer/Explanation

Answer:
Explaination:
(b) and (d). Its molecule is bent.

and Na₂S + H2SO₄ → Na₂SO₄ + H₂S

34. Which of the following orders are correct as per the properties mentioned against each? [NCERT Exemplar]
(a) AS₂O₃ < SiO₂ < P₂O₃ < SO₂ Acid strength.
(b) AsH₃ < PH₃ < NH₃ Enthalpy of vapourisation,
(c) S < O < Cl < F More negative electron gain enthalpy.
(d) H₂O > H₂S > H₂Se > H₂Te Thermal stability.

Answer/Explanation

Answer:
Explaination:
(a) and (d) are correct (b) and (c) are wrong.
(b) PH₃ <ASH₃ <NH₃
(c) Cl > F > S > O

35. Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option. [NCERT Exemplar]

Column I	Column II
(A) Xe F₆	(1) sp³d³ – distorted octahedral
(B) Xe O₃	(2) sp³d² – square planar
(C) Xe OF₄	(3) sp³– pyramidal
(D) Xe F₄	(4) sp³ d² – square pyramidal

Code:
(a) A (1) B (3) C (4) D (2)
(b) A (1) B (2) C (4) D (3)
(c) A (4) B (3) C (1) D (2)
(d) A (4) B (1) C (2) D (3)

Answer/Explanation

Answer:
Explaination:
(a) A (1) B (3) C (4) D (2)

36. Match the items of Columns I and II and mark the correct option. [NCERT Exemplar]

Column I	Column II
(A) H₂SO₄	(1) Highest electron gain enthalpy
(B) CCl₃NO₂	(2) Chalcogen
(C) Cl₂	(3) Tear gas
(D) Sulphur	(4) Storage batteries

Answer/ExplanationCode:
(a) A (4) B (3) C (1) D (2)
(b) A (3) B (4) C (1) D (2)
(c) A (4) B (1) C (2) D (3)
(d) A (2) B (1) C (3) D (4)

Answer:
Explaination:
(a) A (4) B (3) C (1) D (2)

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (Q.27 to Q.29)
(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(c) Assertion is correct, but reason is wrong statement.
(d) Assertion is wrong but reason is correct statement.
(e) Both assertion and reason are wrong statements.
37. Assertion: HI cannot be prepared by the reaction of KI with concentrated H₂SO₄.
Reason: HI has lowest H-X bond strength among halogen acids.[NCERT Exemplar]

Answer/Explanation

Answer: b
Explaination:
(b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. HI gets oxidised to I₂ as H₂SO₄(conc.) is oxidising agent.

38. Assertion: Both rhombic and monoclinic sulphur exist as S₈ but oxygen exists as O₂.
Reason: Oxygen forms pπ – pπ multiple bond due to small size and small bond length but pπ – pπ bonding is not possible in sulphur. [NCERT Exemplar]

Answer/Explanation

Answer: a
Explaination:
(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

39. Assertion: NaCl reacts with concentrated H₂SO₄ to give colourless fumes with pungent smell. But on adding MnO₂ the fumes become greenish yellow.
Reason: Mn02 oxidises HC1 to chlorine gas which is greenish yellow. [NCERT Exemplar]

Answer/Explanation

Answer: a
Explaination:
(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
NaCl + H₂SO₄(conc.) → NaHSO₄ + HCl
4HCl + MnO₂ → MnCl₂ + Cl₂ + 2H₂O

40. The mixture of cone. HCl and anhydrous ZnCl₂ is called ___________ .

Answer/Explanation

Answer:
Explaination: Lucas reagent

41. Out of H₂O and H2S which has higher bond angle? ___________ .

Answer/Explanation

Answer:
Explaination: H₂0

42. Tin reacts with excess of chlorine gas to form ___________ .

Answer/Explanation

Answer:
Explaination: SnCl₄

43. Lead sulphide is heated with air to form ___________ .

Answer/Explanation

Answer:
Explaination: PbO + SO₂

44. I2₂ gets oxidised to ___________ by cone. HNO₃.

Answer/Explanation

Answer:
Explaination: HIO₃

45. Interhalogen compounds are more reactive than helogens except fluorine. [True/False]

Answer/Explanation

Answer:
Explaination: True. It is due to less effective overlapping.

46. C1F is neutral molecule isoelectronic withCIO^–. [True/False]

Answer/Explanation

Answer:
Explaination: True. Both have 17 + 9 = 26 electrons.

47. NaF reacts with SbF₆ to form Na⁺ [SbF₇]^–. [True/False]

Answer/Explanation

Answer:
Explaination:

48. Hydrolysis of XeF6 is redox reaction. [True/False]

Answer/Explanation

Answer:
Explaination: It is false. XeF₆ + 3H₂O > XeO₃ + 6HF.

49. Ozone is thermodynamically less stable than O₂. [True/False]

Answer/Explanation

Answer:
Explaination: True

50. Oxygen does not show +4 and +6 oxidation states like sulphur. Why?

Answer/Explanation

Answer:
Explaination: It is due to the absence of vacant d-orbital in oxygen.

51. Draw the structure of 03 molecule.

Answer/Explanation

Answer:
Explaination:

It is bent molecule.

52. Write one chemical reaction equation to show that SO₂ acts as a reducing agent.

Answer/Explanation

Answer:
Explaination:

53. State reason for the following:
Sulphur has greater tendency for catenation than oxygen. [Delhi 2013; AI2016,12; DoE]
Or
O-O bond is weaker than S-S bond, why? [Chennai 2019]

Answer/Explanation

Answer:
Explaination:
It is because of strong S—S bond than 0—0 due to greater repulsion between valence electrons of smaller atoms of oxygen than sulphur atoms.

54. Write the formulae of any two oxoacids of sulphur. [AI 2015 Allahabad & Dehradun]

Answer/Explanation

Answer:
Explaination:
H₂SO₄ and H₂SO₃.

55. Write one chemical reaction equation to show that cone. H2S04 is a strong oxidising agent.

Answer/Explanation

Answer:
Explaination:
C + 2H₂SO₄(conc.) CO₂+ 2H₂O + 2SO₂

56. Fluorine does not exhibit any positive oxidation state. Why? [Delhi 2013]

Answer/Explanation

Answer:
Explaination:
It is because Fluorine is most electronegative element and best oxidising agent.

57. Despite lower value of its electron gain enthalpy with negative sign, fluorine (F₂) is a stronger oxidising agent than Cl₂. [Delhi 2019; AI 2012; DoE]

Answer/Explanation

Answer:
Explaination:
It is due to higher standard reduction potential of F₂ which is due to low bond dissociation energy of F—F bond due to inter electronic repulsion among small size F atoms, high electron gain enthalpy and highest hydration enthalpy.

58. Although the H-bonding in hydrogen fluoride is much stronger than that in water, yet water has a much higher boiling point than hydrogen fluoride. Why? [Foreign 2012]

Answer/Explanation

Answer:
Explaination:
It is because H₂0 molecules form H-bonds to more extent as compared to HF molecules.

59. Give a chemical reaction involved in chlorine with saturated hydrocarbon.

Answer/Explanation

Answer:
Explaination:

60. Explain giving a reason for the following situation:
In aqueous medium, HCl is a stronger acid than HF. [Foreign 2011]

Answer/Explanation

Answer:
Explaination:
It is because bond dissociation energy of H—Cl is lower than HF due to longer bond length.

61. Give the chemical formula of any one halic acid.

Answer/Explanation

Answer:
Explaination: H0BrO₂ (Bromic acid)

62. Give a chemical reaction involved in the preparation of interhalogen compound.

Answer/Explanation

Answer:
Explaination:

63. Which compound led to the discovery of the compounds of noble gas?

Answer/Explanation

Answer:
Explaination:
which was the first compound of noble gas.

64. Complete the following reaction:
Xe + PtF₆ →

Answer/Explanation

Answer:
Explaination:
Xe + PtF₆ → Xe⁺[PtF₆]^–

65. Sulphur disappears when boiled with an aqueous solution of sodium sulphite. Why?

Answer/Explanation

Answer:
Explaination:
It is due to the formation of sodium thiosulphate which is soluble in water, therefore, sulphur disappears.

P Block Elements MCQ (1-10)

Which one of the following oxides of nitrogen is a solid?
1. NO₂
2. N₂O
3. N₂O₃
4. N₂O₅
Answer/Explanation
1 . (4)
N₂O₅ exists as a solid under the temperature of 273 K. After this temperature, it starts too decompose.
Which one of the following is stable?
1. NCl₃
2. NBr₃
3. NI₃
4. NF₃
Answer/Explanation
2. (4)
The unstable nature of NBr₃, NCl₃ and NI₃ is due to low polarity of the N-X bond and large size difference between nitrogen and halogen atoms. The size of nitrogen and fluorine are comparable.
The BCl₃ is a planar molecule, whereas NCl₃ is pyramidal because
1. N-Cl bond is more covalent than B-Cl bond
2. B-Cl bond is more polar than N-Cl bond
3. BCl₃ has no lone pair but NCl₃ is a lone pair of electrons
4. Nitrogen atom is smaller than boron
Answer/Explanation
3. (3)
Presence of bond pairs and lone pairs affect the shape of the molecule. This is due to the fact that lone pairs and bond pairs repel each other and affect the shape.
BCl₃ has 3 bond pairs and 0 lone pair
NCl₃ has 3 bond pairs and 1 lone pair
Which of the following is tetrabasic acid?
1. Metaphosphoric acid
2. Orthophosphoric acid
3. Hypophosphoric acid
4. Hypophosphorous acid
Answer/Explanation
4. (3)
Tetrabasic acid means an acid having four hydrogens that can be replaced.
Hypophosphorous acid – H₃PO₂
Metaphosphoric acid – HPO₃
Orthophosphoric acid – H₃PO₄
Hypophosphoric acid – H₄P₂O₆
Hypophosphoric acid has 4 replaceable hydrogens.
The reagent used to distinguish between hydrogen peroxide and ozone is
1. PbS
2. Starch and Iodine
3. KMnO₄
4. Bleaching powder
Answer/Explanation
5. (2)
Iodine is used in the quantitative estimation of ozone.
Ozone can be detected by using
1. mercury
2. silver
3. sodium
4. none of these
Answer/Explanation
6. (1)
Ozone reacts with mercury to form mercurous oxide, which sticks to the walls of the container. Therefore, the presence of ozone can be detected by mercury.
The correct sequence of decrease in the bond angle of following hydride is
1. NH₃ > PH₃ > AsH₃ > SbH₃
2. NH₃ > AsH₃ > PH₃ > SbH₃
3. SbH₃ > AsH₃ > PH₃ > NH₃
4. PH₃ > NH₃ > AsH₃ > SbH₃
Answer/Explanation
7. (1)
As we move down the group, the radius of elements increases and electronegativity decreases so bond angle decreases.
Among the following oxides, the least acidic is
1. P₄O₆
2. P₄O₁₀
3. As₄O₆
4. As₄O₁₀
Answer/Explanation
8. (3)
Acidic character of oxides decreases with the decrease in the oxidation state and also decreases down the group.
Which of the following is paramagnetic?
1. N₂O₃
2. NO₂
3. N₂O₅
4. All of these
Answer/Explanation
9. (2)
NO₂ is paramagnetic due to the presence of an unpaired electron in the nitrogen.
Concentrated H₂SO₄ is not used to prepare HBr from KBr because it
1. oxidises HBr
2. reduces HBr
3. causes disproportionation of HBr
4. reacts to slowly with KBr
Answer/Explanation
10. (1)
Conc. H₂SO₄ can not be used to prepare HBr from KBr because conc. H₂SO₄ oxidises HBr to bromine.

P Block Elements MCQ (11-20)

Which of the following is thermally most stable ?
1. H₂S
2. H₂O
3. H₂Se
4. H₂Te
Answer/Explanation
11. (2)
H₂O is the most stable as there is a formation of hydrogen bonds.
The order goes like : H₂Te < H₂Se < H₂S < H₂O
Which of the following oxides react with both HCl and NaOH?
1. N₂O₅
2. ZnO
3. CaO
4. CO₂
Answer/Explanation
12. (2)
ZnO is an amphoteric oxide, therefore it reacts with both HCl and NaOH.
Which gas evolves when urea is treated with sodium hydroxide?
1. Nitrogen
2. Laughing gas
3. Ammonia
4. NO
Answer/Explanation
13. (3)
NH₂CONH₂ + 2NaOH → 2NH₃ + NH₂CO₃
Which of the following is an amphoteric oxide?
1. Cr₂O₃
2. Cl₂O₇
3. SnO₂
4. None of these
Answer/Explanation
14. (1)
Cr₂O₃ is an amphoteric oxide.
Which of the following allotropic forms of sulphur is thermodynamically most stable?
1. β-monoclinic
2. ℽ-monoclinic
3. Plastic sulphur
4. Orthorhombic
Answer/Explanation
15. (4)
Orthorhomic sulphur is the most stable form of sulphur having eight sulphur atoms arranged in an octahedron.
Which of the following does not have a p-o-p bond?
1. P₄O₆
2. P₄O₁₀
3. H₄P₂O₆
4. H₄P₂O₇
Answer/Explanation
16. (3)
H₄P₂O₆ (Hypophosphoric acid) doesn’t have any P-O-P bond.
Which acid on heating produces phosphine?
1. Phosphoric acid
2. Phosphorous acid
3. Peroxymonophosphoric acid
4. Metaphosphoric acid
Answer/Explanation
17. (3)
H₃PO₅ (Peroxymonophosphoric acid) gives H₃PO₄ (orthophoric acid) and PH₃ (phosphine).
Xenon difluoride is
1. linear
2. angular
3. trogonal
4. pyramidal
Answer/Explanation
18. (1)
XeF₂ has a linear shape.
Which of the following is not a reducing agent
1. Sulphur Dioxide
2. Hydrogen Peroxide
3. Carbon Dioxide
4. Nitrogen Dioxide
Answer/Explanation
19. (2)
Carbon dioxide is an oxidising agent.
Which of the following molecules does not possess a permanent dipole moment
1. H₂S
2. SO₂
3. SO₃
4. CS₂
Answer/Explanation
20. (4)
CS₂ has a linear shape, hence doesn’t have a permanent dipole moment.

P Block Elements MCQ (21-30)

Potassium chlorate on heating with concentrated sulphuric acid gives
1. Chlorine dioxide
2. HClO₄
3. KHSO₄
4. All of these
Answer/Explanation
21. (4)
3KClO₃ + 3H₂SO₄ → 2KHSO₄ + HClO₄ + 2ClO₂ + H₂O
The most abundant and common oxidation state of sulphur is
1. -2
2. +4
3. +2
4. +6
Answer/Explanation
22. (2)
Sulphur’s most common oxidation state is +4.
Ozone is
1. an unstable, dark blue, diamagnetic gas
2. an unstable, dark blue, paramagnetic gas
3. a stable, dark blue, paramagnetic gas
4. found in the upper atmosphere where it absorbs UV radiation
Answer/Explanation
23. (1)
Ozone is an unstable, dark blue, diamagnetic gas.
Structure of ClF₃ is
1. T-shape
2. Octahedral
3. Tetrahedral
4. None of these
Answer/Explanation
24. (1)
ClF₃ is of T-shape
In two forms NCl3 whereas P can form both PCl3 and PCl5. Why?
1. N atom has is larger than P in size
2. P is more reactive towards Cl than N
3. P has d orbitals which can be used for bonding but N2 does not have d orbitals
4. None of these
Answer/Explanation
25. (3)
There is no vacant d-orbital in the outermost orbit of nitrogen. Thus nitrogen show valency only 3. There are vacant d-orbital is in the outermost orbit of phosphorus and hence it shows variable covalency 3 and 5 in ground state and excited state respectively. Hence, nitrogen forms only NCl₃ but phosphorus forms PCl₃ and PCl₅ both.
Which characteristic is not correct about sulphuric acid?
1. Sulphonating agent
2. Reducing agent
3. Oxidizing agent
4. Highly Vicious
Answer/Explanation
26. (2)
H₂SO₄ is a strong oxidising agent and can oxidise both metals and non-metals. It’s not a reducing agent.
Which of the following is planar?
1. XeF₄
2. XeO₄
3. XeO₂F₂
4. XeOF₄
Answer/Explanation
27. (1)
XeF₄ has a square planar structure.
Which of the following halide ions is the most basic?
1. Cl^–
2. Br^–
3. I^–
4. F^–
Answer/Explanation
28. (4)
F^– ion is the most electronegative, most unstable and the most basic .
Which of the following gives nitrogen on heating?
1. Ammonium nitride
2. Ammonium nitrate
3. Ammonium nitrite
4. Ammonium hydroxide
Answer/Explanation
29. (3)
Ammonium nitrite on thermal decomposition gives ammonia.
One of the reasons why F-F bond is weak is that
1. The repulsion between the non bonding pairs of electrons of the two fluorine atoms is high
2. The F-F Bond distance is small and hence intermolecular repulsion is small
3. The ionization enthalpy of the fluorine atom is very high
4. The F-F Bond distances large
Answer/Explanation
30. (2)
The F-F Bond distance is small and hence intermolecular repulsion is small.

P Block Elements MCQ (31-40)

The mixture of NaI and NaIO₃ is treated with hot conc. H₂SO₄. The iodine-containing product formed is
1. HIO₃
2. NaIO₄
3. I₂
4. HIO₄
Answer/Explanation
31. (3)
5 NaI + NaIO₃ + 3H₂SO₄ → 3Na₂SO₄ + 3I₂ +3H₂O
When bleaching powder is treated with carbon dioxide
1. Calcium chloride is formed
2. No reaction occurs
3. Chlorine is evolved
4. It absorbs the gas
Answer/Explanation
32. (3)
When bleaching powder is treated with carbon dioxide chlorine is evolved.
PtF₆ converts O₂ to
1. O₂^–PtF₆⁺
2. O₂⁺PtF₆^–
3. F₂⁺PtO₂
4. F₂O⁺Pt
Answer/Explanation
33. (2)
O₂⁺PtF₆^–
Which of the following has the greatest reducing power?
1. HBr
2. HI
3. HCl
4. HF
Answer/Explanation
34. (2)
HI has the strongest reducing agent among halogen acids because of lowest bond dissociation enthalpy.
XeF₆ on complete hydrolysis gives
1. XeO₃
2. Xe
3. XeO₂
4. XeO₄
Answer/Explanation
35. (1)
XeF₂ + 3H₂O → 6HF + XeO₃
Which of the following statements is not correct?
1. Krypton is obtained during radioactive disintegration
2. Argon is used in electric bulbs
3. Half life of radon is only 3.8 days
4. Helium is used in producing very low temperatures
Answer/Explanation
36. (1)
Krypton is not obtained during radioactive disintegration.
Which of the following noble gas is not found in the atmosphere?
1. Rn
2. Kr
3. Ne
4. Ar
Answer/Explanation
37. (1)
Radon and helium is not found in the atmosphere.
What is the colour produced when ammonia is passed through a solution of copper sulphate?
1. Red
2. Orange
3. Yellow
4. Blue
Answer/Explanation
38. (4)
The complex formed is [Cu(NH₃)₄]²⁺ which is square planar and paramagnetic and has a deep blue colour.
What is the correct order of electron affinities among halogens?
1. F₂ > Cl₂ > Br₂ > I₂
2. Cl₂ > Br₂ > F₂ > I₂
3. F₂ < Cl₂ < Br₂ < I₂
4. I₂ < Br₂ < F₂ < Cl₂
Answer/Explanation
39. (2)
The electron affinity of chlorine is maximum in the periodic table and so the bond dissociation enthalpy. Fluorine has lower bond dissociation enthalpy than Br₂ and Cl₂. Due to small size electronic repulsion is very high. I₂ has lowest bond dissociation enthalpy due to its quite larger size it is easiest to break the bond.
What is the hybridisation of Xe in XeF₂?
1. sp³
2. sp²
3. sp²d
4. sp³d
Answer/Explanation
40. (4)
Xe in XeF₂ has sp³d hybridisation.

P Block Elements MCQ (41-50)

Which one of the following undergoes thermal decomposition?
1. XeF₂
2. XeF₄
3. ZeF₆
4. None of these
Answer/Explanation
41. (3)
XeF₆ on thermal decomposition gives XeF₂, XeF₄ and F₂
Which of the following is the least oxidising in nature?
1. HClO₃
2. HClO₄
3. HClO₂
4. HClO
Answer/Explanation
42. (2)
The oxidation state of chlorine in HClO₄, HClO₃, HClO₂ and HClO is 7, 5, 3 and 1. The chlorine with the highest oxidation state has the lowest potential of getting oxidised.
Which one of the following is the most stable thermally?
1. HI
2. HBr
3. HCl
4. HF
Answer/Explanation
43. (4)
The H-X bond strength decreases from HF to HI.
The crystals of ferrous sulphate on heating give
1. FeO + SO₂ + H₂O
2. FeO + SO₃ + H₂SO₄ + H₂O
3. Fe₂O₃ + SO₂ + SO₃ + H₂O
4. Fe₂O₃ + H₂SO₄ + H₂O
Answer/Explanation
44. (3)
FeSO₄ on heating gives Fe₂O₃ + SO₂ + SO₃ + H₂O.
Pure chlorine is obtained
1. By heating MnO₂ with HCl
2. By heating bleaching powder with HCl
3. By heating PtCl₄
4. By heating a mixture of NaCl and MnO₂ with conc. H₂SO₄
Answer/Explanation
45. (3)
Pure chlorine is obtained on heating PtCl₄.
The element which forms oxides in all the oxidation states from +1 to +5 is
1. N
2. P
3. As
4. Sb
Answer/Explanation
46. (1)
Nitrogen has an oxidation state varying from +1 to +5.
Which one of the following acts as an antichlor?
1. MnO₂
2. Na₂S₂O₃
3. K₂Cr₂O₇
4. Na₂SO₄
Answer/Explanation
47. (2)
An antichlor is a substance used to decompose residual HCl or chlorine after chlorine-based bleaching in order to prevent ongoing reaction. Na₂S₂O₃ acts as an antichlor.
N₂O₃ is
1. An acidic oxide and the anhydride of HNO₂
2. An acidic oxide and the anhydride of H₂N₂O₂
3. A neutral oxide and the anhydride of HNO₃
4. A basic oxide and the anhydride of HNO₂
Answer/Explanation
48. (1)
Nitrogen in N₂O₃ has an oxidation state of +3 and so does in HNO₂. Hence, N₂O₃ is an anhydride of HNO₂. N₂O₃ is an acidic oxide.
(NH₄)₂CrO₇ on heating liberates a gas. The same gas will be obtained by
1. Heating NH₄NO₃
2. Treating Mg₃N₂ with H₂O
3. Heating NH₄NO₂
4. Heating H₂O₂ on NaNO₂
Answer/Explanation
49. (3)
(NH₄)₂CrO₇ on heating liberates nitrogen gas which can also be obtained from heating NH₄NO₂.
According to Le Chatlier’s principle, low temperature is required for more yield of ammonia in Haber’s process. What is the temperature commercially kept for the process?
1. 350-450℃
2. 450-550℃
3. 250-350℃
4. 150-250℃
Answer/Explanation
50. (2)
The temperature required for Haber’s process is 450-550℃.

Assertion and Reasoning MCQs

Codes

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true and but R is not a correct explanation of A

(d) A is false, but R is true

1. Assertion (A) Valency of noble gas is 0.

Reason (R) Noble gases possess complete octet.

Answer/Explanation

1. (a)

Noble gases possess the electronic configuration ns¹np⁶ and has 8 electrons in their outer shell, hence there valency is 0.

2. Assertion (A) F₂ has lower bond dissociation enthalpy then Cl₂.

Reason (R) Fluorine is more electronegative than chlorine.

Answer/Explanation

2. (b)

Flourine has low bond dissociation enthalpy due to small atomic size number of electrons create large repulsion in bonded electron.

3. Assertion (A) Acidic character of group 16 hydrides increases from H₂O to H₂Te.

Reason (R) Thermal stability of hydrides decreases down the group.

Answer/Explanation

3. (b)

The acidic character increases down the group and thermal stability of hydrides decreases down the group due to decrease in bond (H-E) dissociation enthalpy down the group.

4. Assertion (A) Interhalogen compounds are more reactive than halogens (except chlorine).

Reason (R) They all undergo hydrolysis giving halide ion derived from the smaller halogen.

Answer/Explanation

4. (b)

Interhalogen compounds are more reactive than halogens because X-X’ bond in interhalogens in weaker than X-X bond in halogen (except F-F bond).

5. Assertion (A) Halogens are not found in free state in nature.

Reason (R) Halogens are highly reactive compound.

Answer/Explanation

5. (a)

Halogens are highly reactive as they have seven electrons in their outermost orbit and they want to stabilize by acquiring an electron. therefore, they do not occur in free state.

6. Assertion (A) F₂ has low reactivity.

Reason (R) F-F bonds Δ_bondH.

Answer/Explanation

6. (d)

F₂ is more reactive that other halogens because its valence electrons are more closer to nucleus and its more electronegative so, bonded electrons repel each other causing low bond dissociation enthalpy.

7. Assertion (A) Ozone layer in the upper region of atmosphere protect earth from UV radiation.

Reason (R) Ozone is a powerful oxidizing agent as compared to oxygen.

Answer/Explanation

7. (b)

Ozone layer filters the radiation coming from sun, hence serves as the protective layers.

8. Assertion (A) S shows paramagnetic nature, when present in vapour state.

Reason (R) S exists as S₂ in vapour state.

Answer/Explanation

8. (a)

In vapor state S exists as S₂ and it behaves like O₂. It has 2 unpaired electrons in antibonding pi orbitals. Presence of these unpaired electrons make S paramagnetic.

9. Assertion (A) F₂ is a strong oxidizing agent.

Reason (R) Electron gain enthalpy of fluorine is less negative.

Answer/Explanation

9. (b)

As F₂ has low bond dissociation energy and high hydration energy than Cl₂ but less electron gain enthalpy due to its smaller size. Due to these factors F₂ wins in getting reduced fastly.

10. Assertion (A) PbI₄ is not a stable compound.

Reason (R) Iodide stabilizes higher oxidizing state.

Answer/Explanation

10. (c)

Small highly elctronegative atoms such as F^– can stabilise higher oxidation state.

11. Assertion (A) Solubility of noble gases in water decreases with increasing size of the noble gas.

Reason (R) Solubility of noble gases in water is due to dipole-induced dipole interaction.

Answer/Explanation

11. (d)

PbI4 is not a stable compound because Pb shows (II) oxidation state more frequently than Pb (IV) due to inert pait effect. Iodide cannot stabilize higher oxidation states.

Frequently Asked Questions (FAQs)

Q1. Why P Block elements called the P Block elements?

They are called so because their valence electrons are in the p-orbital.

Q2. Why group 16 elements called chalcogens?

They are called so because most ores of copper (Greek Chalcos) are oxides or sulphides and also traces of selenium and tellurium.

Q3. Why is nitrogen so inert?

It is due to the presence of the very strong triple bond present between the molecular state of nitrogen N₂.

p- Block Elements : Notes and Study Materials -pdf

Notes and Study Materials

Examples and Exercise

CBSE Class 12th Chemistry Notes: The p-Block Elements

P Block Elements MCQ Chapter 7

The p-Block Elements Class 12 Chemistry MCQs

P Block Elements MCQ (1-10)

P Block Elements MCQ (11-20)

P Block Elements MCQ (21-30)

P Block Elements MCQ (31-40)

P Block Elements MCQ (41-50)

Assertion and Reasoning MCQs

Frequently Asked Questions (FAQs)

Q1. Why P Block elements called the P Block elements?

Q2. Why group 16 elements called chalcogens?

Q3. Why is nitrogen so inert?

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