The p-Block Elements Question Answers: NCERT Class 12 Chemistry

The group 15 elements (nitrogen, phosphorus,arsenic,antimony and bismuth) when reacted with halogen atom ,tend to form halides. The halides formed are of two types:

1) Trihalides (MX₃)

2) Pentahalides (MX₅)        where M -15^TH group element        X- halogen atom

The trihalides are formed by all the elements of group 15 while pentahalides are formed by all the elements except nitrogen because there is absence of vacant d- orbital in its outermost shell.

The oxidation state of +5 in pentahalides is more as compared to +3 oxidation state in trihalides. Due to the higher positive oxidation state of central atom in pentahalide state, these atoms will have larger polarizing power than the halogen atom attached to them. The central atom in pentahalide state will tend to polarize more the halide ion to which it is attached.

But In case of trihalides due to +3 oxidation state the central atom will polarize the halogen atom to a lesser extent as compared to pentahalide state. Therefore, more the polarization, larger will be the covalent character of the bond.

Hence due larger polarization of bond in pentahalide state as compared to trihalide state, the pentahalides are more covalent than trihalides.

The reducing character of the hydrides of Group 15 elements increases from NH₃  to BiH₃(Bismuthine) because the reducing character depends upon the stability of the hydride.The greater the unstability of anhydride,the greater is the its reducing character.Since,the BiH₃ is the least stable (because the size of central atom is greatest & therefore its tendency to form stable covalent bond with small hydrogen atom decreases,as a result the bond strength decreases) in this series,hence the reducing character increases. 

 Dinitrogen (N₂) IS formed by sharing three electron pairs between two nitrogen atoms.The two nitrogen atoms are joined by triple bond(N≡N). The nitrogen atom is very small in size ,therefore the bond length is also quite small(109.8 pm) & as a result the bond dissociation energy is quite high(946Kj / mol).This reason leads N₂ to be very less reactive at room temperature.

Ammonia is prepared using the Haber’s process.

The conditions required to maximize the yield of ammonia are as follows:

1 In accordance to Le Chatelier’s principle ,low temperature will shift the equilibrium to the right because the reaction is exothermic.This gives greater yield of ammonia.Therefore a temperature of about 450°C will be optimal for the preparation of ammonia.

2  High pressure on the reaction at equilibrium favours the shift of the equilibrium to the right because the forward reaction proceeds with a decrease in number of moles.Hence a pressure of about 200 atm will be optimal  for the higher yield of ammonia.

3 A catalyst should be used to increase the rate of reaction & to quickly attain equilibrium.For eg as iron oxide mixed with small amounts of K₂O and Al₂O₃  can be used as catalyst.

4 N₂ & H₂ gases should be pure in nature to increase the yield of ammonia.

Ammonia acts as a Lewis base due to the presence of lone pair of electrons on the nitrogen atom.Therefore,it can form coordinate bond with electron deficient molecules or a number of transition metal cations to from complex compounds.Ammonia reacts with a solution of Cu²⁺ to from a deep blue coloured complex.The reaction for the above complex can be written as follows:

Cu²⁺_(aq) + 4 NH_3(aq)   ⇔  [Cu(NH₃)₄]²⁺_(aq)

Blue                                         Deep Blue

This property of ammonia to form complex compounds is useful in detection of metal ions.

Covalency is defined as the tendency of an atom to form number of covalent bond with other molecule. N₂O₅ (Dinitrogen pentoxide) is one of the form of the nitrogen atom which belongs to group 15 in the periodic table. Nitrogen combines with oxygen to form a number of oxides having different oxidation states. Nitrogen has tendency to form pπ-pπ multiple bonds, which decides the structure of oxides. N₂O₅ has following structure:

From the above structure, it is clear that Nitrogen atom is sharing its electrons with oxygen atom. Nitrogen shares it four pair of electrons with oxygen, therefore nitrogen covalency is four (4).

PH₃ (Phosphine) is a hydride of Phosphorus. Phosphine is a gas and has pyramidal structure. Phosphorus involved is sp³ hybridized.

Now both PH₄⁺ and PH₃ have sp³ hybridization state for phosphorous .In PH₄⁺,all the four orbitals are bonded, whereas in PH₃,there is a lone pair of electrons on Phosphorous .In PH₄⁺, the HPH bond angle is tetrahedral angle of 109.5°.But in PH₃ ,lone pair- bond pair repulsion is more than bond pair-bond pair repulsion so that bond angles become less than normal tetrahedral angle .The bond angle in PH₃ is about 93.6°.

Phosphorus exist in many allotropic forms. The three main forms are:

1) White Phosphorus

2) Red Phosphorus

3) Black Phosphorus

White Phosphorus is obtained from phosphorite rock with coke and sand. It consist of P₄ units.The four P- atoms lie at the corners of a regular tetrahedron. It is less stable and therefore more reactive than other allotropic forms of Phosphorus. On heating with concentrated NaOH in an inert environment of CO₂ following reaction takes place:

Phosphine (PH₃) is formed.This reaction is an example of disproportionation reaction in which oxidation state of phosphorus decreases from 0 in P₄ to -3 in PH₃,while it increases from 0 in P₄ to +1 in NaH₂PO₂.

Phosphorus forms two type of halides i.e phosphorus trihalides (PCl₃) and phosphorus pentahalide PCl₅. In PCl₅ phosphorus undergoes sp³ d hybridization and has trigonal bipyramidal structure .It has three equatorial P-Cl bonds and two axial P-Cl bonds which are different. The axial bonds(219pm) are larger than equatorial bonds(204pm). PCl₅ is thermally less stable than PCl₃.On heating it sublimes but decomposes on stronger heating into trichloride and chlorine.The reaction involved is as under:

          heat 

PCL₅   →   PCL₃ + CL₂

It is a reversible reaction,in which PCl₃ combines with Cl₂ to form PCl₅.

In moist air ,PCl₅ hydrolysis to POCl₃ and finally gets converted to phosphoric acid.The reaction is as follows:

PCL₅    +        H₂O           →             POCl₃      +   2HCl

                                               phosphorus oxychloride

POCl₃    +       H₂O         →            H₃PO₄      +  3HCl              

                                               phosphoric acid

But on treatment of PCl₅ with heavy water (D₂O) only phosphorus oxychloride if formed with the liberation of 2 molecules of deuterium chloride as a side product. The reaction involved is:

PCl₅    +    D₂O        →             POCl₃         +     2DCl      

                                    phosphorus oxychloride

H₃PO₄

Since there are three OH groups present in H₃PO₄, its basicity is three i.e., it is a tribasic acid.

H₃PO₃,on heating, undergoes disproportionation reaction to form PH₃ and H₃PO4. The oxidation numbers of P in H₃PO₃,PH3, and H₃PO₄ are +3, - 3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.

Sulphur mainly exists in combined form in the earth's crust primarily as sulphates [gypsum (CaSO₄.2H₂O), Epsom salt (MgSO₄.7H₂O), baryte (BaSO₄)] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS₂)].

The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H-E) of hydrides on moving down the group.

Therefore,

H₂O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H₂O, which is absent in H₂S. Molecules of H₂S are held together only by weak van der Waal's forces of attraction. Hence, H₂O exists as a liquid while H₂S as a gas.

Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly.

(i) C₂H₄         +       3O₂     →         2CO₂                    +         2H₂O

   Ethene             Oxygen         Carbon dyoxide               Water

(ii)  4Al         +       3O₂        →         2Al₂O₃

     Aluminium       Oxygen               Alumina    

Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive.

O₃                 O₂            +           [O]

Ozone                    Oxygen                Nascent Oxygen

Therefore, ozone acts as a powerful oxidising agent.

Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with 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 using starch as an indicator. The reactions involved in the process are given below.

2I^-     +    H₂O   +   O₃    →   2OH^-    +    I₂    +   O₂

Iodide             Ozone                       Iodine

I₂     +   2Na₂S₂O₃         →      Na₂S₄O₆     +     NaI

              Sodium                     Sodium  

           thiosulphate                tetrathionate                 

SO₂ acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.

2Fe³⁺  +  SO₂ + 2H₂O   →   2Fe²⁺   +  SO₄^2-   +   4H⁺

The electronic configuration of S is 1s² 2s² 2p⁶ 3s² 3p⁴.

During the formation of SO₂, one electron from 3p orbital goes to the 3d orbital and S undergoes sp² hybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and d-orbital contain an unpaired electron each. One of these electrons forms p π- p π bond with one oxygen atom and the other forms p π- d π bond with the other oxygen. This is the reason SO₂ has a bent structure. Also, it is a resonance hybrid of structures I and II.

Both S-O bonds are equal in length (143 pm) and have a multiple bond character.

SO₂ is a colourless and pungent smelling gas.

It can be detected with the help of potassium permanganate solution. When SO₂ is passed through an acidified potassium permanganate solution, it decolonizes the solution as it reduces MnO₄^- ions to Mn²⁺ ions 

5SO₂    +     2MnO₄^-     +   2H₂O    →   5SO₄^2-    +   4H⁺   +  2Mn²⁺

Sulphuric acid is an important industrial chemicaland is used for a lot of purposes. Some important uses of sulphuric acid are given below.

(i) It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate.

(ii) It is used in the manufacture of pigments, paints, and detergents.

(iii) It is used in the manufacture of storage batteries.

Manufacture of sulphuric acid by Contact process involves three steps.

1. Burning of ores to form SO₂

2. Conversion of SO₂ to SO₃ by the reaction of the former with O₂ (V₂O₅ is used in this process as a catalyst.)

3. Absorption of SO₃ in H₂SO₄ to give oleum (H₂S₂O₇)

The key step in this process is the second step. In this step, two moles of gaseous reactants combine to give one mole of gaseous product. Also, this reaction is exothermic. Thus, in accordance with Le Chatelier's principle, to obtain the maximum amount of SO₃ gas, temperature should be low and pressure should be high.

It can be noticed that  K_a1 >>  K_a2

This is because a neutral H₂SO₄ has a much higher tendency to lose a proton than the negatively charged HSO₄^- . Thus, the former is a much stronger acid than the latter.

Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors.

1. Bond dissociation energy

2. Electron gain enthalpy

3. Hydration enthalpy

The electron gain enthalpy of chlorine is more negative than that of fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine. Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine. Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine.

Anomalous behaviour of fluorine

(i) It forms only one oxoacid as compared to other halogens that form a number of oxoacids.

(ii) Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expected.

Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite, KCl.MgCl₂.6H₂O. Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.

When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances.

Cl₂   +  H₂O   →   2HCl   +  [O]

Coloured substances + [O] →  Oxidized colourless substance

Two poisonous gases that can be prepared from chlorine gas are

(i) Phosgene (COCl₂)

(ii) Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

ICl is more reactive than I₂ because I-Cl bond in ICl is weaker than I-I bond in I₂.

Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.

Balanced equation:

XeF₆    +    2 H₂O    →    XeO₂F₂   +   4 HF

It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF₂ have not been isolated. They have only been identified.

General trends in group 15 elements

(i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is ns2 np3.

(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds. In addition to the -3 state, N and P also show -1 and -2 oxidation states.

All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.

(iii) Ionization energy and electronegativity:

First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.

(iv) Atomic size: On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.

Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N₂. In N₂, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen's small size that it is able to form pπ-pπ bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.

General trends in chemical properties of group - 15

(i) Reactivity towards hydrogen: The elements of group 15 react with hydrogen to form hydrides of type EH₃, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH₃ to BiH₃.

(ii) Reactivity towards oxygen: The elements of group 15 form two types of oxides: E₂O₃and E₂O₅, where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.

(iii) Reactivity towards halogens: The group 15 elements react with halogens to form two series of salts: EX₃ and EX₅. However, nitrogen does not form NX₅ as it lacks the d-orbital. All trihalides (except NX₃) are stable.

(iv) Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit -3 oxidation states.

Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in NH₃ than towards phosphorus in PH₃. Hence, the extent of hydrogen bonding in PH₃ is very less as compared to NH₃.

An aqueous solution of ammonium chloride is treated with sodium nitrite.

NO and HNO₃ are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Ammonia is prepared on a large-scale by the Haber's process.

Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation.

Hydride NH₃ PH₃ AsH₃ SbH₃

H-M-H angle 107° 92° 91° 90°

The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H-M-H bond angle.

N(unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, R₃N=O does not exist.

NH₃ is distinctly basic while BiH₃ is feebly basic.

Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.

Nitrogen owing to its small size has a tendency to form pπ-π multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, N₂. On moving down a group, the tendency to form pπ-pπ bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the P₄ state.

Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N-N single bond as compared to the P-P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N-N single bond.

On heating, orthophosphorus acid (H₃PO₃) disproportionates to give orthophosphoric acid (H₃PO₄) and phosphine (PH₃). The oxidation states of P in various species involved in the reaction are mentioned below.

PCl₅ can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl₅, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent.

The elements of group 16 are collectively called chalcogens.

(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns² np⁴, where nvaries from 2 to 6.

(ii) Oxidation state: Asthese elements have six valence electrons (ns² np⁴), they should display an oxidation state of -2. However, only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits the oxidation state of -1 (H₂O₂), zero (O₂), and +2 (OF₂). However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides: These elements form hydrides of formula H₂E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H₂E₂. These hydrides are quite volatile in nature.

Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form pπ-pπ bonds and form O₂(O==O) molecule. Also, the intermolecular forces in oxygen are weak van der Wall's, which cause it to exist as gas. On the other hand, sulphur does not form M₂ molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.

Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O^2-ion is much more than the oxide involving O^-ion. Hence, the oxide having O^2-ions are more stable than oxides having O^-. Hence, we can say that formation of O^2-is energetically more favourable than formation of O^-.

Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.

Sulphuric acid is manufactured by the contact process. It involves the following steps:

Step(i):

Sulphur or sulphide ores are burnt in air to form SO₂.

Step(ii):

By a reaction with oxygen, SO₂ is converted into SO₃ in the presence of V₂O₅ as a catalyst.

Step (iii):

SO₃ produced is absorbed on H₂SO₄ to give H₂S₂O₇ (oleum). This oleum is then diluted to obtain H₂SO₄ of the desired concentration. In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.

Sulphur dioxide causes harm to the environment in many ways:

1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.

2. Even in very low concentrations, SO² causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.

3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

The general electronic configuration of halogens is np⁵, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.

Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size.

Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.

Uses of ClO2:

(i) It is used for purifying water.

(ii) It is used as a bleaching agent.

Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.

(1)   Cl₂        +       H₂O        →          HCl              +          HOCl

                                               Hydrochloric acid         Hypochlorous acid

(2)   2F_2(g)    +     2H₂O_(I)    →     4H⁺_(aq)  +   4F^-_(aq) +   O_2(g) + 4HF_(aq)    

(i) Cl₂ can be prepared from HCl by Deacon's process.

(ii) HCl can be prepared from Cl₂ on treating it with water.

Cl₂        +       H₂O        →                 HCl              +          HOCl

                                               Hydrochloric acid         Hypochlorous acid

Neil Bartlett initially carried out a reaction between oxygen and PtF₆. This resulted in the formation of a red compound, O⁺₂[PtF₆]^-.

Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. Thus, he tried to prepare a compound with Xe and PtF₆. He was successful and a red-coloured compound, Xe⁺[PtF₆]^- was formed.

Let the oxidation state of p be x.

(i) H₃PO₃

3 + x + 3(-2) = 0

3 + x - 6 = 0

x - 3 = 0

x = 3

(ii) PCl₃

x + 3(-1) = 0

x - 3 = 0

x  = 3

(iii) Ca₃P₂

3(+2)  +  2(x)   =  0

6 + 2x = 0

2x  =  -6

x  = -6 / 2

x = -3

(iv) Na₃PO₄

3(+1) + x + 4(-2)  = 0

3 + x - 8 = 0

x - 5 = 0

x  = 5

(v) POF₃ :

x  + (-2)  + 3(-1) = 0

x - 2 - 3  = 0

x  - 5  = 0 

x = 5

XeF₂, XeF₄ and XeF₆ are obtained by a direct reaction between Xe and F2. The condition under which the reaction is carried out determines the product.

ClO^- is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown.

Total electrons ClO^- = 17 + 8 + 1 = 26

In ClF = 17 + 9 = 26

ClF acts like a Lewis base as it accepts electrons from F to form ClF₃.

(i) XeO₃ can be prepared in two ways as shown.

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

XeF₆  +  3H₂O      →   XeOF₃  +  6HF

(ii) XeOF₄ can be prepared using XeF₆.

XeF₆  +  H₂O      →   XeOF₄  +  2HF

(i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of F2 is lower than that of Cl₂ and Br₂. This is due to the small atomic size of fluorine. Thus, the increasing order for bond dissociation energy among halogens is as follows:

I₂< F₂< Br₂< Cl₂

(ii) HF < HCl < HBr < HI

The bond dissociation energy of H-X molecules where X = F, Cl, Br, I, decreases with an increase in the atomic size. Since H-I bond is the weakest, HI is the strongest acid.

(iii) BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃

On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.

NeF₂ does not exist.

(i) XeF₄ is isoelectronic with  ICl^-₄ and has square planar geometry.

(ii) XeF₂ is isoelectronic to  IBr^-₂ and has a linear structure.

(iii) XeO₃ is isostructural to BrO^-₃ and has a pyramidal molecular structure.

Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waal's radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waal's radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Uses of neon gas:

(i) It is mixed with helium to protect electrical equipments from high voltage.

(ii) It is filled in discharge tubes with characteristic colours.

(iii) It is used in beacon lights.

Uses of Argon gas:

(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.

(ii) It is usually used to provide an inert temperature in a high metallurgical process.

(iii) It is also used in laboratories to handle air-sensitive substances.