The p-Block Elements Important Questions

Ques. How many groups of elements are there in P-Block elements?
Ans. In a Periodic Table, there are six groups of elements residing in the P-Block numbered from 13 to 18.

Ques. Why does I-Cl react more vigorously than I₂? (All India Competition 2012)

Ans. The bond between I and Cl is weaker in I-Cl than the I-I bond in I₂. Hence, it breaks easily and is more reactive.

Ques. What do you mean by the Inert Pair Effect?
Ans. The inert pair effect is sometimes referred to as the occurrence of oxidation states which is two units less than the group of oxidation states.

Ques. How does metallic and non metallic character of an element vary in a group of elements?
Ans. It is only the P-Block elements of the periodic table that contain metals and nonmetals collectively. The non-metallic group of elements can reduce as the group progresses. Most metallic elements in the P-Block are heaviest.

Ques. Why do the 3rd period elements expand their covalence character above four?
Ans. The primary reason why 3rd period elements expand their covalence character is due to the presence of d-orbital in p-block elements. Their basic properties is that they can expand their octet and have a covalence more than four.

Ques. Why do the heavier elements of the periodic table form \( \pi\) - bonds?
Ans. Due to the combined effects of elements in terms of its size and the availability, D-orbitals can significantly determine the capacity of an electron to create a π-bonds. It is in fact the heavier elements of the p-block elements that form π-bonds.

Ques. Why does CCl₄ behave like an electron-precise molecule?
Ans. The total number of electrons that can surround the core atom in a CCl₄ molecule is eight, which makes it an electron-precise molecule.

Ques. How are fullerenes obtained?
Ans. Fullerenes were first created by heating graphite in the presence of inert gasses such as helium or argon in an electric arc.

Ques. Why is the boiling point of PH₃ less than NH₃?

Ans. Boiling point of NH₃ is higher than PH₃ because of the strong intermolecular hydrogen bonding between the nitrogen and hydrogen atoms.

Ques. Show the disproportionate reaction of H₃PO₃.

Ans. 4H₃PO₃ ­→ 3H₃PO₄ + PH₃.

Ques. Among SbH₃ and BiH₃, which is a stronger reducing agent and why?

Ans. BiH3 is less stable in comparison to SbH₃. Hence, it is a stronger reducing agent than BiH₃.

Ques. Among PH₃ and H₂S which is more acidic and why?

Ans. PH₃ has a lone pair of electrons and it acts as Lewis acid. H₂S is more acidic because of the weaker S-H bond.

Ques. The element Fluorine does not have any positive oxidation state. Why? (All India 2010)

Ans. In the periodic table, fluorine is the most electronegative element. It does not have d-orbitals in its valence shell. Hence, it is unable to expand its octet and hence does not show any positive oxidation state.

Ques. Why does N₂O dimerises to form N­₂O₄?

Ans. N₂O has an unpaired electron on nitrogen. This makes it very reactive. Hence it tries to form dimerized N₂O₄, a stable compound with an even number of electrons.

Ques.  Nitrogen can show a maximum covalency of 4, while the heavier elements of group 15 show covalence greater than 4. Why?

Ans. Nitrogen has only 4 orbitals in the valence shell while other heavier elements in group 15 have vacant d orbitals. So the maximum covalency of nitrogen is 4 while the heavier elements show a covalence of more than 4.

Ques. On moving from top to bottom, the tendency to exhibit -3 oxidation state, of the group 15 elements decreases. Explain.

Ans. On moving from top to bottom in group 15 the size of the atom increases along with the metallic character while the electronegativity decreases. This causes a decrease in the tendency to exhibit an oxidation state of +3.

Ques. The group 15 elements have higher ionisation enthalpies than those of corresponding members of group 14 and 16 elements. Why?

Ans. Due to the extra stable half-filled p orbitals electronic configuration and smaller size in group 15 elements they have higher ionisation enthalpy than the corresponding members of group 14 and 16.

Ques. Halogens are strong oxidising agents. Justify.

Ans. Halogens have high negative electron gain enthalpies and are more electronegative with less dissociation energies. Hence, due to the high tendency to gain electrons they act as strong oxidising agents.

Ques. Why is HF the weakest acid and HI is the strongest?

Ans. Iodine is less electronegative and has a large size. Thus, in HI the bonding between H and I is weaker which will easily split in comparison to HF. Hence, due to more liberation of H+ ions, HI acts as a stronger acid than HF.

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Short Answer Questions [2 Marks Questions]

Ques. Explain the nature of a boric acid as a Lewis acid in water?
Ans. Boric acid is considered as the weak monobasic acid that helps to take an electron from a hydroxyl ion to operate it as a Lewis acid. Boric acid can take OH ions and form a hydroxyl ion as a result of the reaction.

B(OH)₃ + 2HOH → [B (OH)₄] + H3O

Ques. Why is CCl? immiscible in water, whereas SiCl₄ can easily hydrolysed?
Ans. CCl₄ is usually insoluble in water due the nature of water and the compound as water is polar in nature and CCl₄ is nonpolar. The carbon atoms of a compound have no vacant orbitals to absorb an electron given off by the oxygen in the water. Whereas SiCl₄ can be easily hydrolysed due to empty d-orbitals in Si to accommodate the electron from the oxygen atom in water. 

Ques. Elaborate the reason why Silicon atoms do not form analogous graphite.

Ans. In graphite, C is sp2 hybridized and every C is connected to three other C atoms shaping hexagonal rings. Subsequently, graphite has a two-layered sheet-like structure. Silicon, then again, doesn't shape an analogue of C in view of the accompanying two reasons:

• Silicon has a much lesser inclination for catenation than C as Si-Si bonds are much weaker than C-C bonds.
• Silicon due to its bigger size than C goes through sp3 hybridisation.

Ques. Why does only carbon form covalent compounds whereas lead forms ionic compounds?

Ans. Carbon can't lose electrons to frame C4 because the sum of the amount of four ionization enthalpies is exceptionally high. It can't acquire four electrons to shape C4 as it is not unfavorable. Henceforth C structures are just covalent compounds. Down group 14, ionization enthalpies decrease, Pb being the last element has a low ionization enthalpy, that is, it can lose electrons to shape ionic compounds.

Ques. How can borax be prepared from Colemanite ores?

Ans. Borax can also be called sodium tetraborate decahydrate. It can be written as Na₂B₄O₇.10H₂O. This is prepared from Colemanite ores in the following manner.

Borax (Sodium Borate)

Powdered minerals are boiled with the sodium carbonate solution and filtered. Afterwards, the filter is concentrated and cooled with the crystals of borax. 

Ca₂B₆O₁₁ + 2Na₂CO₃ \( \to\)Na₂B₄O₇ + 2NaBO₂ + 2CaCO₃

The mother liquor that may contain sodium metaborate is treated with the current of CO₂, to convert it into borax that separates out.

4NaBO₂ + CO2→Na₂B₄O₇ + Na₂CO₃

Ques. Explain the formation of interhalogen compounds. What is the general compositions assigned to them? (CBSE AI 2013)

Ans. The compounds which contain two or more halogen atoms are called interhalogen compounds. They are formed due to the combination of two or more halogen atoms. They can be assigned the general composition of XX’n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X. For example, CIF, ICl₃, IF₇.

Ques. Explain the following occurrences.

1. XeF₂ has linear structure and not bent structure.
2. Phosphorous shows tendency for Catenation

Ans: i. According to VSEPR theory, the total number of electron pairs in the central atom's valence shell determines the molecule's shape. The linear structure of XeF₂ has 2 pairs of bond electrons and 3 pairs of lone electrons which causes the shape to be linear and not bent.

Ii. Due to greater P-P bond energy, several Phosphorous atoms exhibit both cyclic and open chain compounds which shows marked tendency for catenation.

Ques.  Arrange the following trichlorides in decreasing order of bond angle.

NCl₃, PCl₃, AsCl₃, SbCl₃

Ans. The decreasing order of bond angles of the trichlorides are as follows:

NCl₃< PCl₃< AsCl₃<SbCl₃

Ques. Explain why Dioxygen is a gas while sulphur (S₈) is a solid.

Ans. In an Oxygen molecule the bonding between the O atoms is pπ−pπ. Hence, the intermolecular forces existing between the atoms are weak Van Der Waals forces. As the atoms are small in size with high electron density and they tend to repel other molecules in it. So, Dioxygen is a gas.

In the Sulphur molecule (S₈), the molecular force is strong covalent bonds and it is a polyatomic molecule with strong intermolecular forces. So, it exists in solid form.

Ques.  Give reasons why phosphorus forms pentachloride whereas nitrogen and bismuth do not?

Ans: Nitrogen does not have d-orbitals. Hence it is unable to form pentachloride. Bi does not form pentachloride due to the inert pair effect. Phosphorus has vacant d-orbitals and can expand its octet easily. Hence it forms pentachloride.

Ques.  Complete the following chemical reactions and balance them.

(i) Na₂SO₃ + Cl₂ + H₂O → Na₂SO₃ + Cl₂ + H₂O

(ii) NaHCO₃ + HCl → NaHCO₃ + HCl

Ans: i)Na₂SO₃ + Cl₂ + H₂O → 2HCl + Na₂SO₄Na₂SO₃ + Cl₂ + H₂O → 2HCl + Na₂SO₄

ii)NaHCO₃ + HCl → NaCl + CO₂ + H₂O

Ques. Under what conditions the formation of ammonia gas is favoured along with reactions involved in Haber’s process.

Ans. Haber’s process is given by the following chemical equation:

N₂(g)+3H₂(g)→·FeMo₂NH₃(g)  N₂(g)+3H₂(g)→·FeMo₂NH₃(g)

Under the following conditions the formation of ammonia gas is favoured:

• High pressure (200 atm)
• Moderate temperature (700K)
• Presence of catalyst Fe/FeO with small amount of K₂OK₂O and Al₂O3Al₂O₃ increases the rate of attainment of equilibrium.

Ques. What is the chemical formula of bleaching powder?

Ans. The chemical formula of bleaching powder is given by Ca(OCl)₂. CaCl₂. Ca(OH)₂. 2H₂O.

Ques. What happens when NaCl is heated with conc. H₂SO₄ in the presence of MnO₂.

Ans: When NaCl is heated with concentrated sulphuric acid then the following reaction occurs:

4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂.

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Long Answer Questions [3 Marks Questions]

Ques. What are the important uses of borax?

Ans. Some of the important uses of borax are 

• In the assembling of borosilicate glass (or pyrex glass).
• In making polishes and coatings.
• In hardening of candle wicks.
• In a subjective investigation for a borax bead test in the laboratory.

Ques. What do you mean by halides? Explain it with examples.

Ans. Carbon joins with halogens atoms to frame both simple and blended tetrahalides. On account of basic halides, every one of the four anticipated tetrahalides (for example CF₄, CCl₄, CBr₄ and Cl₄) are known to exist. The stability of the simple tetrahalides diminishes with the rising nuclear mass of the halogen.

(CF₄ > CCl₄ > CBr₄ > Cl₄)

Among the blended halides the better-known compounds are (CFCl₃, CF₂Cl₂ and CCl₃Br).

Ques. What is an allotrope? Give some examples.

Ans. The existence of one or more than one forms of typical chemical element which is seen to occur in the same physical state is known as an allotrope. Carbon, phosphorus and sulfur are a few components that display allotropy.

1. Diamond and graphite are allotropic types of carbon.
2. Red phosphorus and white phosphorus are both allotropes of phosphorus itself.
3. Rhombic sulfur, monoclinic sulfur and plastic sulfur are allotropic types of sulfur.

Ques. Why is the N₂ molecule less reactive at room temperature? (CBSE Al 2015, H.P.S.B. 2015, 2016, Meghalaya S.B. 2017)

Ans. Molecular nitrogen has a triple bond present between the two nitrogen atoms (NsN). It also has some polar characters. The presence of a triple bond gives the nitrogen molecule a very high bond dissociation energy (941.4 kJ mol^-1). Therefore, it becomes difficult to break the bond between the nitrogen molecule and hence nitrogen does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Ques. Give reasons for the following.

1. Sulphur exhibits paramagnetic characteristics in vapour state.
2. F2 is the strongest oxidising agent among halogens.
3. In spite of having the same electronegativity, oxygen forms hydrogen bonds while chlorine does not.

Ans:

1. Sulphur partially exists as S₂ molecule in the vapour state. It has two unpaired electrons in the antibonding π∗π∗ orbitals. Hence, Sulphur in vapor state exhibits paramagnetic characteristics.

2. Due to its small size, F₂ accepts electrons readily. It also has high electronegativity and low dissociation energy.

3. In comparison to chlorine, oxygen has a smaller size and as a result higher electron density per unit volume. Hence, oxygen forms hydrogen bonds and chlorine does not.

Ques. Explain:

1. Helium does not form chemical compounds.
2. Bond dissociation energy of fluorine is less than that of chlorine.
3. Two S-O bonds in SO₂ are identical.

Ans: 

1. Helium is an inert gas and is extremely stable. It does not react with any other element. Hence, there are no known compounds of Helium.
2. There is an inter electronic repulsion between F atoms in a fluorine molecule. Due to this repulsion between F atoms, F-F bond length in fluorine molecules is higher than Cl-Cl bond length in chlorine molecules.
3. While SO₂ is being formed, one electron from 3p orbital goes to the 3d orbital and the S undergoes sp² hybridization. Two such orbitals form sigma bonds with two oxygen atoms which indicates two S-O bonds are identical.

Ques. Out of the following hydrides (H₂O, H₂S, H₂Te) of group 16 elements, which will have Highest bond angle, lowest boiling point, highest electropositive hydrogen, 

i. H₂O

ii. H₂Te

iii. H₂S

Ans:

1. highest bond angle
2. lowest boiling point
3. highest electropositive hydrogen

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Very Long Answer Type Questions [5 Marks Questions]

Ques. Highlight some general trends of the Group 13 elements of the periodic table.

1. Atomic size
2. Ionization enthalpy
3. Metallic character
4. Oxidation states
5. Nature of halides

Ans. Atomic Size: Group-13 elements have a bigger nuclear size since it develops going down a group, hence an additional shell with electrons is added, bringing about a shielding effect. The presence of the shielding effect is high because that the d and f orbitals are totally involved in Group-14, which has a covalent radius and a small radius.

Ionization enthalpy: The group meaningfully affects the ionization enthalpy, which diminishes as the size of the group develops. Since the atomic charge is supplemented by the screening effect, Group 13 has flighty ionization enthalpy. Due to the low shielding impact, Group 14 has a high ionization enthalpy, which then diminishes.

Metallic character: Group-13 elements are known to display a variation in metallic character. It is usually higher in Al than B. Ideally, the reduction potential also increases from Al to Tl, which results in the decrement of electropositivity, leading the metallic character to decrease as well.

Oxidation states: Because of the lone pairs, Group 13 has a great deal of +3 oxidation states, despite the fact that they continuously change to +1 as the grouping request advances. Group 14 has oxidation conditions of +4, +2, and +2, with high enthalpies and covalent nature.

Nature of halides: Halides are covalent mixtures framed by Group 14 accepting or sharing electrons.

Ques. Give reasons for the following

1. AlCl₃ is a Lewis acid.
2. PbO₂ is a stronger oxidizing agent rather than SnO₂.
3. The +1 oxidation state of thallium atoms are more stable than its +3 state.

Ans. 

AlCl₃ is a Lewis acid

Since the Al atoms in AlCl₃ have just six electrons in their valence shell, they can take a couple of octets. Thus, it has capacities as a Lewis acid.

PbO₂ is a stronger oxidizing agent rather than SnO2

Lead and tin are both in the +4 oxidation state in PbO2 and SnO2. Pb2+ particles are more steady than Sn2+ particles as a result of the more grounded inert pair effect. The oxidation state of Sn+4 is more steady than the oxidation state of Sn+2. However, because the +2 oxidation state of Pb is more steady because of the inert pair effect, Pb⁺⁴ effectively converts to Pb⁺² and functions as a good oxidizer.

The +1 oxidation state of thallium atoms is more stable than its +3 state

The Tl +1 oxidation state is more steady than the Th +3 oxidation state because of the inert pair effect. It has a high viable atomic charge because of the presence of d and f orbitals in inward shells, which have a weak shielding effect. Thus, the ns² electron pair turns out to be strongly attracted to the nucleus, avoiding participation in bonding.

Ques. Give reasons for the question, why BCl₃ exist as a monomer whereas AlCl₃ is dimerised through a halogen bridging atoms?

Ans. BCl₃ is an electron-deficient particle as it exists as a monomer. Also, it obtains strength through the deformation of pπ−pπ back holding. Chlorine gives two electrons to Boron's empty 2p-orbital. Accordingly, the backbonding that gives the particle its twofold bond property is delocalized. AlCl₃ is an electron-poor compound too. 

Back bonding is not possible in AlCl₃ in light of the fact that aluminum is more prominent than boron. The octet of aluminum metal is completed by making a planning association with the chlorine particle of another atom, AlCl₃. Therefore, a dimer particle is framed when chlorine atoms bridge between two Al molecules.

Ques. Give the chemical equations involved in the following:

1. Preparation of PH₃ in the laboratory
2. Purification process of PH₃
3. Solution of PH₃ in water on irradiation with light
4. Phosphine on absorption in CuSO₄
5. Prove that PH₃ is basic in nature

Ans:

1. P4 + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
2. Phosphine is absorbed in HI to form PH4I which on treating with KOH gives off phosphine.

PH₄I+KOH→KI+H₂O+PH₃

3. Phosphine in water decomposes into red phosphorus and hydrogen in the presence of light.

PH₃→P₄+6H₂

4. Reaction with CuSO₄:

3CuSO₃+3PH₃→Cu₃P₂+3H₂SO₄

5. PH₃ reacts with acids like HI to produce PH₄I, which is a salt. So PH₃ is basic in nature.

Ques.

1. Explain the anomalous behaviour of nitrogen?
2. Discuss the trend of chemical reactivity of group 15 elements with oxygen, halogens, and metals.
3. H₃PO₃ is a dibasic. Why?

Ans:

1. Nitrogen shows anomalous behaviour because the distance between the lone pair to lone pair electron is minimised.
2. Trends of the chemical reactivity of group 15 elements with oxygen, halogens and metals are discussed below:

Reactivity towards oxygen: group 15 elements form oxides E₂O₃ & E₂O₅. In these types of oxides, the higher oxidation state is more acidic than the lower oxidation state. Moving down the group, acidic character of the elements decreases.

Reactivity towards halogens: these group elements form salts of the form EX₃ and EX₅. Except N, all elements in this group form EX₅. Due to lack of d-orbital in nitrogen it is not able to form these types of oxides. All trihalides formed by these group elements are stable except nitrogen trihalide.

Reactivity towards metals: these group elements form binary compounds with metals with oxidation state of -3

3. The ionised bond of P-H in H₃PO₃ causes the dibasic character. It has two P-OH bonds.

Ques. (i) Considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, draw a comparison between oxidising actions of F₂ and Cl₂.

(ii) How to maximise the yield of H₂SO₄ in the contact process.

(iii) Arrange the following in the increasing order of property mentioned:

(a) H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)

(b) NH₃, PH₃, AsH₃, SbH₃, BiH₃ (Base strength) (Delhi 2016)

Ans. (i) If we consider the standard reduction potential value of fluorine, then we find that it is more than that of chlorine, so fluorine is a stronger oxidising agent.

• The bond dissociation enthalpy of Fluorine is less as compared to that of chlorine.
• The negative electron gain enthalpy of fluorine is slightly less than that of chlorine.
• The hydration enthalpy of fluoride ion is much higher than that of Cl– ion due to smaller size.

(ii) The conditions to maximise the yield of sulphuric acid by contact process are:

• high pressure of about 2 bars
• low temperature 720 K
• presence of V₂O₅ catalyst

(iii) (a) H₃PO₄ < H₃PO₃ < H₃PO₂ (Reducing character)

(b) BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃ (Basic strength)

Ques. Explain the following:

(i) Ozone is thermodynamically unstable.

(ii) Solid PCl₅ is ionic in nature.

(iii) Fluorine forms only one oxoacid HOF.

(b) Draw the structure of (i) BrF₅ (ii) XeF₄ (Delhi 2016)

Ans. (i) The reasons for Ozone being thermodynamically very unstable are as follows:

• The decomposition of ozone into oxygen is exothermic in nature. (ΔH = -ve)
• There is also an increase in entropy which in turn makes ΔG -ve and reaction spontaneous.

(ii) In the solid state, PCl₅ exists as [PCl₃] + [PCl₆]^– in which the cation is tetrahedral and anion is octahedral. Due to the presence of strong attractive forces, it is a solid.

(iii) Due to absence of d-orbitals in fluorine, it can only form one oxoacid i.e., HOF.

b) (i) Structure of BrF₅

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