NCERT Solutions Class 12 Chemistry Chapter 7
NCERT Solutions Class 12 Chemistry Chapter 7 The p Block Elements
The p Block Elements is a widely suggested topic for CBSE students of Class 12. The topic has been explained clearly by the experts at Extramarks to enable the students to get an understanding of the various concepts that come under the chapter.
These solutions prepare an excellent approach to master the subject. Also, these solutions answer any concerns regarding the concept by giving in-depth knowledge through study material on p block elements. In addition to NCERT Solutions Class 12 Chemistry Chapter 7, Extramarks provides important questions, question papers examples and sample solution papers.
Key Topics Covered In NCERT Solutions Class 12 Chemistry Chapter 7
The key topics covered in NCERT Solutions Class 12 Chemistry Chapter 7- Solutions include
|7.2||Groups 15 elements physical and chemical properties|
|7.5||Phosphorus Allotropic forms|
|7.6||Group 16 elements properties|
|7.7||Group 17 elements properties|
|7.8||Group 18 elements properties|
Students can refer to the topics under NCERT Solutions For Class 12 Chemistry Chapter 7 by clicking on the respective exercise.
Below is a brief of the various exercises under NCERT Solutions Class 12 Chemistry Chapter 7.
The classification of elements is as per the Periodic Table, and all these elements are further classified into different groups. In this context, we will study more about p block elements. The studies are under Group 15, 16 and 17 elements with a detailed discussion of properties. The study of the elements and processes is applied in wide applications in industries and various other fields. The discussions here will start from the elements such as N, P, As, Sb and Bi. Here, you will study electronegativity, physical and chemical properties, Ionisationenthalpy, and other crucial properties.
This section of NCERT Solutions Class 12 Chemistry Chapter 7 offers a detailed introduction to – p Block Elements as mentioned above. With these solutions, you can now solve exercise problems with much ease. Besides that, you will also be able to understand and revise the concepts better.
p block elements: The elements from groups 13 to 18 in the periodic table are called p-Block elements. In this p block, elements have a valence shell electronic configuration ns2np1–6. Already we have discussed group13 and 14 elements in the previous class.
7.2 Group 15 Elements
Under NCERT Solutions Class 12 Chemistry Chapter 7, we have detailed Group 15 Elements. The group 15 elements have a valence shell electronic configuration as ns2np3. And the element present in these groups is Nitrogen, Phosphorus, Arsenic, Antimony, and Bismuth. The nitrogen element is different from other elements of this group as it is tiny in size. It makes multiple pπ–pπ bonds with itself and highly electronegative atoms. Elements of this group represent a series of properties similarly.
- They show two crucial oxidation states, +5 and +3.
- They interact with halogen, oxygen, and hydrogen.
Group 15 elements are also called elements of the Nitrogen family and include Nitrogen, Phosphorus, arsenic, antimony and Bismuth. The p-block elements are also called the Representative Elements, placed on the right side of the main periodic table.
As conceived by Dimitri Mendeleev, the modern periodic table arranges all the elements known to man based on their atomic number, rare to every element. The conclusion of this arrangement was the periodic table. The elements with the same properties were rearranged into a column known as a group.
Periodic Trends in Group 15 Elements: In Group 15 elements, as you move down a group starting with the lightest element and finishing with the heaviest, you’d notice a general flow in properties as you move down the order.e.g.,, Nitrogen is a gas and non-metal; as you move down the group, we encounter metalloids then, at the bottom, metal, that is Bismuth. More examples are shared in NCERT Solutions Class 12 Chemistry Chapter 7. These trends are shown in the periodic table to help us better understand atoms’ behaviour and allow us to predict new elements.
|Atomic mass (AMU)||14.01||30.97||74.92||121.76||209.98|
|Valence electron configuration||[He]2s2 2p3||[Ne]3s2 3p3||[Ar]3d10 4s24p3||[Kr]4d10 5s25p3||[Xe]4f14 5d106s26p3|
Boiling point (°C)
|Density (g/cm3) at 25°C||1.15(g/L)||1.8||5.7||6.68||9.79|
|Atomic radius (pm)||56||98||114||133||143|
|First, Ionization energy (kJ/mol)||1402||1012||947||834||703|
|Common Oxidation state(s)||-3 to +5||+5, +3, -3||+5, +3||+5, +3||+3|
|Ionic radius (pm)||146(-3)||212(-3)||58(+3)||76(+3)||103(+3)|
Some of the modern periodic table trends for group 15 elements of the p-Block are shown below. The same has been explained in detail in NCERT Solutions Class 12 Chemistry Chapter 7 study material.
- The valence shell electronic configuration plays a significant role in how an element performs. The electronic configuration of group 15 elements is ns2np3.
- In group 15, all elements have the same arrangement, so they’re similar.
- The s-orbital in this group is filled, and the p-orbitals are half-filled, making their configuration extra stable.
Atomic and Ionic Radii
- When you see the electronic configuration of elements in the table below, you will observe that new orbitals are added to the atom with every step you move downwards.
- This addition of new orbitals will increase both the atomic and the ionic radii of group 15 elements.
- However, from Arsenic to Bismuth, only a tiny increase in ionic radius is observed.
- Filled d and f orbitals in heavier members.
- is the amount of energy required to eliminate an electron from the outermost orbit of the atom.
- Ionisation enthalpy measures how difficult the nucleus is carrying on to the electron.
- The value decreases down the group with the increasing atomic size.
- Due to an increase in distance between the nucleus and the valence shell as we move down the group.
- All the elements of the group survive in a polyatomic state.
- The first, Nitrogen, is gas; however, as you go down, there is a remarkable increase in the metallic character of the Nitrogen element.
- Nitrogen and Phosphorus are non-metals, Arsenic and Antimony are metalloids, and Bismuth is a metal.
- Except for Nitrogen, all the other elements have allotropes.
- The valence shells of the p block elements have an electronic configuration of ns2 np3.
- So the p block elements here can either lose 5 electrons or gain 3.
- The standard oxidation states of these elements are -3, +3, and +5.
- With a decrease in the Ionization enthalpy and electronegativity, therefore, to the increasing atomic radius, the tendency to gain 3 electrons to create a -3 oxidation state decreases down the group.
- Bismuth hardly forms any compound with a -3 oxidation state.
- As we move down, the stability decreases by +5 and that of +3 increases due to the inert pair effect.
7.3 Dinitrogen, N2:
Under NCERT Solutions Class 12 Chemistry Chapter 7, students learn about the following.
Preparation: Air’s liquefaction and fractional distillation produce commercial dinitrogen when N2 distils out of the first (b.p.=77.2K).
NH4Cl (aq) + NaNO2 (aq) → N2 (g) + 2H2O (l) + NaCl (aq)
Very pure nitrogen can be collected by the thermal decomposition of sodium and barium azide solution.
Properties: Dinitrogen is colourless, odourless, without taste and non-toxic gas. It has two stable isotopes: 14N and 15 N. It has very less solubility in water.
As of the high bond enthalpy of N ≡ N, dinitrogen is rather inert at room temperature. At high temperatures, it directly combines with some metals to form primarily ionic nitrides and, with non-metals, covalent nitrides.
7.4 Ammonia Preparation:
In this section of NCERT Solutions Class 12 Chemistry Chapter 7, students learn about ammonia preparation. On a large scale, Ammonia is manufactured by Haber’s process:
N2 (g)+ H2 (g) 3(g)
On a small scale, Ammonia is obtained from ammonium salts which decompose when treated with caustic soda or lime
2NH4Cl + Ca(OH)2—2NH3 +2H2O + CaCl2
Properties: Ammonia gas is a colourless gas with a pungent smell. Its freezing and boiling points are 198.4 and 239.7 K, respectively.
7.5 Phosphorus Allotropes
Phosphorus is formed in many allotropic forms. The important ones are white, red and black. In this section of NCERT Solutions Class 12 Chemistry Chapter 7, students are given more details on Phosphorus allotropes.
- White Phosphorus: It’s a translucent white waxy solid. It is poisonous, not soluble in water, but soluble in CS2 and glows in the dark (Chemiluminescence). This White Phosphorus consists of discrete tetrahedral molecules.
- Red Phosphorus is generated by heating white Phosphorus at 573 K in an inert atmosphere for several days. It is less reactive than White Phosphorus. It is polymeric, consisting of P4 tetrahedral linked together.
- Black Phosphorus has two forms α-black Phosphorus and β-black Phosphorus. α-Black Phosphorus is obtained when red Phosphorus is heated in a sealed tube at 803 K. It can be sublimed in air and has opaque monoclinic or rhombohedral crystals. It does not oxidise in the air. β-Black Phosphorus is prepared by heating white Phosphorus at 473 K under high pressure. It does not burn in the air up to 673 K.
7.6 Group 16 Elements
The elements of group 16 have an electronic configuration of ns2np4 and have a maximum oxidation state of +6. It has varying chemical and physical properties. The preparation of dioxygen in the laboratory is carried out by heating KClO3 with a MnO2 catalyst.
Students may understand the clear concept of p blocks elements by referring to NCERT solutions class 12 chemistry chapter 7.
7.7 Group 17 Elements
Under Exercise 7.7 of NCERT Solutions Class 12 Chemistry Chapter 7, students learn about group 17 elements. The elements of group 17 are found only in the combined state and are highly reactive. The elements that belong to this group are Fluorine, Chlorine, Bromine, Iodine, and Astatine. Their highest oxidation state is +7, whereas the common oxidation state is -1.
7.8 Group 18 Elements
The elements of group 18 are noble gases. Their valence shell electronic configuration except Helium is ns2np6. The electronic configuration of He is 1s2. All the noble gases except radon occur in the atmosphere.
Exercise 7.9 includes frequently asked questions that students may benefit from. Students may refer to NCERT solutions class 12 chemistry chapter 7 study material provided by Extramarks while preparing for their examination.
NCERT solutions class 12 chemistry chapter 7 Exercise & Answer Solutions
NCERT solutions class 12 chemistry chapter 7 is explained in detail by the experts of Extramarks. Students can benefit from the detailed solutions provided by the experts. The Solutions cover all the concepts that fall under Chapter 7, The p Block Elements, such as Dinitrogen preparation, Phosphorus Allotropic forms, Dinitrogen preparation, Group 16 elements properties and more. Students may refer to Extramarks NCERT solutions class 12 chemistry chapter 7 by registering on the website.
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Q.1 Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.
The general characteristics of Group 15 elements:
All the elements in group 15(N, P, As, Sb, Bi) have 5 valence electrons.
The general electronic configuration: ns2 np3
(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. The gain of electron is very difficult as the nucleus have to attract three more electrons. Only nitrogen accepts electrons in its valance shell due to its small 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 have −2 oxidation states (in NH2NH2 and P2H4 respectively). N only shows oxidation state of –1 in NH2OH.
All the elements present in this group show +3 and +5 oxidation states. The stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases due to the inert pair effect.
(iii) Ionization energy and electronegativity:
First ionization energy decreases on moving down a group due to the increase in atomic size. As we move down a group, electronegativity decreases, owing to an increase in atomic radii.
(iv)Atomic size: On moving down a group, the atomic size increases. As the number of shells increases down the group, atomic size also increases.
Q.2 Why does the reactivity of nitrogen differ from phosphorus?
Nitrogen is less reactive than phosphorus. Nitrogen exists as a diatomic molecule. In N2, the two nitrogen atoms are attached via triple bond (N≡N). This triple bond has very high bond strength, which is very difficult to break. It is because of small size of nitrogen, it is able to form pπ−pπ bonds with itself. As a result, nitrogen is inert and unreactive in its elemental state.
On the other hand, phosphorus exists as a tetra-atomic molecule (P4). Since the P—P single bond is much weaker than N≡N triple bond; therefore, phosphorus is much more reactive than nitrogen in its elemental state.
Q.3 Discuss the trends in chemical reactivity of group 15 elements.
General trends in chemical reactivity of group-15 are as follows:
(i) Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit −3 oxidation states.
(ii) 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.
(iii) Reactivity towards halogens: The group 15 elements react with halogens (Cl, Br, I) to form two series of salts: EX3 and EX5. However, nitrogen does not form NX5 as it lacks the d-orbital. The trihalides of N are unstable, however NF3 is stable. Other trihalides are known only as their unstable ammoniates, i.e., NBr3.NH3 and NI3.NH3. The instability of NCl3, NBr3 and NI3 is because of weak N–X bond due to the large difference in size.
(iv) Reactivity towards oxygen: The elements of group 15 form two types of oxides: E2O3 and E2O5, where E = N, P, As, Sb, or Bi. The acidic character decreases on moving down a group. Oxides of N are acidic in nature; Oxides of As and Sb are amphoteric, while oxide of Bi is basic in nature.
Q.4 Why does NH3 form hydrogen bond but PH3 does not?
The electronegativity of N (3.0) is much higher than that of H (2.1). Thus N–H bond is polar and NH3 undergoes intermolecular H-bonding.
In contrast, both P and H both have an electronegativity of 2.1. Thus P–H bond is not-polar and hence PH3 does not undergo H-bonding because one main condition of H-bonding is H atom should be attached to an electronegative atom.
Q.5 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
Small amounts of NO and HNO3 are also formed in this reaction; these impurities can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.
Q.6 How is ammonia manufactured industrially?
On a large scale, ammonia is manufactured by Haber’s process.
The optimum conditions for manufacturing ammonia are:
- Pressure: 200 × 105 Pa.
- Temperature: ~700 K.
- Catalyst is used such as iron oxide with small amounts of A2O3 and K2O.
Q.7 Illustrate how copper metal can give different products on reaction with HNO3.
Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum.
Cu gives different products on reaction with HNO3 depending upon the conc. of nitric acid.
Q.8 Give the resonating structures of NO2 and N2O5.
The resonating structure of NO2 is:
The resonating structure of N2O5 are as follows:
Q.9 The H-N-H angle value is higher than H-P-H, H-As-H and H-Sb-H angles. Why?
|Hydride||H-E-H bond angle|
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. As we move down the group size of the central atom goes on increasing and its electronegativity goes on decreasing. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H−E−H bond angle.
Q.10 Why does R3P=O exist but R3N=O does not (R = alkyl group)?
Due to the absence of d-orbital N cannot participate in pπ-dπ bonding. For this reason nitrogen cannot expand its coordination number beyond four. In R3N=O, N has a covalency of 5.
The compound R3N=O does not exist.
In contrast, P due to presence of d-orbitals form pπ-dπ bonds and hence can expand its covalency beyond 4.
P form R3P=O in which the covalency of P is 5.
Q.11 Explain why NH3 is basic while BiH3 is only feebly basic.
Among the group 15 elements, N has 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. So, NH3 is highly basic but BiH3 is feebly basic.
Q.12 Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
Nitrogen owing to its small size has a tendency to form pπ−pπ multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule (NºN).
On the other hand, due to the larger size of P and low electronegativity, its pπ−pπ bonding is not feasible. Instead it prefers to form P–P single bonds; hence it exists as P4 molecules.
Q.13 Write main differences between the properties of white phosphorus and red phosphorus.
|White Phosphorus||Red Phosphorus|
Q.14 Why does nitrogen show catenation properties less than phosphorus?
Catenation property depends upon the strength of the element-element bond (same elements). Since the N–N bond strength is much weaker than P–P bond strength, N does not show catenation property. Again, due to the smaller size of nitrogen atom, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond.
Q.15 Give the disproportionation reaction of H3PO3.
Orthophosphorus acid (H3PO3) on heating undergoes self-oxidation reduction to give orthophosphoric acid (H3PO4) and phosphine (PH3). This kind of reaction is called disproportionation reaction. The oxidation states of P in various species involved in the reaction are mentioned below.
Q.16 Can PCl5 act as an oxidising as well as a reducing agent? Justify.
The oxidation state of P in PCl5 is +5. Since P has five electrons in its valence shell thus it can donate its 5 valence electrons and cannot increase its oxidation state beyond +5. Therefore, PCl5 can only act as an oxidising agent (which can donate electrons). It can decrease its oxidation state from +5 to +3 and act as an oxidising agent. For example,
In the above reactions, oxidation number of P decreases from +5 in PCl5 to +3 in PCl3 and that of Ag and H increases from 0 to +1. Thus PCl5 acts as an oxidising agent.
Q.17 Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
The elements of group 16 are collectively called chalcogens.
(i) Electronic configuration: All these elements have the same ns2np4 (n=2 to 6) valence shell electronic configuration and hence are justified to place in group 16 of the periodic table.
(ii)Oxidation states: As these elements have six valence electrons (ns2np4), they accept 2 electrons to complete their octets and they can 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 (Like H2O, H2S, H2Se, H2Te and H2Po).
Q.18 Why is di-oxygen a gas but sulphur a solid?
Due to small size and high electronegativity, oxygen forms pπ-pπ multiple bonds. As a result oxygen can exist as diatomic (O2) molecules. These molecules are held together by weak van der Waals forces of attraction which can easily overcome by the collision of the molecule at room temperature. So O2 exist as gas at room temperature.
On the other hand, due to the bigger size and lower electronegativity, S does not form multiple bonds. Instead, it prefers S–S single bond. Again, S–S single bond is stronger than O–O single bond; more energy is required to break this bond. S has a much greater tendency to undergo catenation and lower tendency for pπ–pπ multiple bonds. It exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid at room temperature.
Q.19 Knowing the electron gain enthalpy values for
O →O− and O →O2− as −141 and 702 kJmol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O−?
Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable will be the ionic compound.
Consider the following reaction of a divalent metal (M) with oxygen. The formation of MO2 and MO involves the following steps:
If electron gain-enthalpy were the only factor involved in the formation of O2– ions, we would expect that O will prefer to form O– rather than O2– ions. But a large number of oxides have O2– species and not O– is present. The reasons behind it are
- O2– is more stable than O– because it has stable noble gas configuration.
- Due to the higher charge on O2− than on O–ion, the lattice energy released during the formation of oxides containing O2− species in the solid state is much higher than lattice energy produced during the formation of oxides containing O– species. The lattice energy of formation of MO type of species is quite higher that can compensates the higher electron gain enthalpy (ΔegH2) needed during the formation of O2– to O– ions. Thus the formation of MO variety is energetically more feasible than the compound of O–ions. For that reason, oxygen forms a large number of oxides having O2– species and not O–.
Q.20 Which aerosols deplete ozone?
Aerosols such as chlorofluorocarbons (CFCs) and Freon (CCl2F2) 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. The ozone layer depletes By the free radical mechanism, initiated by chlorine free radicals. This causes hole in the layer.
Q.21 Describe the manufacture of H2SO4 by contact process?
Sulphuric acid is manufactured by the Contact Process which involves three steps:
(i) Production of SO2 by burning sulphur or roasting of sulphur pyrites.
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.
Q.22 How is SO2 an air pollutant?
SO2 can act as an air pollutant because of the following reasons:
(i) Even in very low concentrations (5 ppm), SO2 causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
(ii) It combines with rain water present to form sulphurous acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble (CaCO3).
(iii) Even in low concentration (0.03 ppm) of SO2 is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves (loss of green colour). This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.
Q.23 Why are halogens strong oxidising agents?
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.
Due to low bond dissociation enthalpy, high electronegetivity and large electron gain enthalpy halogens have a strong tendency to accept electrons and get reduced. They only can act as strong oxidising agents.
According to the electrode potentials of halogen atoms it is evident that F2 is the strongest oxidising agent and I2 is the weakest oxidising agent.
Q.24 Explain why fluorine forms only one oxoacid, HOF.
Fluorine forms only one oxoacid (HOF), because of its high electronegativity and small size and in this compound oxidation state of F is +1.
Again, due to the absence of d-orbitals, F can’t expand its valency beyond +1 (i.e. +3, +5, and +7); hence F does not form higher oxoacids such as HOFO, HOFO2, HOFO3.
Q.25 Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Both chlorine and oxygen have almost the same electronegativity values, but the size of chlorine is larger than oxygen atom. Thus electron density per unit volume on oxygen is higher than that of chlorine. So oxygen is able to form hydrogen bond but chlorine cannot.
Q.26 Write two uses of ClO2.
The uses of ClO2:
(i) It is an excellent bleaching agent.
(ii) It is a powerful oxidizing agent and chlorinating agent.
Q.27 Why are halogens coloured?
The halogens are coloured because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region while the remaining light is transmitted. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.
For example, F2 absorbs violet light (higher excitation energy) and hence appears pale yellow while I2 absorbs yellow and green light (lower excitation energy), so it appears deep violet. For the same reason Br2 shows orange red colour and Cl2 shows greenish yellow colour.
Q.28 Write the reactions of F2 and Cl2 with water.
Reactions are as follows:
Q.29 How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only.
(a) HCl can be oxidized to Cl2 by a number of oxidizing agents such as MnO2, KMnO4, K2Cr2O7 etc.
MnO2 + 4 HCl → MnCl2 + Cl2 2H2O
(b) HCl can be prepared from Cl2 by dissolving it in water.
Q.30 What inspired N. Bartlett for carrying out reaction between Xe and PtF6?
Neil Bartlett initially carried out a reaction in which oxygen reacts with PtF6 to yield an ionic solid, O2+PtF6–.
In this reaction, O2 gets oxidized to O2+ by PtF6.
O2(g) + PtF6(g) → O2+[PtF6]–
Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. He thought that PtF6 should oxidize Xe to Xe+. Thus, he tried to prepare a compound with Xe and PtF6. When Xe and PtF6 were mixed, a rapid reaction took place and a red solid with formula, Xe+PtF6– was formed.
Q.31 What are the oxidation states of phosphorus in the following:
(i)H3PO3, (ii) PCl3, (iii) Ca3P2, (iv)Na3PO4 , (v)POF3?
Let the oxidation state of P be x.
(i) H3PO3 :
3(+1) + x +3(–2) = 0
Or, x = +3
(ii) PCl3 :
x + 3(–1) = 0
Or, x = +3
3(+2) + 2 x = 0Or, x = –3
(iv) Na3PO4 :
3(+1) + x + 4 (–2) = 0
Or, x = +5
(v) POF3 :
x + 1(–2) + 3(–1) = 0
Or, x = +5
Q.32 Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO2.
(ii) Chlorine gas is passed into a solution of NaI in water.
The required reactions are as follows:
Q.33 How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
XeF2, XeF4, and XeF6 are obtained by a direct reaction between Xe and F2. The products obtained are dependent on the reaction conditions.
Q.34 With what neutral molecule is ClO− isoelectronic? Is that molecule a Lewis base?
ClO– has (17 + 8 + 1) = 26 electrons and OF2 has (8 + 2 x 9) = 26 electron system. ClO− is isoelectronic to OF2.
If we replace O–(9 elecrtons) in ClO– by F (9 electrons) atom, the resulting neutral molecule is ClF.
Since Cl in ClF can further donate electrons to two more F atoms to form ClF3. Therefore, ClF acts as a Lewis base.
Q.35 How are XeO3 and XeOF4 prepared?
(i) XeO3 can be prepared in two ways as shown
6XeF4 + 12H2O → 4Xe + 2XeO3 +24HF + 3O2
XeF6 + 3H2O → XeO3 + 6HF
(ii) XeOF4 can be prepared using XeF6.
XeF6 + H2O → XeOF4 +2HF
Q.36 Arrange the following in the order of property indicated for each set:
(i) F2, Cl2, Br2, I2 : increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI : increasing acid strength.
(iii) NH3, PH3, AsH3, SbH3, BiH3 − increasing base strength.
(i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. The bond dissociation energy is decreased as it requires less energy to break the bonds as we moving down the group (bond distance increases from F2 to I2). The bond dissociation energy of F2 molecule is very less than Br2 and Cl2, it is due to the fact that F-atom is very small in size and hence the three lone pairs of electrons on each F atom repel the bond pairs holding the F-atom in the F2 molecule.
Thus, the increasing order for bond dissociation energy among halogens is as follows:
I2< F2< Br2< Cl2
(ii) The relative acid strength of HF, HCl, HBr, and HI depends on the bond dissociation enthalpies. Since the bond dissociation enthalpies of H–X bond decreases from H–F to H–I as the size of the atom increases from F to I. Thus the acid strength increases in the following order.
HF < HCl < HBr < HI
(iii)Due to the presence of lone pair of electrons on the central atom in AH3, all behave as Lewis bases. On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom and as well electronegativity also decreases and hence the basic strength decreases as we move from NH3 to BiH3.
BiH3< SbH3< AsH3< PH3< NH3
Q.37 Which one of the following does not exist?
(i)XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6
The sum of 1st and 2nd ionization enthalpies of Ne is much higher than those of Xe. Thus F2 can oxidize Xe to Xe2+ but cannot oxidize Ne to Ne2+. So,NeF2 does not exist but all the xenon fluorides (XeF4 and XeF6) and xenon oxyfluoride (XeOF4) do exist.
Q.38 Give the formula and describe the structure of a noble gas species which is iso-structural with:
(i)ICl4– (ii) IBr2– (iii) BrO3–
(i) ICl4– is iso-electronic with XeF4 and a square planar geometry
(ii) XeF2 is iso-electronic to IBr2– has a linear structure.
(iii)XeO3 is iso-structural to BrO3– and has a pyramidal molecular structure.
Q.39 Why do noble gases have comparatively large atomic sizes?
Noble gases are inert in nature; they don’t form molecules so the atomic radii of the noble gasses correspond to van der Waal’s radii. By definition, van der Waal’s radii are larger than covalent radii. That is why noble gases have comparatively large atomic sizes.
Q.40 List the uses of Neon and argon gases.
Followings are the uses of neon gas:
(i) It is mainly used in discharge tubes and fluorescent lamps which are used in advertising purposes.
(ii) It is used in beacon lights.
(iii) It is filled in discharge tubes with characteristic colours.
Followings are the uses of argon gas:
(i) It is used to provide an inert temperature in a high metallurgical process. (ii) It is also used in laboratories to handle air-sensitive substances.
(iii)Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.
FAQs (Frequently Asked Questions)
1. How important are the NCERT solutions class 12 chemistry chapter 7?
The class 12 chemistry chapter 7 NCERT solutions help students with various concepts of The p Block Elements. The Solutions give a detailed explanation of the elements that fall under the chapter. The NCERT Solution allows students to study for the exams without depending on someone to explain. Students may refer to NCERT solutions class 12 chemistry chapter 7 for free on Extramarks.
2. Do NCERT solutions class 12 chemistry chapter 7 cover all exercises of the chapter?
Yes, NCERT solutions class 12 chemistry chapter 7 cover all exercises in detail and hence students can depend on the same to study. In addition to understanding the chapter, students can use the Solution – The p Block Elements as revision notes.