NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements- In this chapter, students will get to know the general introduction of p-block elements. P-block elements are arranged in six groups of the periodic table which are 13, 14, 15, 16, 17, and 18. This chapter covers only two groups 13 and 14, the rest groups will be covered in the class 12th chapter 7 The p-Block elements. In this chapter, there are 38 questions in the exercise. The NCERT solutions for Class 11 Chemistry Chapter 11 The p-Block Elements are designed and explained by our subject experts. These NCERT solutions for class 11 help students in their preparation of both CBSE class 11 final examination and in the competitive exams like JEE Main, NEET etc. It is recommended to students to go through CBSE NCERT solutions for Class 11 Chemistry Chapter 11 The p-Block Elements thoroughly to maximise their marks in examination. Since this is a theoritical chapter, hence most of the concepts will be learnt directly from solutions of NCERT for Class 11 Chemistry Chapter 11 The p-Block Elements. This is perhaps the longest chapter of inorganic chemistry with lot of things to mug up. Boron family and carbon family are very important topics of this chapter. Hence It is highly recommended to go through these two topics very thoroughly. If you are looking for the answers of any other class from 6-12 then NCERT solutions are there for you as it's the easiest way to get all the solutions of NCERT.
The NCERT Class 11 chemistry chapter 11 p-Block Elements is an important chapter for the class 11 students because various questions are asked from this topic in class 11 final examination. After completing the theory as well as NCERT solutions for Class 11 Chemistry Chapter 11 The p-Block Elements students will be able to explain the properties and electronic configuration of p-block elements; general trends in the p-block elements like atomic radius, electronegativity, ionisation enthalpy and electron gain enthalpy; able to describe the trends in the chemical and physical properties of group 13 and group 14 elements; explain anomalous behaviour of boron and carbon; describe allotropic forms of carbon and explain explain properties of important compounds of boron, carbon and silicon.
In p-block elements, the last electron enters the p-orbital of the outermost energy shell. The electronic configuration of p-block elements is (except for helium He). The number of p-orbitals is three and the maximum number of electrons that can be filled in a set of p-orbitals is six. There are a total of six groups from 13 to 18 in the p-block elements and boron, carbon, nitrogen, oxygen, fluorine and helium head the groups. The general electronic configuration and the important oxidation states of p-block elements are shown in the table given below-
General electronic configuration
The first member of the group
Group Oxidation state
Other oxidation states
||+4, +2, -2
||+5, +3, +1, -1
+6, +4, +2
Topics of NCERT Grade 11 Chemistry Chapter 11 The p-Block Elements
11.1 Group 13 Elements: The Boron Family
11.2 Important Trends and Anomalous Properties of Boron
11.3 Some Important Compounds of Boron
11.4 Uses of Boron and Aluminium and their Compounds
11.5 Group 14 Elements: The Carbon Family
11.6 Important Trends and Anomalous Behaviour of Carbon
11.7 Allotropes of Carbon
11.8 Some Important Compounds of Carbon and Silicon
NCERT Solutions for Class 11 Chemistry Chapter 11 The P-Block Elements- Exercise Questions
Question 11.1(i) Discuss the pattern of variation in the oxidation states of
These elements belong to group 13 of the general electronic configuration. These elements have three valence electrons, therefore, they can exhibit a maximum only +3 oxidation state. Boron shows only +3 oxidation state but other elements also show +1 oxidation state. On moving down the group from, its stability increases due to inert pair effect. After removal of the last electron from p- orbital, the remaining two electrons behave like inert gases and don't take part in bonding. They are strongly attracted by the nucleus. This reluctance of the s- electron pair to take part in bond formation is called inert pair effect.
Question 11.1(ii) Discuss the pattern of variation in the oxidation states of
These elements belong to group 14, of the general electronic configuration . These elements have four valence electrons, therefore, they can exhibit commonly +4 oxidation state. On moving down the group from +2 oxidation state becomes more and more common. Carbon and silicon mostly show +4 oxidation state, its stability decreases on moving down the group. It is due to the inert pair effect. Though shows both +2 and +4 oxidation states. The stability of the lower oxidation state decreases down the group.
Question 11.2 How can you explain higher stability of as compared to ?
Both boron and Thallium belong to the 13th group of the periodic table. We know that the stability of lower oxidation state (+1) is increases down the group. Thus is more stable than . It is due to the +3 of Boron is more stable than +3 of Thallium. is highly oxidising in nature it reverts back to its +1 oxidation state by gaining an electron.
Question 11.3 Why does boron triflouride behave as a Lewis acid ?
The electronic configuration of boron is . It has three valance shell electrons. It can form only three covalent bonds. It means Boron has only 6 electrons and its octet is not complete, boron needs 2 more electrons to complete its octet. In , its octet is incomplete. Hence it is an electron deficient molecules so it acts as a Lewis acid.
Question 11.4 Consider the compounds, and . How will they behave with water ? Justify.
is a Lewis acid, it is an electron deficient molecules. So, it readily undergoes to hydrolysis. On hydrolysis gives boric acid.
does not undergo hydrolysis because carbon has no vacant orbital, so it cannot accept an electron from the water molecule to form intermediate.
+ water no reaction
Question 11.5 Is boric acid a protic acid ? Explain.
Boric acid is not a protic acid. It is a weak monobasic acid, behaves as a Lewis acid by accepting a pair of an electron from the water molecules()
Question 11.9 What are electron deficient compounds ? Are and electron deficient species ? Explain.
Electron-deficient compounds are those in which the octet of the central atom is not completed. Therefore, the metal wants to complete its octet by accepting an electron from a donor molecule.
is an electron-deficient compound because of the central atom, Boron has only 6 electrons, two electrons are needed to complete its octet.
silicon has 4 valence electrons and by covalent bonding, it gets 4 electrons from four chlorine atom. Hence, is not an electron-deficient molecule.
Question 11.15. If B–Cl bond has a dipole moment, explain why molecule has zero dipole moment.
Due to differences in electronegativities of Boron and chlorine, the partial charges develop in B-Cl bond. is a non-polar compound. This molecule is trigonal planner shape (symmetrical) and we know that dipole moment is a vector quantity. In fig. we can see that, they can cancel out each other.
Question 11.17. Suggest a reason as to why is poisonous.
carbon monoxide is more poisonous than carbon dioxide, (non-toxic in nature). has the ability to block delivery of oxygen to the organ and tissue. Also, it binds to haemoglobin to form a complex carboxyhaemoglobin, which is 300 times more stable than the oxygen-haemoglobin complex. If the concentration of these complex reaches to 3-4% then the capacity of blood to carry oxygen is reduced. On the other hand, is not poisonous, it is harmful only at high concentration.
Question 11.18. How is excessive content of responsible for global warming ?
Carbon dioxide can ability to trap the heat emitted by the sun radiation. Higher the amount of , higher the amount of heat trapped. Therefore, this results in an increase in temperature of the atmosphere causing global warming. Hence, excessive content of carbon dioxide is a threat to us.
Question 11.19. Explain structures of diborane and boric acid.
The four terminal H atom and Boron lie in the same plane but two bridge H atom is below the plane. The four terminal B-H bonds are regular two centre-two electron bonds. The two (B-H-B) bonds are known as banana bonds(3-centre-2-electron bonds).
It is a weak monobasic acid. It is Lewis acid, not protic acid. It has a layered structure, the molecule is linked by hydrogen-bonding with another molecule. The dotted line represents H-bonding.
Question 11.20(a) What happens when
Borax is heated strongly?
The molecular formula of borax is. On heating, it loses water molecules and converts into sodium metaborate() and if we continue the heating process it becomes anhydride, called as boric anhydride.
Question 11.22(iii) Give reasons :
Graphite is used as lubricant.
Graphite has a layered structure and each layer is bonded by the weak van der Waals forces. These layers can slide over each other that's why graphite can be used as a lubricant.
Question 11.22(iv) Give reasons :
Diamond is used as an abrasive.
The carbons of the diamond are in hybridisation. Each carbon is attached with the four neighbour carbon atom by a strong covalent bond. This covalent bond gives the diamond a very rigid structure and it is very difficult to break this covalent bond and due to this reason it is the hardest substance and used as an abrasive
Question 11.22(v) Give reasons :
Aluminium alloys are used to make aircraft body.
Aluminium has very good tensile strength and is very light in weight also. It can be alloyed with the various metals such as copper, manganese and Silicon etc. Al is malleable and ductile. Because of these properties of Aluminium, it can be used in making aircrafts bodies.
Question 11.22(vi) Give reasons :
Aluminium utensils should not be kept in water overnight.
Aluminium utensils should not be kept in water overnight because it reacts with water with the help of atmospheric oxygen and forms an oxide layer on the surface of utensils. When we kept water on this utensils for a long period of time, some amount of oxide dissolved in water which is harmful(water become basic in nature).
Question 11.24 How would you explain the lower atomic radius of as compared to ?
The atomic radius of the gallium is less than the atomic radius of aluminium. This is because of the weak shielding effect of the 3d electron of gallium. The shielding effect of d-electrons are feeble so that the effective nuclear charge experienced by the valence electron in is quite higher than the
Question 11.25 What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?
The various forms of the same element are known as allotropes. They have the same chemical properties but different physical properties.
Allotropes of carbon -
In diamond, all the carbon is hybridised and each of the carbon is attached with the four neighbour carbon atom. by a strong covalent bond. This provides it with a very rigid structure and makes it the hardest substance.
In Graphite, carbon is hybridised and attached with three neighbours carbon. It has a layered structure, which is attached with the weak van der Waal forces. These layers can slide over one another so that's why they are soft and used as a lubricant.
Question 11.26 Classify the given oxide as neutral, acidic, basic or amphoteric. Write suitable chemical equations to show their nature.
is a neutral
Being acidic, the oxides react with bases to form a salt. So take base = sodium hydroxide ()
these compounds react with a base to form salts. ex.- sodium metaborate, sodium silicate and sodium carbonate respectively.
It reacts with an acid to form salts. let acid = hydrochloric acid (). for example thallium chloride.
These oxides have a tendency to react with acid as well as with the base.
The reaction of lead oxide with acid and base-
The reaction of aluminium oxide with base and acid-
Question 11.28 When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of to give soluble complex (B). Compound (A) is soluble in dilute to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.
In given information, when X is treated with the sodium hydroxide it gives A(white ppt) and it is soluble in excess of . Thus X must be aluminium. So, the white ppt of A is aluminium hydroxide.
The complex B is soluble in excess of sodium hydroxide So it means it is a sodium tetrahydroxy aluminate(II) complex.
Now, when we add dilute HCl to aluminium hydroxide it gives the compound C (Aluminum chloride).
And when compound (A) is heated strongly, it gives D and this compound used to extract metal X
Question 11.29(a) What do you understand by
inert pair effect
Inert pair effect-
On moving down the group in the periodic table, the tendency of -orbital electron to participate in bonding is decreased. This effect is known as the inert pair effect. For example, in group 13 element ()the stability of +1 oxidation is more than the +2 oxidation state due to the poor shielding of the electrons by the and electron, as a result, electrons are strongly held by the nucleus.