Welcome to Swastik Classes’ NCERT Solutions for Class 11 Chemistry Chapter 10 – S-Block Elements. This chapter delves into the intriguing world of s-block elements, which occupy the leftmost two columns of the periodic table. Understanding the properties and characteristics of these elements is vital for building a strong foundation in chemistry.
Our meticulously crafted solutions aim to simplify the complex concepts and principles covered in this chapter. Whether you’re grappling with the trends in the properties of s-block elements or the reactions and compounds they form, our detailed explanations and step-by-step solutions will guide you towards a deeper comprehension.
The team of experienced educators at Swastik Classes has designed these solutions in accordance with the NCERT (National Council of Educational Research and Training) syllabus. Each solution is carefully structured to provide clarity and accuracy, helping students excel in their examinations and develop a profound understanding of the subject matter.
By utilizing these NCERT solutions, you can enhance your problem-solving skills, reinforce your conceptual knowledge, and improve your overall performance in chemistry. We encourage you to make the most of this resource and embark on a journey towards excellence in the realm of s-block elements.
Swastik Classes is committed to providing quality education and empowering students to succeed academically. We believe that by offering comprehensive and reliable solutions, we can contribute to your educational journey and foster a love for learning.
So, let’s dive into the captivating world of s-block elements with Swastik Classes’ NCERT Solutions for Class 11 Chemistry Chapter 10. Get ready to unravel the mysteries of these elements and unlock your true potential in the fascinating field of chemistry!
NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 10-S-BLOCK ELEMENTS – Exercises
Chapter-10 S-Block Elements
Answer the following Questions.
1. Mention the general chemical and physical features of the alkali metals?
(1) The alkali metal is soft and so we can cut them easily. We can able to cut the sodium metal even by using the knife.
(2) Generally the alkali metal is lightly coloured and mostly they appear as silvery white.
(3) Its atomic size is larger and so their density is low. The density of the alkali metal increases down in the group from Li to Cs except K which has low density than sodium.
(4) Alkali metal is weak in its metallic bonding and so they are having low boiling and melting points.
(5) The salts present in alkali metals exposes colour to flames because, heat of the flame excites electron which is located on the outer orbital to higher energy level. In this excited state of electron getting reversed back to the ground level, the emission of excess energy in the form of radiation falls in visible region.
(6) Metals like K and Cs loses electrons when they get irradiated with light and also displays photoelectric effect.
(1) Alkali metal reacts with water and forms oxides and hydroxides. So, the reaction will be more spontaneous while moving down the group.
(2) Alkali metal reacts with water and forms dihydrogens and hydroxides.
General reaction :
2M + 2H20 → 2M+ + 2OH– + H2
(3) Dihydrogen reacts with alkali metals and forms metal hydrides. The hydrides from this have higher melting points and they are solids which are ionic.
2M + H2 → 2M+ H–
(4) Alkali metals directly reacts with halogens and forms ionic halides except Li.
2M + CL2 → 2MCI (M = Li ,K, Rb, Cs)
It has the ability to easily distort the cloud of the electron which is around the –ve halide ion because, lithium ion is smaller in size . Hence, Lithium halide is naturally covalent.
(5) Alkali metals are very stronger reducing agents. This increases as we move down the group except lithium. Due to its high hydration energy it results in strong reducing agent among all alkali metals.
(6) To result in blue coloured solution(deep blue) which are naturally conducting, they get dissolved in liquid ammonia.
M +(x + y) NH3 → [ M ( NH3 )x ]+ + [ M ( NH3)y ]–
2. Classify the gradation properties and the general characteristics of the alkaline earth metals.
(i)(Noble gas) ns2 is the electronic configuration of alkaline earth metal.
(ii) To occupy the nearest inert gas configuration, these metals lose two of their electrons; and so its oxidation state is +2.
(iii) The ionic radii and atomic radii is smaller than alkali metals. When they moved down towards the group, there is an increase in ionic radii and atomic radii due to decrease in effective nuclear charge.
(iv) The ionization enthalpy is low because the alkaline earth metals are larger in size. The first ionization enthalpy is higher than metals of group 1.
(v) They appear in lustrous and silvery white. They are soft as alkali metals.
(vi)Factors that cause alkaline earth metals to contain high boiling point and melting point :
(*) Atoms of alkali metals are larger than that of alkaline earth metals.
(*) The strong metallic bonds are formed by two valence electrons.
(vii) Ca- brick red, Sr- crimson red, Ba-apple green results in colours to flames.
Electrons are bounded strongly to get excited in Be and Mg. Therefore, they do not expose any colours to the flame.
The alkali metals are more reactive than alkaline earth metals.
(i) Reaction with water air and: Due to the formation of oxide layer on their surface, beryllium and Mg are most inert to water and air.
(a) Beo and Be3 N2 is formed when powdered Be is burnt in air.
(b) For the formation of MgO and Mg3 N2, Mg is burnt in air with dazzling sparkle. since, Mg is more electropositive.
(c) The formation of respective nitrides and oxides is by instant reaction of Sr,Ca, and Ba with air.
(d) Ca,Sr, and Ba can able to react vigorously even with water which is cold.
(ii) when they react with halogens, halides is formed at high temperature.
M +X2 → MX2 (X = F, CL, Br, I)
(iii) Except Be, all the alkaline earth metals react with hydrogen to form hydrides.
(iv) alkaline earth metals instantly react with acids to form salts with the liberation of hydrogen gas.
M +2HCL→ MCL2 + H2(X)
3. Alkali metals are not found naturally. Why ?
Sodium, cesium, lithium, francium, potassium, rubidium all together comprises of alkali metals. they consist of only one electron on its valence shell, which gets loosed easily due to their low ionizing energies. Alkali metals are not found naturally in their elemental state and also they are highly reactive.
4. Mention the oxidation state of Na in Na2O2.
Let the oxidation state of Na be y.
In case of peroxides, the oxidation state of oxygen is -1.
2(y) + 2(-1) = 0
2y – 2 = 0
2y = 2
Y = +1
Therefore, the oxidation state of Na is +1.
5. Why Na reacts lesser than potassium?
on moving down the group in the alkali metals, the size of the atom increases and the effect of the nuclear charge gets decreased. Due to this factors, the electron of potassium which is located outer gets lost easily as compared to Na. therefore, potassium reacts higher than Na.
6. State comparison between alkaline earth metals and alkali metals w.r.t.
1.Solubility of hydroxide
2.Ionization enthalpy and
3.Basicity of oxides
|Sr. No||Alkaline earth metals||Alkali Metals|
|a)||Solubility of hydroxide:They are less soluble compared to alkali metals as it has high lattice energy and is having higher charge densities account for higher lattice.||Solubility of hydroxide:They are more soluble compared to alkaline earth metals.|
|b)||Ionization Enthalpy:They have smaller atomic size and higher effective nuclear charge compared to alkali metals, which causes their 1st ionization enthalpy higher than that in alkali metals, but the 2nd ionization enthalpy is less than that of alkali metals||Ionization Enthalpy:They have large atomic size compared to alkaline earth metals, so they are having less 1st ionization enthalpy, so they lose valance electrons very easily.|
|c)||Basicity of oxides:Their oxides are quite basic but less as compared to those of alkali metals as they are less electro positive than alkali metals.||Basicity of oxides:Their oxides are basic in nature as they are highly electropositive, which makes their oxides highly ionic.|
7. Mention the similarities of lithium and magnesium irrespective of its chemical behaviour.
Similarities between lithium and magnesium:
(i) lithium and magnesium reacts slow with cold water.
(ii) oxides of lithium and magnesium are less soluble in H2O. also the hydroxides of both decompose at high temperature.
2LiOH → Li2O + H2O Mg(OH)2 → MgO + H2O
(iii) Nitrides is formed from both the lithium and magnesium when they react with N2.
6Li + N2 → 2Li3N 3Mg + N2→ Mg3N2
(iv) Neither Li nor Mg form superoxides or peroxides.
(v) Both the carbonates of lithium and magnesium are naturally covalent. They decomposes on heating.
Li2CO3 → Li2O + CO2 MgCO3 → MgO + CO2
(vi) They do not form bicarbonates which are solid.
(vii) Both MgCl2 and LiCl are soluble in ethanol because they are naturally covalent.
(viii) Both MgCl2 and LiCl are naturally deliquescent . They crystallize as hydrates from aqueous solutions.
LiCL.2H2O and MgCL2 . 8H2O
8. Alkaline earth metals and alkali cannot be obtained by chemical reduction methods. Explain why?
By using a stronger reducing agent, the oxides of metals gets reduced by the process called chemical reduction. Alkaline earth metals and alkali metals are strong among the reducing agents. No stronger reducing agent is available than them. Therefore alkaline earth metals and alkali cannot be obtained by chemical reduction of their oxides.
9. Potassium and Cesium is used in photoelectric cells than lithium. Explain why?
Lithium, potassium, and cesium, are all alkali metals. But still, potassium and cesium are used in photoelectric cell and not Lithium because Li is smaller in size when compared to the other two.
On the other hand, cesium and potassium have low ionization energy. Therefore, they lose electrons easily. This property is utilized in photoelectric cells.
10. The solution of dissolved alkali metal in liquid ammonia acquire different colours. Explain the reason for change of colour.
When the alkali metal is dissolved in liquid ammonia, a deep blue coloured solution is formed.
M+ (x+y) NH3 → M+ (NH3)x +e−3 (NH3)y
The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is deep blue in colour.
Clusters of metal ions are formed at higher concentration (3M) which causes the solution to attain a copper-bronze colour and a metallic lustre.
11. Other alkaline earth metals give colour to flame whereas beryllium and magnesium wont. Explain why?
The valence electrons get excited to a higher energy level when an alkaline earth metal is heated.
It radiates energy which belongs to the visible region when this excited electron comes back to its energy level which is low. The colour is observed here. The electrons are strongly bound in the beryllium and magnesium. The energy required to excite these electrons is very high. When the electron reverts back to its original position, the energy released does not fall in the visible region. Hence, no colour is seen in the flame.
12. Explain the different reactions that occur in the solvay process.
The process of preparing sodium carbonate is called solvay process. Sodium hydrogen carbonate is formed when carbon dioxide gas is bubbled through a brine solution saturated with ammonia. The obtained sodium hydrogen carbonate is then converted into sodium carbonate.
(i) Brine solution is saturated with ammonia.
2NH3 + H2O + CO2 → (NH4)2 CO3
This ammoniated brine is filtered for purity removal.
(ii) When carbondioixide reacts with ammoniated brine, it results in the formation of insoluble sodium hydrogen carbonate.
NH3 + H2O + CO2 → NH4 HCO3 NACL + NH4 HCO3 → NH4 HCO3 + NH4CL
(iii) NaHCO3 is obtained by the solution which contains crystals of NaHCO3 is filtered.
(iv) NaHCO3 is heated strongly to convert it into NaHCO3.
2Na HCO3→ Na2 CO3 +CO2 + H2O
(v) Ammonia is recovered when the filterate which is removed after NaHCO3 is mixed with Ca (OH)2 and heated.
Ca(OH)2 + 2NH4CL → LNH3 + 2H2O + CaCL2
The overall reaction taking place in Solvay process is
2NaCL+ CaCO3 → Na2CO3 + CaCL2
13. Potassium carbonate cannot be prepared by Solvay process. Explain why?
Solvay process is not applicable for the preparation of potassium carbonate because potassium carbonate is soluable in water and it doesn’t precipitates out like sodium bicarbonate.
14. Li2 CO3 gets decomposed at lower temperature but Na2 CO3 decomposes at high temperature. Explain why?
The electropositive character increases while moving down in the group of alkali metal which results in increase in stability of alkali carbonates. Generally, lithium carbonate is not stable when it reacts to heat because lithium carbonate is covalent. Due to the smaller size of lithium ion it polarizes large carbonate ion which results in the formation of stable lithium oxide.
Li2 Co3 → Li2O + CO2
This is why sodium carbonate decomposes at high temperature and lithium carbonate decomposes at low temperature.
15. Compare thermal stability vs solubility of compounds of alkaline earth metals with alkali metals.
The alkali metals carbonates are very stable to heat. But carbonates of lithium decomposes and results in the formation of lithium oxide while heating. The carbonates of alkaline earth metals also decompose which results in the formation of carbon dioxide and oxide while heating.
Na2CO3 → No Effect Li2 Co3 → Li2O + Co2 MgCo3 → MgO + Co2
Exception of Li2 CO3, the carbonates of alkali metals are soluble in water and also while we move down the group, the solubility increases. Carbonates of alkaline earth metals are insoluble in water.
Except LiNO3, the nitrates of alkali metals gets decomposed while heating strong which results in the formation of nitrites.
2KNO3 → 2KNO2 + O2(g)
LiNO3, on decomposition, gives oxide.
2LiNO3 → Li2O + 2NO2(g) + O2(g)
Like lithium nitrate, alkaline earth metal nitrates also decomposes to give oxides.
2Ca(NO3) → 2CaO + 4NO2(g) + O2(g)
Nitrates of both group 1 and group 2 metals are soluble in water.
Sulphates of both group 1 and group 2 metals are stable towards heat.
Sulphates of alkali metals are soluble in water. But the sulphates of alkaline earth metals shows various activities,
CaSO4 Sparingly soluble
BeSO4 Fairly soluble
16. How would you convert Sodium metal into
(i) sodium carbonate?
(ii) sodium peroxide
(iii) sodium hydroxide
(iv) sodium metal
(i) Sodium carbonate
Sodium hydrogen carbonate is obtained as a precipitate by reacting sodium chloride with ammonium hydrogen carbonate. The resultant crystals can be heated to obtain Sodium Carbonate.
2NH2 + H2O → (NH4)2 CO3
(NH4)2 CO3 + H2O + CO2 → 2NH4 HCO3
2NH4 HCO3 + NaCl → NH4Cl + Na HCO3
The resultant crystals can be heated to obtain Sodium Carbonate.
2Na HCO3 → Na2 CO3 + CO2 + H2O
(ii) Sodium peroxide
After Na metal is gotten from Downs process, it is heated on Aluminium trays in presence of air(without CO2) to form Sodium peroxide.
2Na + O2(air) → Na2O2
(iii) By electrolyzing a solution of sodium chloride, we can get Sodium hydroxide. This process is commonly known as Castner- Kellner process.
The process is carried out using a mercury cathode and a carbon anode.
Sodium metal, deposited at cathode forms an Amalgam by combining with Mercury.
1.Cathode: Na+ +e– → Hg Na-Amalgam
2.Anode; Cl– → 1/2Cl2 + e–
(iv) Sodium chloride can be converted into sodium by Downs process.
It can be achieved by electrolysis of fused CaCl2 (60 %) and NaCl (40%) at
1123 K in a special apparatus (Downs cell).
A graphite block is the anode while steel is made the cathode. Metallic Ca and Na are formed at the cathode. Molten Na is supported by dipping in kerosene.
NaCl− Electrolysis →Na++Cl−
At Cathode: Na+ +e– → Na
At Anode: Cl– + e– → Cl
Cl + Cl → Cl2
17. Explain the reaction that occurs when
(i) quick lime is heated in the presence of silica
(ii) magnesium is burnt in presence of air
(iii) calcium nitrate is heated
(iv) slaked lime reacts with chlorine?
(i) Silica (SiO2) combines with Quick lime (CaO) resulting in formation of Slag.
CaO + SiO2 →Heat CaSiO3
(ii) When Magnesium is burnt in air, it does so with a dazzling bright light resulting in the formation of Mg3N2 and MgO.
2Mg + O2 → Burning 2MgO
3Mg + N2 → Burning Mg3N2
(iii) Calcium nitrate, when heated, undergoes decomposition to form calcium oxide.
2Ca (NO3)2(s) → Heat 2CaO(s) + 4NO2(g) + O2(g)
(iv) Bleaching powder is formed when chlorine is made to react with slaked lime.
Ca(OH)2 + Cl2 → Heat CaOCl2 + H2O
18. Mention any two important uses of the below given compounds:
(ii) caustic soda
(iii) sodium carbonate
(i) Caustic soda
(a) Heavily used in soap industries.
(b) Common reagent in laboratories.
(ii) Sodium carbonate
(a) Finds uses in both soap and glass industries.
(b) Also finds use as a water softener.
(iii) Quick lime
(a)Finds use as a primary material for manufacturing slaked lime.
(b) It helps in the manufacture of cement and glass.
19. Sketch the following structures:
i) BeCl2 (Solid)
ii) BeCl2 (Vapour)
In solid phase, BeCl2 is a polymer.
BeCl2 has a linear structure and exists as a monomer in vapour state.
20. How do you explain the fact that the carbonates and hydroxides of potassium and sodium get easily dissolved in water while those of calcium and magnesium are only slightly soluble in water?
Since the atomic sizes of magnesium and calcium are smaller than that of sodium and potassium, calcium and magnesium form carbonates and hydroxides with higher lattice energies. Thus, they are only sparingly soluble whereas those of potassium and sodium are readily soluble due to low lattice energies.
21. Mention the uses of the following substances:
(ii) plaster of paris.
Uses of cement:
3.Most important ingredient in concrete
Uses of Plaster of Paris:
1.Used to make casts and moulds
2.Used to make surgical bandages
Uses of limestone:
1.Preparation of cement and lime
2.As a flux in iron ore smelting
22. Why do you think only lithium salts are commonly hydrated while salts of other alkali metals are anhydrous in nature?
Since Lithium has the smallest size among all the alkali metals, it can easily polarized water molecules. Thus, smaller the size of the ion, greater is its ability to polarize water molecules.
Hence, trihydrated Lithium Chloride and other Lithium salts can be easily polarized. Due to this reason, other alkali metal ions can only form anhydrous salts.
23. Why do you think is LiCl soluble in not only water, but also acetone, whereas LiF is barely soluble in water?
LiF has a greater ionic character than LiCl which disturbs the balance between hydration energy and lattice energy. This balance is crucial for the solvability of ions in solution. Due to greater covalent character and lower lattice energy, dissolution of LiCl is more exothermic in nature than that of LiF.
24. Mention the importance of potassium, Magnesium, sodium and calcium in biological fluids.
(i) Sodium (Na):
They are found in our blood plasma and the interstitial fluids around the cells. They help in
(a) Transmission of nerve signals.
(b) They regulate the flow of water across the membranes of the neighboring cells.
(c) Transport sugars and amino acids from and to cells.
(ii) Potassium (K):
They are found mostly in the cell fluids in greater quantities.
They help in
(a) Activating enzymes.
(b) Oxidising glucose to form ATP.
(c) Transmitting nerve signals.
(iii) Magnesium (Mg) and calcium (Ca):
They are also called as macro-minerals named so because of their abundance in our body. Mg helps in
(a) Relaxing nerves and muscles.
(b) Building and strengthening bones.
(c) Maintaining blood circulation in our body.
Ca helps in
(a) coagulation of blood
(b) Maintaining homeostasis.
25. What do you think will happen when
(i) Sodium metal is immersed in water?
(ii) Sodium metal is heated in abundance of air?
(iii) Sodium peroxide gets dissolved in water?
(i) Sodium reacts to form NaOH and H2 gas when it is dropped in water. The reaction occurs as shown below:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
(ii) Sodium peroxide is formed when sodium reacts with oxygen while heating it in presence of air. The reaction proceeds as shown below:
2Na(s) + O2(s) → Na2O2(s)
(iii) NaOH and water are formed as a result of hydrolysis of Sodium peroxide when it is dissolved in water.
Na2O 2(s) + 2H2O(l) → 2NaOH(aq) + H2O2(aq)
26. Explain the following observations:
(a) The ionic mobility of the alkali metal ions are in the following order:
Li+ < Na+ < K+ < Rb+ < Cs+
(b) The only metal that can form a nitride directly is Lithium.
(c) E0 for M2+(aq) + 2e– → M(s) (where M = Ca, Sr or Ba) is nearly constant.
(a) The ionic and atomic sizes of the metals tend to increase while going down the alkali group.
The increasing order of the ionic sizes of the alkali metal ions is as shown below:
Li+ < Na+ < K+ < Rb+ < Cs+
Smaller the size of an ion, greater is its ability to get hydrated. Li+ ion gets heavily hydrated since it is the smallest in size whereas Cs+ has the largest size and is the least hydrated ion.
The alkali metal ions when arranged in the decreasing order of their hydrations is as shown below:
Li+ > Na+ > K+ > Rb+ > Cs+
Higher the mass of a hydrated ion, the lesser is its ionic mobility. Thus, hydrated Li+ is the least mobile ion whereas hydrated Cs+ is the most mobile ion.
The ionic mobility of the alkali metal ions are in the following order:
Li+ < Na+ < K+ < Rb+ < Cs+
(b) The only metal that can form a nitride directly is Lithium because Li+ has a smaller size and is easily compatible with the N3– ion. Thus, the lattice energy released is very high which is enough to overcome the amount of energy needed to form N3– ion.
(c) Electrode potential (E°) of any M2+/ M electrode is decided by three factors:
(i) Enthalpy of hydration
(ii) Enthalpy of vaporisation
(iii) Ionisation enthalpy
The cumulative effect of these factors on Ba, Sr, and Ca is almost the same.
As a result, their electrode potentials are also same.
27. Why do you think
(a) is a solution of Na2CO3 alkaline in nature?
(b) are alkali metals mostly prepared by electrolysis of their fused chlorides?
(c) is Sodium more important for our survival than potassium?
(a) Sodium bicarbonate and sodium hydroxide are the end products when Na2CO3 is hydrolyzed. Since, the product are alkaline in nature, a solution of Na2CO3 is considered to be alkaline in nature.
(b) Chemical reduction cannot be used to prepare alkali metals since they themselves are reducing in nature. Alkali metals are highly electropositive and thus cannot be prepared by displacement reactions. Since they also react with water, these alkali metals cannot be prepared by electrolysis of their aqueous solutions. Thus, alkali metals are mostly prepared by electrolysis of their fused chlorides.
(c) Sodium ions are primarily found in the Blood plasma and the interstitial fluids around the cells whereas Potassium ions are found within the cell fluids. Sodium
ions help in the transmission of nerve signals and also regulate the flow of water and transport sugars and amino acids into the cells.
Thus, Sodium is more important for our survival than potassium.
28. Write down the balanced equations for the following reactions?
(a) Water and Na2O2
(b) KO2 and Water
(c) Na2O and CO2
(a) 2Na2O2(s) + 2H2O(l) → 4NaOH(aq) + O2(aq)
(b) 2KO2(s) + 2H2O(l) → 2KOH(aq) + H2O2(aq) + O2(aq)
(c) Na2O(s) + CO2(g) → Na2CO3
29. Why do you think the following phenomena occur?
(i) BeSO4 in soluble in water while BeO is barely soluble in water.
(ii) BaO in soluble in water while BaSO4 is barely soluble in water.
(iii) Solubility of LiI is more than that of KI in ethanol.
(i) The sizes of Be2+ and O2- are small and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BeO is barely soluble in water.
Whereas the size of an SO42- is large compared to Be2+ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BeSO4 is easily soluble in water.
(ii) The sizes of Ba2+ and SO42- are large and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BaSO4 is barely soluble in water.
Whereas the size of an O2- is small compared to Be2+ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BaO is easily soluble in water.
(iii) The lithium ion has a smaller size and as a result of that, it has a higher polarizing capability. This enables it to polarize the electron cloud around an iodide ion thus resulting in a greater covalent character in LiI than KI. Thus, LiI is easily soluble in ethanol.
30. Which of the following alkali metals has the least melting point?
Cs has the least melting point of the given alkali metals since it has the largest size. Due to a larger size, the binding capability of Cs is limited and the lattice energy released during the formation of its compounds is less and can be easily broken.
31. Which of the given below alkali metal is capable of forming hydrated salt?
Li is capable of forming hydrated salts because of its size. Since it is smaller in size, it has a higher charge density and can easily attract water molecules around it and for hydrated salts like LiCl.2H2O. The other alkali metals have a bigger size and lesser charge density and thus aren’t capable of forming hydrated salts.
32. Which of the following carbonates of alkali earth metals has the most stable state thermally?
Thermal stability is directly proportional to the size of the cation i.e., larger the size of the atom, greater is its thermal stability. The biggest cation among the given compounds is Ba.
Thus, BaCO3 will be the most thermal carbonate among the given compounds followed by SrCO3, CaCO3 and MgCO3.
Conclusions for NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 10-S-BLOCK ELEMENTS
Swastik Classes’ NCERT Solutions for Class 11 Chemistry Chapter 10 – S-Block Elements provide a comprehensive and valuable resource for students seeking to master the properties and characteristics of s-block elements.
Throughout this chapter, we have offered detailed solutions, explanations, and step-by-step guidance to ensure a thorough understanding of the concepts covered. By utilizing these solutions, students can strengthen their problem-solving skills, reinforce their conceptual knowledge, and enhance their performance in examinations.
Our team of experienced educators has carefully crafted these solutions, aligning them with the NCERT syllabus. This ensures that students receive accurate and reliable information that is essential for their academic success. We strive to simplify complex topics such as the trends in the properties of s-block elements and the reactions and compounds they form, making them more accessible and easier to comprehend.
Swastik Classes is committed to providing quality education and empowering students in their educational journey. By offering these NCERT solutions, we aim to facilitate a deep appreciation for chemistry and instill a love for learning in our students.
We encourage you to make the most of these solutions and embrace the challenges and wonders of s-block elements. With Swastik Classes’ NCERT Solutions for Class 11 Chemistry Chapter 10, you can confidently navigate the properties and characteristics of s-block elements, paving the way for a strong foundation and future success in the subject.
We wish you all the best in your studies and hope that our solutions contribute to your growth and achievements in the field of chemistry.
Q1: What is the importance of s-block elements in chemistry?
A1: S-block elements are of great importance in chemistry due to their wide range of applications. They are highly reactive and play crucial roles in various biological processes, industrial applications, and everyday life. For example, alkali metals from Group 1 are used in the production of batteries, while alkaline earth metals from Group 2 are utilized in the manufacturing of cement and steel.
Q2: What are the general characteristics of s-block elements?
A2: The s-block elements are located on the left side of the periodic table and include Groups 1 and 2. Some of the general characteristics of s-block elements include:
- They have low ionization energies, making them highly reactive.
- They exhibit metallic properties such as good conductivity and malleability.
- They tend to form ionic compounds with non-metals.
- They generally have low melting and boiling points.
- They show a gradual increase in atomic size and metallic character down the group.
Q3: What are the differences between alkali metals and alkaline earth metals?
A3: Alkali metals (Group 1) and alkaline earth metals (Group 2) have several differences, including:
- Alkali metals have one valence electron, while alkaline earth metals have two valence electrons.
- Alkali metals are more reactive than alkaline earth metals.
- Alkali metals have lower melting and boiling points compared to alkaline earth metals.
- Alkali metals form predominantly ionic compounds, while alkaline earth metals tend to form both ionic and covalent compounds.
- Alkali metals are softer and have lower densities compared to alkaline earth metals.
Q4: What are the common uses of alkali metals?
A4: Alkali metals have various applications, including:
- Sodium (Na) and potassium (K) are used in the production of soaps and detergents.
- Lithium (Li) is utilized in batteries for electronic devices and electric vehicles.
- Potassium compounds are employed as fertilizers in agriculture.
- Sodium is used in the manufacturing of glass and as a coolant in nuclear reactors.
- Rubidium and cesium find applications in atomic clocks and specialized scientific research.
Q5: What are the properties and uses of alkaline earth metals?
A5: Alkaline earth metals possess the following properties and uses:
- They have higher melting and boiling points compared to alkali metals.
- They exhibit low electron affinities and relatively low ionization energies.
- Calcium (Ca) is essential for the development and maintenance of strong bones and teeth in humans.
- Magnesium (Mg) is used in the production of lightweight alloys, such as those used in aerospace industries.
- Barium (Ba) compounds are employed in medical imaging procedures, such as X-rays and CT scans.
- Strontium (Sr) is used in the manufacture of fireworks due to its vibrant red color when burned.