The alkali metals are low melting. Which of the following alkali metal is expected to melt if the room temperature rises to 30°C?
A. Na
B. K
C. Rb
D. Cs
The alkali metals are physically soft, silvery white and light metals. Due to the large atomic size, these elements have low density which increases down the group from Li to Cs. Caesium has the largest atomic size out of the given examples, and had weak metallic bonding ability due to the presence of a single valence electron. Hence it has the lowest melting point of 302°K or 28°C, hence it will melt if the room temperature rises to 30°C. The correct answer is (iv).
Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously?
A. Li
B. Na
C. K
D. Cs
Alkali metals react with water and form hydroxides and dihydrogen.
2M + 2H2O → 2M+ + 2OH- + H2 where M is an alkali metal. Lithium has the most negative EӨ value, however it has the least vigorous reaction with water. It is even less vigorous than sodium, which has the least negative EӨ value. This is because of the small size of Li and its high hydration energy. It is also because Li has the highest melting point. The correct answer is (i).
The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.
A. Sublimation enthalpy
B. Ionisation enthalpy
C. Hydration enthalpy
D. Electron-gain enthalpy
Alkali metals are strong reducing agents, with lithium being the strongest and sodium being the weakest reducing agent. Li is has the smallest atomic size and the highest hydration enthalpy. Hydration enthalpy is the amount of energy released when one mole of ions undergo hydration. Due to this, Li is the strongest reducing agent.
Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?
A. MgCO3
B. CaCO3
C. SrCO3
D. BaCO3
Carbonates of all alkaline earth metals decompose to give metal oxide and carbon dioxide. The thermal stability of a metal carbonate depends on the cationic size, it increases with increasing cationic size. The carbonates form the positively charged metal cation and negatively charged CO32-. The larger the cationic size, the more electropositive is the ion. The electropositivity of the alkaline earth metals increases down the group from Be to Ba. Due to these reason, BaCO3 is the most stable metal carbonate.
The correct option is (iv).
Which of the carbonates given below is unstable in air and is kept in CO2 atmosphere to avoid decomposition.
A. BeCO3
B. MgCO3
C. CaCO3
D. BaCO3
Beryllium carbonate is unstable and is kept in CO2 atmosphere to avoid decomposition. Beryllium is also the least electropositive element out of the group due to the atomic size. BeCO3 decomposes to form BeO and CO2 as carbonates of all alkaline earth metals decompose to give metal oxide and carbon dioxide. This reaction is reversible, hence to shift the equilibrium to the right, BeCO3 is placed in CO2 atmosphere.
Metals form basic hydroxides. Which of the following metal hydroxide is the least basic?
A. Mg(OH)2
B. Ca(OH)2
C. Sr(OH)2
D. Ba(OH)2
Alkaline earth metals burn in oxygen to form monoxides which are basic in nature (except BeO) and react with water to form sparingly soluble hydroxides. The solubility, thermal stability and the basic character of these hydroxides increase with increasing atomic number from Mg(OH)2 to Ba(OH)2. Ionization enthalpy increases from Mg to Ba, hence the M—O bond becomes weaker from Mg to Ba, and hence Mg will form the weakest base. Hence, Mg(OH)2 is the least basic in nature. The correct option is (i).
Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is
A. BeCl2
B. MgCl2
C. CaCl2
D. SrCl2
Group 2, or alkaline earth metals form halides and except Beryllium, the metal halides are ionic in nature. Beryllium chloride however is covalent in nature due to the small size, high ionization enthalpy and high electronegativity. Due to this, BeCl2 is soluble in organic solvents such as ethanol. The correct answer is (i).
The order of decreasing ionisation enthalpy in alkali metals is
A. Na > Li > K > Rb
B. Rb < Na < K < Li
C. Li > Na > K > Rb
D. K < Li < Na < Rb
Ionization enthalpy is the energy required to displace an electron from the valence shell of an atom in ground state. The smaller the size of the atomic radii, more energy is required to separate the valence electron. In the given metals, the size of the atomic radii in decreasing order is Li > N > K > Rb. The correct option is (iii).
The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to
A. Ionic nature of lithium fluoride
B. High lattice enthalpy
C. High hydration enthalpy for lithium ion.
D. Low ionisation enthalpy of lithium atom
Since lattice enthalpy decreases much more than the hydration enthalpy with increasing ionic size for alkali metals, the solubility of LiF is the lowest in water due to highest lattice enthalpy.
Amphoteric hydroxides react with both alkalies and acids. Which of the following Group 2 metal hydroxides is soluble in sodium hydroxide?
A. Be(OH)2
B. Mg(OH)2
C. Ca(OH)2
D. Ba(OH)2
Amphoteric hydroxides are compounds which can behave both as an acid and base. Beryllium hydroxide is amphoteric in nature as it reacts with acid and alkali both. Beryllium hydroxide reacts with sodium hydroxide to form beryllate ion which is soluble in the same. The reaction is given as
Be(OH)2 + 2OH–→ [Be(OH)4]2–
Hence, the correct answer is (i).
In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH4Cl with Ca(OH)2. The by-product obtained in this process is
A. CaCl2
B. NaCl
C. NaOH
D. NaHCO3
Sodium carbonate is prepared from sodium by Solvay process. The equations in the process involved are as follows:
2NH3 + H2O + CO2→ (NH4)2CO3
(NH4)2CO3 + H2O + CO2→ 2NH4HCO3
NH4CO3 + NaCl → NH4Cl + NaHCO3
2NaHCO3→ Na2CO3 + CO2 + H2O
2NH4Cl + Ca(OH)2→ 2NH3 + CaCl2 + H2O
Carbon dioxide is passed through a concentrated solution of sodium chloride saturated with ammonia, which forms ammonium carbonate followed by ammonium hydrogen carbonate. Ammonium hydrogen carbonate crystals separate and they are heated to form sodium carbonate. NH3 is recovered from the solution which contains NH4Cl is heated and treated with Ca(OH)2. Calcium chloride is obtained as a by-product. The correct answer is (i).
When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to
A. ammoniated electron
B. sodium ion
C. sodium amide
D. ammoniated sodium ion
Alkali metals like sodium dissolve in liquid ammonia to give a deep blue colour. The reaction can be given as
M + (x+y)NH3 → [M(NH3)x]+ + [e(NH3)y]-
The deep blue colour is due to the ammoniated electron. It absorbs energy in the visible region of light and thus imparts blue colour to the solution. The correct answer is (i).
By adding gypsum to cement
A. setting time of cement becomes less.
B. setting time of cement increases.
C. colour of cement becomes light.
D. shining surface is obtained.
Gypsum is CaSO4.2H2O. Gypsum is added in 2-3% of weight to cement along with CaO (lime) and clay. Gypsum is added in order to increase the setting time of cement. Slowing the setting time of cement makes it harden sufficiently. The correct answer is (ii).
Dead burnt plaster is
A. CaSO4
B. CaSO4. 1/2H2O
C. CaSO4.H2O
D. CaSO4.2H2O
Dead burnt plaster is formed when gypsum CaSO4.2H2O is heated to 393K. This causes loss of water of crystallisation and forms anhydrous calcium sulphate.
2(CaSO4.2H2O) → 2(CaSO4).H2O + 3H2O
Suspension of slaked lime in water is known as
A. lime water
B. quick lime
C. milk of lime
D. aqueous solution of slaked lime
Suspension of slaked lime in water is known as milk of lime. It is a suspension of calcium hydroxide in water. The correct answer is (iii).
Which of the following elements does not form hydride by direct heating with dihydrogen?
A. Be
B. Mg
C. Sr
D. Ba
Alkali metals react with dihyrogen at high temperatures to form hydrides. Beryllium has a small atomic size and has a high ionisation enthalpy. Due to this, it does not react on direct heating with dihydrogen. The correct answer is (i).
The formula of soda ash is
A. Na2CO3 .10H2O
B. Na2CO3.2H2O
C. Na2CO3.H2O
D. Na2CO3
Soda ash is formed when sodium carbonate decahydrate, washing soda, is heated above 373K. This causes Na2CO3.10H2O to lose all water of crystallisation to form anhydrous sodium carbonate or soda ash. The correct answer is (iv).
A substance which gives brick red flame and breaks down on heating to give oxygen and a brown gas is
A. Magnesium nitrate
B. Calcium nitrate
C. Barium nitrate
D. Strontium nitrate
Compounds containing calcium burn with a brick red flame. Hence Calcium nitrate will break down on heating to give a brick red flame and a brown gas which is NO2.
2Ca(NO3)2 → 2CaO + 4NO2 + O2
Which of the following statements is true about Ca(OH)2?
A. It is used in the preparation of bleaching powder
B. It is a light blue solid
C. It does not possess disinfectant property.
D. It is used in the manufacture of cement.
Calcium hydroxide or Ca(OH)2 reacts with chlorine to form hypochlorite, which is a component of bleaching powder. The reaction is as follows.
2Ca(OH)2 + 2Cl2 → CaCl2 + Ca(OCl)2 + 2H2O
The correct answer is (i).
A chemical A is used for the preparation of washing soda to recover ammonia. When CO2 is bubbled through an aqueous solution of A, the solution turns milky. It is used in white washing due to disinfectant nature. What is the chemical formula of A?
A. Ca (HCO3)2
B. CaO
C. Ca(OH)2
D. CaCO3
Chemical A used in the preparation of washing soda Na2CO3 to recover ammonia is Ca(OH)2. It is a reaction involved in the Solvay process for preparation of sodium carbonate.
2NH4Cl + Ca(OH)2→ 2NH3 + CaCl2 + H2O
When CO2 is bubbled through a solution of Ca(OH)2, it forms milky precipitate of CaCO3.
Ca(OH)2 + CO2→ CaCO3 + H2O
Ca(OH)2 is used as slaked lime as white wash due its disinfectant nature.
Dehydration of hydrates of halides of calcium, barium and strontium i.e., CaCl26H2O, BaCl2 .2H2 O, SrCl2.2H2O, can be achieved by heating. These become wet on keeping in air. Which of the following statements is correct about these halides?
A. act as dehydrating agent
B. can absorb moisture from air
C. Tendency to form hydrate decreases from calcium to barium
D. All of the above
The substances get wet on contact with air as they are hygroscopic in nature and can absorb moisture around them. The tendency to form halide hydrates gradually decreases down the group. For example, MgCl2 .8H2O, CaCl2.6H2O, SrCl2.6H2O and BaCl2.2H2O. Hence, all the properties are correct and the correct answer is (iv).
In the following question two or more options may be correct.
Metallic elements are described by their standard electrode potential, fusion enthalpy, atomic size, etc. The alkali metals are characterised by which of the following properties?
A. High boiling point
B. High negative standard electrode potential
C. High density
D. Large atomic size
Out of the given properties, alkali metals are characterised by their high negative standard electrode potential and large atomic size. Alkali metals are characterised by being soft and light. They have low boiling point and also low density due to the large atomic size.
In the following question two or more options may be correct.
Several sodium compounds find use in industries. Which of the following compounds are used for textile industry?
A. Na2 CO3
B. NaHCO3
C. NaOH
D. NaCl
NaOH is used in textile industries as a mercerising agent for cotton fabrics, Na2CO3 is also used in textiles and as water softening, laundering and cleaning agents.
In the following question two or more options may be correct.
Which of the following compounds are readily soluble in water?
A. BeSO4
B. MgSO4
C. BaSO4
D. SrSO4
Out of the given compounds, BeSO4 and MgSO4 are readily soluble in water. The solubility of the sulphates decreases from CaSO4 to BaSO4. The compounds are readily soluble because the greater hydration enthalpies of Be2+ and Mg2+ ions overcome the lattice enthalpy factor.
In the following question two or more options may be correct.
When Zeolite, which is hydrated sodium aluminium silicate is treated with hard water, the sodium ions are exchanged with which of the following ion(s)?
A. H+ ions
B. Mg2+ ions
C. Ca2+ ions
D. SO42- ions
Zeolite is hydrated sodium aluminium silicate, or Al2Si2O8. Hard water consists mainly of calcium and magnesium salts, hence sodium ions are exchanged with mainly Mg2+ and Ca2+ ions. The reaction is given as
Na2Z + M2+→ MZ + 2Na+
Where M is Mg2+ or Ca2+ and Z is Zeolite.
In the following question two or more options may be correct.
Identify the correct formula of halides of alkaline earth metals from the following.
A. BaCl2.2H2O
B. BaCl2 .4H2O
C. CaCl2 .6H2O
D. SrCl2.4H2O
The tendency to make halide hydrates decreases down the group of alkaline earth metals. Examples are MgCl2 .8H2O, CaCl2.6H2O, SrCl2.6H2O and BaCl2.2H2O. Hence the correct answers are (i) and (iii).
In the following question two or more options may be correct.
Choose the correct statements from the following.
A. Beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
B. Beryllium sulphate is readily soluble in water as the greater hydration enthalpy of Be2+ overcomes the lattice enthalpy factor.
C. Beryllium exhibits coordination number more than four.
D. Beryllium oxide is purely acidic in nature.
Beryllium is chemically inert to air and water and is not readily attacked by acids because of the presence of a BeO film on the surface of the metal. BeSO4 is readily soluble in water. This is because Be2+ ions overcome the lattice enthalpy factor. Beryllium does not have a coordination number of more than 4. Beryllium oxide is not purely acidic but amphoteric in nature.
In the following question two or more options may be correct.
Which of the following are the correct reasons for anomalous behaviour of lithium?
A. Exceptionally small size of its atom
B. Its high polarising power
C. It has high degree of hydration
D. Exceptionally low ionisation enthalpy
Lithium shows anomalous behaviour where it shows similar properties to alkaline earth metals, known as diagonal relationship. This is due to the exceptionally small size of the atom. Anomalous behaviour is also due to the high polarising power of the lithium ion. Polarisation is the distortion of electron cloud of the anion by the cation. Options (iii) and (iv) are also properties of lithium but they are also properties of alkali metals. The correct answers are (i) and (ii).
How do you account for the strong reducing power of lithium in aqueous solution?
Alkali metals are strong reducing agents, with lithium being the strongest of the group. The reducing power of an element can be determined by its standard electrode potential, which is a measure of the tendency of the element to lose electrons in aqueous solution. It is determined by three factors: (i) Sublimation enthalpy (ii) Ionization enthalpy and (iii) Enthalpy of hydration. Sublimation enthalpy of alkali metals is similar. Lithium has highest negative EӨ value, which is –3.04V. Lithium has a small atomic size, the highest ionization enthalpy but it is compensated by its high hydration enthalpy. Due to this, the reducing power of lithium is highest in an aqueous solution.
When heated in air, the alkali metals form various oxides. Mention the oxides formed by Li, Na and K.
Alkali metals are highly reactive and their chemical reactivity increases down the group. Alkali metals burn vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms peroxide, and potassium form superoxide. The reactions are as follows:
4LiO +O2→ 2Li2O (oxide)
2Na + O2→ Na2O2 (peroxide) with Na2O as a minor product
K + O2→ KO2 (superoxide)
The formation of oxide to superoxide is
O2- + 1/2O2→ O22- + O2→ 2O2-
Oxide peroxide superoxide
The increasing stability of the peroxide or superoxide, as the size of the metal ion increases, is due to the stabilisation of large anions by larger cations through lattice energy effects.
Complete the following reactions
(i) O2-2 + H2O →
(ii) O-2 + H2O →
(i) Peroxide ions react with H2O to form hydrogen peroxide.
O22- + 2H2O → H2O2 + 2OH-
(ii) Superoxide ions react with H2O to form hydrogen peroxide and O2.
O2- + 2H2O → H2O2 + O2 + 2OH-
Lithium resembles magnesium in some of its properties. Mention two such properties and give reasons for this resemblance.
Lithium has some properties which resemble the properties of the second element of Group 2, magnesium. This is known as a diagonal relationship in the periodic table.
(i) Lithium and magnesium are both lighter and harder than the other metals in their respective groups.
(ii) Halides of both elements, LiCl and MgCl2 are soluble in ethanol.
(iii) Both lithium and magnesium form a nitride Li3N and Mg3N2 respectively.
These two elements have similar properties because of their similar atomic and ionic radii.
Name an element from Group 2 which forms an amphoteric oxide and a water soluble sulphate.
The element from Group 2 is Beryllium. Beryllium reacts with oxygen to give BeO which is covalent in nature, unlike other elements of Group 2 which form ionic compounds. Beryllium oxide is amphoteric in nature unlike the other compounds which are basic in nature. Sulphates of Group 2 are soluble in water and BeSO4 is readily soluble in water. This is because of the hydration enthalpy of Be2+ ions being much higher than the lattice enthalpy of BeSO4.
Discuss the trend of the following:
(i) Thermal stability of carbonates of Group 2 elements.
(ii) The solubility and the nature of oxides of Group 2 elements.
(i) All Group 2 elements form carbonates. All carbonates of alkaline earth metals are insoluble in water. These carbonates decompose on heating to give carbon dioxide and the metal oxide. The thermal stability of these carbonates increases down the group, as the solubility reduces. The order is BeCO3 < MgCO3 < CaCO3 < SrCO3 < BaCO3. Beryllium carbonate is unstable and requires to be in an atmosphere of CO2. The thermal stability of the carbonates increases with increases with increasing cationic size. The more stable the oxide of an alkaline earth metal, the less stable is the carbonate of the same. Hence BeCO3 is highly unstable as BeO is stable.
(ii) All alkaline earth metals form oxides with oxygen in the formula MO where M is the metal. The alkaline earth metals burn in oxygen to give metal oxides. The compounds are ionic in nature except BeO which is covalent. The oxides are basic in nature except BeO which is amphoteric in nature and behaves as both acid and base. The basic nature increases down the group due to the increase in ionic nature. The enthalpy of formation of these oxides are high and hence they are thermally stable. They also react with water to form sparingly soluble hydroxides. As the size of the cations increase, the lattice energy decreases. BeO and MgO have the highest lattice energy and they are insoluble in water.
Why are BeSO4 and MgSO4 readily soluble in water while CaSO4, SrSO4 and BaSO4 are insoluble?
The solubility of the alkaline earth metal sulphates decrease down the group. While BeSO4 and MgSO4 are readily soluble in water, CaSO4, SrSO4, and BaSO4 are insoluble. This is because the greater hydration enthalpies of Be2+ and Mg2+ ions overcome the lattice enthalpy factor and therefore their sulphates are soluble in water. The hydration enthalpies of the rest compounds do not match the lattice enthalpy factor, thus making these sulphates insoluble in water.
All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.
All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents because the alkali metal compounds form ionic compounds due to their large ionic size and low ionization enthalpy, while lithium forms compounds of covalent nature due to their small ionic size, high ionization enthalpy, and high electronegativity. Ionic compounds dissolve in water, and covalent compounds dissolve in organic solvents.
In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing (NH4) 2CO3 with sodium chloride? Explain.
We cannot obtain sodium carbonate directly by treating the (NH4)2CO3 solution with NaCl, sodium chloride. In the Solvay process, carbon dioxide is passed through a concentrated solution of sodium chloride saturated with ammonia, which forms ammonium carbonate followed by ammonium hydrogen carbonate. Ammonium hydrogen carbonate crystals separate and they are heated to form sodium carbonate. NH3 is recovered from the solution which contains NH4Cl is heated and treated with Ca(OH)2. The reaction of (NH4)2CO3 with NaCl gives two products, Na2CO3 and NH4Cl which are both soluble in water which does not shift the equilibrium to the right.
(NH4)2CO3⇌ NH4Cl + Na2CO3
Write Lewis strucure of O-2 ion and find out oxidation state of each oxygen atom? What is the average oxidation state of oxygen in this ion?
The Lewis structure of O2- is . Oxygen atom with zero charges has 6 electrons, therefore the oxidation state is 0. When the oxygen atom has a negative charge, it has 7 electrons. Hence, the oxidation state is -1. The average oxidation state is 0 + (-1)/2 = -1/2.
Why do beryllium and magnesium not impart colour to the flame in the flame test?
Alkaline earth metals except Be and Mg impart a characteristic colour to the Bunsen burner flame in the flame test. The different colours arise due to the different energies (provided by heat in this case) for the states of electron excitation and de-excitation. Be and Mg electrons are tightly bound to the atom due to the small atomic and ionic size. The electrons do not gain excitation from the energy provided from the flame, hence they do not impart any colour to the flame.
What is the structure of BeCl2 molecule in gaseous and solid state?
Beryllium chloride is covalent in nature.
The structure of BeCl2 in solid state is a polymeric chain structure and each Be atom is surrounded by four Cl atoms. Two Cl atoms are covalently bonded and two are bonded by coordinate bonds. The gaseous/vapour state is different than the solid state. BeCl2 tends to form a chloro-bridged dimer at temperatures below 1200K and dissociates into a linear monomer at high temperatures of the order of 1200 K.
Match the elements given in Column I with the properties mentioned in Column II.
(i) → (c)
As Li has the most negative E⊖ value due to the high hydration energy.
(ii) → (b)
As Alkali metals are more acidic than the given alkaline earth metals, while Li base is covalent in nature.
(iii) → (d)
Calcium oxalate is insoluble in water
(iv) → (a)
Due to the high hydration energy.
(iv) → (e)
Electronic configuration of Ba is 1s22s22p63s23p63d104s24p64d105s25p66s2.
Match the compounds given in Column I with their uses mentioned in Column II.
(i) → (c)
Calcium carbonate used in manufacturing high quality paper because the paper and cardboard industries use lime-based coating pigments and filters such as GCC(Ground Calcium Carbonate). GCC is made from concentrated and fine-ground calcium carbonate and used to make fine paper, cardboard packaging and pulp-based paper.
(ii) → (d)
Calcium Hydroxide is used in white washing because for white washing walls as it slowly reacts with CO2in air to form a thin layer of Calcium Carbonate on the walls. The layer of Calcium Carbonate gives a shiny finish to the walls.
Ca(OH)2 + CO2→ CaCO3 + H2O
(iii) → (b)
CaO is used in manufacturing of sodium carbonate from caustic soda.
Above process is also known as Solvay Process. This reaction occurs in two steps.
First step,
NaCl + CO2 + NH3 + H2O → NaHCO3 + NH4Cl(l)
First Ammonia converts into Sodium Hydrogen Carbonate and Ammonium Chloride.
Second Step,
2NaCl + CaCO3→Na2CO3 + CaCl2
So, in the next or second step, there is the formation of Sodium Carbonate.
(iv) → (a)
CaSO4 is used in Dentistry, Ornamental work because CaSO4with adequate quantity of water, it forms a plastic mass that sets into a hard solid.
Match the elements given in Column I with the colour they impart to the flame given in Column II.
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
(i) → (f)
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
So, Cs imparts blue colour.
(ii) → (d)
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Thus, Na imparts yellow colour.
(iii) → (b)
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Thus, Potassium imparts violet colour.
(iv) → (c)
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Thus, Ca imparts brick red colour.
(v) → (e)
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Thus, Sr imparts crimson red colour.
(vi) → (a)
The colour of the flame imparted depends on the energy required for electron excitation and deexcitation. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Thus, Barium imparts apple green colour.
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): The carbonate of lithium decomposes easily on heating to form lithium oxide and CO2.
Reason (R) : Lithium being very small in size polarises large carbonate ion leading to the formation of more stable Li2O and CO2 .
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct
(iv) A is not correct but R is correct.
(i) Both A and R are correct and R is the correct explanation of A.
Explanation: Unlike other alkali metal carbonates which are stable in heat Lithium being very small in size polarises a large CO32– ion leading to the formation of more stable Li2O and CO2.
Li2CO3→ Li2O + CO2
Polarisation is the distortion of electron cloud of the anion CO32- by the cation Li2+.
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Beryllium carbonate is kept in the atmosphere of carbon dioxide.
Reason (R) : Beryllium carbonate is unstable and decomposes to give beryllium oxide and carbon dioxide.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
(i) Both A and R are correct and R is the correct explanation of A.
Explanation: All alkaline earth metal carbonates are stable at room temperature except beryllium carbonate. It decomposes to give BeO and CO2, hence it is kept in a CO2 atmosphere so that the equilibrium shifts to the right of the equation.
BeCO3⇌ BeO + CO2
The s-block elements are characterised by their larger atomic sizes, lower ionisation enthalpies, invariable +1 oxidation state and solubilities of their oxosalts. In the light of these features describe the nature of their oxides, halides and oxosalts.
The s-block elements contain two groups on the periodic table, the alkali metals (Group 1) and the alkaline earth metals (Group 2). The nature of their oxides, halides and oxosalts is given as follows:
1) Alkali metals:
(i) Oxides: Alkali metals form oxides on combustion in excess air. Lithium forms oxide Li2O and peroxide Li2O2 in minor quantities. Sodium forms the peroxide, Na2O2 (and superoxide NaO2 in minor quantities) whilst potassium, rubidium and caesium form the superoxides, MO2. Going down the group, the size of the atoms increases and the ability to form peroxides and superoxides also increases. This happens due to the stabilisation of large anions by larger cations through lattice energy effects.
(ii) Halides: Alkali metal halides can be produced by reaction with the appropriate carbonate, oxide or hydroxide and hydrohalic acid (HX). The melting and boiling points follow the trend flouride > chloride > bromide > iodide. All alkali metal fluorides have high negative enthalpy of formation. The values for fluorides become less negative down the group and for chlorides, bromides and iodides the values become more negative down the group. All of these halides are highly soluble in water except LiF which has high lattice energy.
(iii) Oxosalts: All alkali metals react with oxo-acids like carbonic acid H2CO3 and sulphuric acid H2SO4 to form oxo-salts like carbonate and sulphate respectively. They are soluble in water and thermally stable. The carbonates and hydrogencarbonates, M2CO3 and MHCO3 respectively are thermally stable. The stability of these compounds increases with increasing electropositive character down the group. The exception is Li2CO3 which is not thermally stable due to the small size of the Li atom, which polarizes the large CO23- ion. This leads to the decomposition and formation of LiO and CO2. The hydrogencarbonate does not exist as a solid.
2) Alkaline Earth Metals
(i) Oxides: Alkaline earth metals burn in oxygen to form metal oxides in the form of MO. The oxides are ionic in nature except BeO which is covalent in nature. The enthalpies of formation of oxides is high. The oxides are basic in nature except BeO which is amphoteric.
(ii) Halides: Halides of alkaline earth metals are ionic in nature with the exception of BeCl2 which is covalent and is soluble in organic solvents. The fluorides are relatively less soluble than the chlorides owing to their high lattice energies.
(iii) Oxosalts: Alkaline earth metals form oxosalts with oxo-acids. They form carbonates with carbonic acid, sulphates with sulphuric acid and nitrates with nitric acid. Carbonates are insoluble in water and they decompose on heating to form the oxides and carbon dioxide. Beryllium carbonate is unstable and has to be placed in a CO2 atmosphere to stop it from decomposing. The sulphates are stable in heat. BeSO4 and MgSO4 are readily soluble in water because of hydration enthalpies of Be2+ and Mg2+ overcoming lattice enthalpy, and the solubility reduces from CaSO4 to BaSO4. Nitrates show decreasing tendency to form hydrates with increasing size and decreasing hydration enthalpy in the group. All of them decompose to give metal oxide MO and NO2 and O2.
Present a comparative account of the alkali and alkaline earth metals with respect to the following characteristics:
(i) Tendency to form ionic / covalent compounds
(ii) Nature of oxides and their solubility in water
(iii) Formation of oxosalts
(iv) Solubility of oxosalts
(v) Thermal stability of oxosalts
When a metal of group 1 was dissolved in liquid ammonia, the following observations were obtained:
(i) Blue solution was obtained initially.
(ii) On concentrating the solution, blue colour changed to bronze colour.
How do you account for the blue colour of the solution? Give the name of the product formed on keeping the solution for some time.
Alkali metals dissolve in liquid ammonia to give a deep blue colour. These solutions are also conducting in nature. The blue colour is obtained because of ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The reaction is given as
M + (x+y)NH3 → [M(NH3)x]+ + [e(NH3)y]-
As the concentration of the metal increases, the colour of the solution become bronze in colour, which is due to the formation of metal ion clusters.
If the product formed is kept standing for some time, it slowly liberates hydrogen resulting in the formation of amide.
M+(am) + e- +NH3(l) → MNH2(am) +1/2H2(g)
The stability of peroxide and superoxide of alkali metals increase as we go down the group. Explain giving reason.
The equation of formation of oxide, peroxide, and superoxide is
O2- + 1/2O2→ O22- + O2→ 2O2-
Oxide peroxide superoxide
The increasing stability of the peroxide or superoxide, as the size of the metal ion increases, is due to the stabilisation of large anions by larger cations through lattice energy effects.
The size of the cations down group 1 increases from Li to Cs. Li has the smallest atomic and ionic size and has a strong positive field around it. Li+ combines with small anion called oxide ion (O2-) with strong negative field and form Lithium oxide, Li2O and Lithium peroxide in very small amounts.
Na+ ion is larger than Li+ ion and hence with less positive field around it. It combines with peroxide ion (O22-) with less negative field and form sodium peroxide and sodium superoxide in small quantity.
K+, Rb+ and Cs+ ions are bigger in size with even lesser positive field around them. Hence they can stabilize the bigger superoxide ion (O2-) with less negative field and form super oxides.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkiness disappears in the last step.
Compound (A) is Calcium oxide, CaO. Addition of water to CaO forms Compound (B), which is Ca(OH)2.
CaO + H2O → Ca(OH)2
When CO2 is passed through the Ca(OH)2 solution, it turns milky and forms a white precipitate which is Compound (C) which is calcium carbonate, CaCO3.
Ca(OH)2 + CO2→ CaCO3 + H2O
On passing excess of CO2, the precipitate dissolves to form calcium hydrogencarbonate, which is compound (D).
CaCO3 + CO2 + H2O → Ca(HCO3)2
The milkiness dissolves due to formation of a soluble compound calcium hydrogencarbonate.
Lithium hydride can be used to prepare other useful hydrides. Beryllium hydride is one of them. Suggest a route for the preparation of beryllium hydride starting from lithium hydride. Write chemical equations involved in the process.
Beryllium cannot react with dihydrogen like other alkaline earth metals to form beryllium hydride. Lithium hydride on the other hand is formed by heating with dihydrogen at 1073K. Steps to form beryllium hydride from lithium hydride are as follows.
Lithium hydride reacts with anhydrous aluminium chloride in presence of ether to form LiAlH4, lithium aluminium hydride.
LiH + Al2Cl6→ 2LiAlH4 + 6LiCl
LiAlH4 reacts with BeCl2 to form BeH2.
2BeCl2 + LiAlH4→ 2BeH2 + LiCl + AlCl3
An element of group 2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide. Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid.
The only element from Group 2 which forms a covalent oxide and is amphoteric in nature is Beryllium, Be. The covalent oxide is BeO which is amphoteric. When dissolved in water, it forms Be(OH)2. The reactions of Be(OH)2 with an acid and a base are as follows.
Be(OH)2 reacts with a base to forms beryllate ion.
Be(OH)2 + 2OH–→ [Be(OH)4]2–
Be(OH)2 reacts with an acid to form beryllate chloride.
Be(OH)2 + 2HCl + 2H2O → [Be(OH)4]Cl2
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and aminoacids into cells. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of its peroxide. Why does the element impart colour to the flame?
Sodium ions participate in transfer of nerve signals, regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells. Sodium imparts a golden yellow colour to the flame in a flame test. This happens because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum, and for sodium is yellow at 589.2 λ/nm.
The reaction of sodium with oxygen to form sodium peroxide is given as
2Na + O2→ Na2O2 (peroxide)
The stability of the sodium peroxide is due to the stabilisation of large anion O22- by larger cation Na+.