Author Archives: Priyanka

In This Post we are  providing Chapter-4 THE p-BLOCK ELEMENT NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON THE p-BLOCK ELEMENT

1.Why the elements of group 13 are called p-block elements?

Ans.Group 13 elements are called p-block elements because the last electron is present in the p-orbital (np¹). The valence shell configurations are B (2s² 2p¹), Al (3s², 3p¹), Ga (4s², 4p¹), In (5s² 5p¹) Tl (6s² 6p¹)

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2.The elements B, Al, Ca, In and Tl are placed in the same group of the periodic table. Give reason.

Ans. The elements B, Al, Ga, In and Tl are placed in the same group of the periodic table because each one has the same number of electrons (ns² np¹) in its valance shell.

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3.Aluminium forms in not formed why?

Ans. Due to presence of vacant d-orbital’s, Al can expand its octet to form bonds with six fluoride ions whereas B cannot. Boron does not have d-orbital’s.

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4.The atomic radius of Ca is less than that of Al. Why?

Ans. This is due to the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium.

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5.C and S are always tetravalent but Ge, Sn And Pb show divalency. Why?

Ans. Inert pair is more prominent as we move down the group in p – block elements. Ge, Sn and Pb show divalency due to inert pair effect.

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6.Some halides of group 14 elements form complexes of the type . Give reason.

Ans. The halides of the elements having vacant d-orbital’s can form complexes like because in such a case the central atom can increase its coordination number from 4 to 6 due to availability of vacant d–orbital’s.

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7.is lead known whereas not. Give reason.

Ans. The main reasons are that

(i) Six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.

(ii) Interaction between lone pair of chloride ion and Si⁴⁺ is not very strong.

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8.PbI₄ does not exist. Why?

Ans.PbI₄ does not exist because Pb – I bond initially formed during the reaction does not release enough energy to unpair 6s² electrons and excite one of them to higher orbital to have four unpaired electrons around lead atom.

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9.Why is carbon different from other member of the group?

Ans. Carbon differs from rest of the members of its group due to its smaller size, higher electro negativity, higher ionization enthalpy and unavailability of d-orbital’s.

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10.Why does the covalence of carbon not expand beyond four?

Ans. In carbon, only s and p orbital’s are available for bonding and therefore it can accommodate only four pairs of electrons around it. This limit the maximum covalence to four whereas other members can expand their covalence due to the presence of d-orbital’s.

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11.Why does carbon show different allotropic forms?

Ans. Due to property of catenation and pπ – pπ bond formation Carbon is able to show different allotropic forms.

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12.Silicon has no allotropic form analogous to graphite. Why?

Ans. Due to large size. Si has little or no tendency for pπ – pπ bonding. Whereas carbon atom forms easily pπ – pπ bonds due to smaller size in graphite structure. Hence, Si does not exhibit graphite structure.

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13.Why does graphite conduct electricity?

Ans. Graphite forms hexagonal ring and undergoes sp² hybridization. The electrons are delocalized over the whole sheet. Electrons are mobile and therefore graphite conducts electricity over the sheet.

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14.Graphite is used as lubricant. Give reason.

Ans. Graphite has sp² hybridized carbon with a layer structure due to wide separation and weak inter – layer bonds the two adjacent layers can easily slide over each other. This makes graphite act as a lubricant.

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15.How are silicones manufactured?

Ans.They are manufactured by hydrolysis of chlorosilanes –

where R is a methyl or phenyl group.

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In This Post we are  providing Chapter-3 THE s-BLOCK ELEMENTS NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON THE s-BLOCK ELEMENTS

Question 1.
Why the solubility of alkaline metal hydroxides increases down the group?

Answer:
If the anion and the cation are of comparable size, the cationic radius ‘vill influence the lattice energy. Since lattice energy decreases much more than the hydration energy with increasing ionic size, solubility will increases as we go down the group. This is the case of alkaline earth metal hydroxides.

Question 2.
Why the solubility of alkaline earth metal carbonates and sulphates decreases down the group?

Answer:
If the anion is large compared to the cation, the lattice; energy will remain almost constant within a particular group. Since the hydration energies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Question 3.
Why cannot potassium carbonate be prepared by the SOLVAY process?

Answer:
Potassium carbonate cannot be prepared by the SOLVAY process because potassium bicarbonate (KHCO₃) is highly soluble in water, unlike NaHCO₃ which was separated as crystals. Due to its high solubility KHCO₃ cannot be precipitated by the addition of ammonium bicarbonate to a saturated solution of KCl.

Question 4.
What are the main uses of calcium and magnesium?

Answer:
Main uses of calcium:

1. Calcium is used in the extraction of metals from oxides which are difficult to reduce with carbon.
2. Calcium, due to its affinity for O₂ and N₂ at elevated temperatures, has often been used to remove air from vacuum tubes.

Main uses of Magnesium:

1. Magnesium forms alloys with Al, Zn, Mn and Sn. Mg-Al alloys being light in mass are used in aircraft construction.
2. Magnesium (powder and ribbon) is used in flashbulbs, powders incendiary bombs and signals.
3. A suspension of Mg(OH)₂ in water is used as an antacid in medicine.
4. Magnesium carbonate is an ingredient of toothpaste.

Question 5.
What is meant by the diagonal relationship in the periodic table? What is it due to?

Answer:
It has been observed that some elements of the second period show similarities with the elements of the third period situated diagonally to each other, though belonging to different groups. This is called a diagonal relationship.

The cause of the diagonal relationship is due to the similarities in properties such as electronegativity, ionisation energy, size etc. between the diagonal elements. For example on moving from left to right across a period, the electronegativity increases, which on moving down a group, electronegativity decreases. Therefore on moving diagonally, two opposing tendencies almost cancel out and the electronegativity values remain almost the same as we move diagonally.

Question 6.
Why is the density of potassium less than that of sodium?

Answer:
Generally, in a group density increases with the atomic number, but potassium is an exception. It is due to the reason that the atomic volume of K is nearly twice Na, but its mass (39) is not exactly double of Na (23). Thus the density of potassium is less than that of sodium.

Question 7.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
Sodium and potassium ions (Na⁺ and K⁺) are larger than the corresponding Mg²⁺ and Ca²⁺ ions. Due to this lattice energy of Mg(OH)₂, Ca(OH)₂, MgCO₃ and CaCO₃. (Lattice energy is defined as the energy required to convert one mole of the ionic lattice into gaseous ions.

Thus lattices with smaller ions have higher lattice energies). The hydration energies of Mg²⁺ and Ca²⁺ are higher than Na⁺ and K⁺ because of their smaller sizes. But the difference in lattice energies is much more. Therefore, the hydroxides and carbonates of Mg²⁺ and Ca²⁺ are insoluble in water because of their higher lattice energies.

Question 8.
Why is it that the s-block elements never occur in free state/nature? What are their usual modes of occurrence and how are they generally prepared?

Answer:
The elements belonging to the s-block in the periodic table (i.e. alkali and alkaline earth metals) are highly reactive because of their low ionisation energy. They are highly electropositive forming positive ions. So they are never found in a free state.

They are widely distributed in nature in the combined state. They occur in the earth’s crust in the form of oxides, chlorides, silicates and carbonates.

Generally, a group I metals are prepared by the electrolysis of fused solution.
For example:

1. 
At cathode: Na⁺ + e^– → Na
At anode: Cl^– → Cl + e^–
Cl + Cl → Cl₂

2. KOH ⇌ K⁺ + OH^–
At cathode: K+ + e^– → K
At anode: 4OH → 4OH + 4e
4OH → 2H₂O + O₂
or
4OH^– → 2H₂O + O₂ + 4e^–
These metals are highly reactive and therefore cannot be extracted by the usual methods, because they are strong reducing agents.

Question 9.
Explain what happens when

(i) Sodium hydrogen carbonate is heated.
Answer:

(ii) Sodium with mercury reacts with water.
Answer:
2Na-Hg + 2H₂O → 2NaOH + H₂ ↑ + 2Hg

(iii) Fused sodium metal reacts with ammonia.
Answer:

Question 10.
What is the effect of heat on the following compounds?

(a) Calcium carbonate
Answer:

(b) Magnesium chloride hexahydrate
Answer:

(c) Gypsum
Answer:

(d) Magnesium sulphate heptahydrate.
Answer:

In This Post we are  providing Chapter-2 Hydrogen NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON HYDROGEN

1.Why does hydrogen occupy unique position in the periodic table?

Ans. Inspite of the fact that hydrogen, to a certain extent resembles both with alkali metals (ns’) and halogens (ns² np⁵), it differs from them as well. Hydrogen has very small size as a consequence H⁺ does not exist freely and is always associated with other atoms or molecules. Thus, it is unique in behaviors and is therefore, best placed separately in the periodic table.

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2.Give the main characteristics of isotopes.

Ans. Since, the isotopes have the same electronic configuration, they have almost the same chemical properties. The only difference is in their rates of reactions, mainly due to their different enthalpy of bond dissociation. However, in physical properly of these isotopes differ considerably due to their large mass differences.

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3.How can the production of dlhydrogen obtained from ‘coal gasification be increased’?

Ans. By reacting carbon monoxide of syngas mixtures with steam in the presence of iron chromate as catalyst

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4.Why is dihydrogen used an fuel cells for generating electrical energy?

Ans. Because it does not produce any pollution and releases greater energy per unit mass of fuel in comparison to gasoline or any other fuel.

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5.What is understood by hydrogenation?

Ans. Hydrogenation is used for the conversion of polyunsaturated oils into edible fats.

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6.Which fuel is used as a rocket fuel?

Ans. Dihydrogen is used as a rocket fuel in space research.

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7.What happens when sodium hydride reacts with water?

Ans. Saline hydride (sodium hydride) react violently with water producing dihydrogen gas 

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8.What is the geometry of the compound formed by group 14 to form molecular hydride?

Ans. Tetrahedral in structure.

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9.What are the characteristic features of ionic or saline hydrides?

Ans. The ionic hydrides are crystalline, non – volatile non – conducting in solid state. However their melts conduct electricity.

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10.Which gas is produced on electrolysis of ionic hydride?

Ans. Dihydrogen gas is produced at the anode on electrolysis of ionic hydride.

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11.How does H⁺ ion forms hydronium ion (OH₃⁺) in water?

Ans. In water H⁺ ion forms a covalent bond with H₂O and forms hydronium ion, (H₃O⁺).

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12.Show with reaction the amphoteric nature of water.

Ans. Water acts as an acid with NH₃ and base with H₂S

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13.Why is ice less dense then water and what kind of attractive forces must be overcome to melt ice?

Ans. The structure of ice is an open structure having a number of vacant spaces. Therefore, the density of ice is less than water. When ice melts the hydrogen bonds are broken and the water molecules go in between the vacant spaces. As a result, the structure of liquid water is less open than structure of ice. Thus ice is less dense than water                                                                                     

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14.Why does hard water not form lather with soap?

Ans. Hard water does not produce lather with soap readily because the cations (Ca²⁺ and Mg²⁺) present in hard water react with soap to precipitate of tcalcium and magnesium salts of fatly acids.

From hard water sodium stearate form Ca/Mg stearate

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15.Why is water an excellent solvent for ionic or polar substances?

Ans. Water is a polar solvent with a high dielectric constant. Due to high dielectric constant of water the force of attraction between cation and anion gets weakened. Thus water molecules are able to remove ions from the lattice site using in dipole forces easily.

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In This Post we are  providing Chapter-1 REDOX REACTION NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON REDOX REACTION

Question 1.
What are redox reactions? Give an example.

Answer:
Redox reaction is a reaction in which oxidation and reduction take place simultaneously, e.g.
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Question 2.
Define oxidation and reduction in terms of electrons.

Answer:
Oxidation involves loss and reduction involves the gain of electrons.

Question 3.
How many grams of K₂Cr₂O₇ is required to oxidize Fe²⁺ present in 15.2 gm of FeSO₄ to Fe³⁺ if the reaction is carried out in an acidic medium.

Answer:
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O
from the balanced equation, it is clear that 6 moles of FeSO₄ = 1

a mole of K₂Cr₂O₇ or 6 × 152 gm of FeSO₄ are oxidized by
K₂Cr₂O₇ = 294 gm
or
15.2 gm of FeSO₄ are oxidized by K₂Cr₂O₇

Question 4.
15.0 cm3 of 0.12 M KMnO₄ solution are required to oxidize 20 ml of FeSO₄ solution in an acidic medium. What is the concentration of FeSO₄ solution?

Answer:
The balanced chemical equation for the redox reaction is
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8 H₂O

Applying molarity equation to the above redox reaction

Question 5.
16.6 gm of pure KI was dissolved in water and the solution was made up to one liter, cm³ of this solution was acidified with 20 cm³ of 2 MHCl the resulting solution required 10 cm³ of decinormal KIO₃ for complete oxidation of I- ions to ICl. Find out the value of v.

Answer:
The chemical equation for the redox reaction is
IO- + 2I^– + 6HCl → 3ICl + 3Cl^– + 3H₂O

Molarity of KI solution = 16.6166 = 0.1 m

Applying molarity equation
0.1×v2 (KI) = 10×0.11(KIO3)
v = 20 cm³

Question 6.
Calculate the cone of hypo (Na₂S₂O₃ 5H₂O) solution in g dm-3 if 10.0 of this solution decolorized 15 ml of M/40 iodine solution.

Answer:
The balanced equation for the redox reaction is
2S₂O₃^2- + I₂ → 2I^– + S₄O₆^2-
from the balanced equation, it is evident that
2 moles of Na₂S₂O₃ = 1 mole of I₂

Applying the molarity equation we have,

Thus, the molarity of the hypo solution = 3/40 M
mol. mass of Na₂S₂O₃.5H₂O = 248 g mol^-1
cone, of Na₂S₂O₃.5H₂O = 248×340
= 18.6 gdm^-3

Question 7.
How many millimoles of potassium dichromate is required to oxidize 24 cm³ of 0.5 M mohr’s salt solution in an acidic medium.

Answer:
No. of millimoles of K₂Cr₂O₇ present in 24 cm³ of 0.5 m solution = 24 × 0.5 = 12
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6(NH₄)₂SO₄.FeSO₄.6H₂O + 7H₂SO₄ → K₂SO₄ + 6(NH₄)₂SO₄ + 3Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + 43H₂O from the balanced equations.

6 moles mohr’s salt are oxidised by K₂Cr₂O₇ = 1 moles
∴ 12 millimoles of mohr’s salt will be oxidised by
K₂Cr₂O₇ = 16 × 12 = 2 millimoies.

Question 8.
2.48 gm of Na₂S₂O₃.xH₂O was dissolved per liter of the solution. 20 cm³ of this solution required 10 cm³ of 0.01 M iodine solution. Find out the value of x?

Answer:
The balanced equation for the redox reaction is
2Na₂S₂O₃ + I2 → Na₂S₄O₆ + 2NaI
Let the molarity of Na₂S₂O₃ .xH₂O solution = M₁.

Applying molarity equation to the above redox reaction, we have
M1×202(Na2S2O3) = 10×.011(I2)
∴ M₁ = 0.01 M

mol wt. of Na₂S₂O₃.xH₂O
= 2 × 23 + 2 × 32 + 3 × 16 + x × 18
= 158 + 18x

Amount of Na₂S₂O₃.xH₂O present per litre
= (158 + 18x) × 0.01 g

But the actual amount dissolved = 2.48 g. equating these values, we have
(158 + 18x) × 0.01 = 2.48
or
x = 5.

Question 9.
The half cell reactions with their oxidation potentials are:
Pb(s) → Pb²⁺(aq) + 2e^–, E°_oxi = + 0.13 v.
Ag(s) → Ag⁺(aq) + e^–, E°_oxi = – 0.80 v.
Write the cell reaction and calculate its EMF.

Answer:
The equations are as
Pb²⁺(aq) + 2e^– → Pb(s), E°_oxi = – 0.13 v
Ag⁺(aq) + e^– → Ag(s), (E° = + 0.80 v) …(2)

To obtain the equation for the cell reaction, multiply equation (2) by 2 and subtract equation (1) from it we get
Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s)
E°_cell = + 0.80 – ( – 0.13) = + 0.93 v

Question 10.
Predict whether zinc or silver reacts with 1 M H₂SO₄ to give out hydrogen or not. Given that the standard potential of zinc & silver are – 0.76 v & + 0.80 v respectively.

Answer:
(a) To predict the reaction of zinc with H₂SO₄ If Zn reacts, the following reactions should take place.
Zn + H₂SO₄ → ZnSO₄ + H₂
i.e. Zn + 2H⁺ → Zn²⁺ + H₂

By conventions, the cell will be represented as-
Zn / Zn²⁺ ∥ H⁺ / H₂

standard EMF of the cell,
E°cell = E°_H⁺/H2 – E°_zn²⁺/zn
= 0 – (- 0.76) = + 0.76 v

Thus the EMF of the cells comes out to be positive. Hence the reaction takes place.

(b) To predict the reaction, of silver with H₂SO₄.
If Ag reacts, the following reactions should take place.
2Ag + H₂SO₄ → Ag₂SO₄ + H₂
2Ag + 2H⁺ → 2Ag⁺ + H₂

By convention, the cell may be written as
Ag / Ag⁺ ∥ H⁺ / H₂
E°_cell = E°_H⁺,H2 – E°_Ag⁺,Ag
= 0 – 0.80 = – 0.80 v.
The EMF of the cell is negative
Hence, this reaction does not take place.

In This Post we are  providing Chapter-7 EQULIBRIUM NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON EQUILIBRIUM

Question 1.
Justify the statement that water behaves like acid as well as a base on the basis of the protonic concept.

Answer:
Water ionizes as H₂O + H₂O ⇌ H₃O⁺ + OH^–
With strong acids, water behaves as a base by accepting a proton from an acid.
HCl + H₂O ⇌ H₃O⁺ (aq) + Cl^– (aq)
While with bases, water behaves as an acid by liberating a proton
NH₃ + H₂O ⇌ NH₄⁺ (aq) + OH (aq).

Question 2.
What is p^OH? What is its value for pure water at 298 K?

Answer:
p^OH = – log [OH^–]
p^H + p^OH = 14 for pure water at 298 K
p^H = 7
or
p^OH of water at 298 = 7.

Question 3.
Calculate the p^H of a buffer solution containing 0.1 moles of acetic acid and 0.15 mole of sodium acetate. The ionization constant for acetic acid is 1.75 × 10^-5.

Answer:

Question 4.
An aqueous solution of CuSO₄ is acidic while that of Na₂SO₄ is neutral. Explain.

Answer:
CuSO₄ + 2H₂O ⇌ Cu(OH)₂ + H₂SO₄ (weak base strong acid)
CuS04 is the salt of weak base Cu(OH)₂ and a strong acid H₂SO₄.
Thus the solution will have free H⁺ ions and will, therefore, be acidic.

Na₂SO₄, being the salt of a strong acid H₂SO₄ and a strong base.
NaOH does not undergo hydrolysis. The solution is, therefore, neutral.

Question 5.
The dissociation constants of HCN, CH₃COOH, and HF are 7.2 × 10^-10, 1.8 × 10^-5, and 6.7 × 10^-4 respectively. Arrange them in increasing order of acid strength.

Answer:
More the value of Ka, the stronger the acid
Their Ka1S are 6.7 × 10^-4 > 1.8 × 10^-5 > 7.2 × 10^-10
∴ HCN < CH₃COOH < HF.

Question 6.
The dissociation of PCl₅ decreases in presence of Cl₂. Why?

Answer:
For PCl₅ ⇌ PCl₃ + Cl₂.
According to Le Chatelier’s principle, an increase in the concentration of Cl₂ (one of the products) at equilibrium will favor the backward reaction, and thus the dissociation of PCl₅ into PCl₃ and Cl₂ decreases.

Question 7.
The dissociation of HI is independent of pressure while dissociation of PCl₅ depends upon the pressure applied. Why?•

Answer:
For 2HI ⇌ H₂ + I₂
K_c = x24(1−x)2
where x = degree of dissociation

For PCl₅ ⇌ PCl₃ + Cl₂
K = x2V(1−x)2; V = Volume of container.

K_c for HI does not have a volume factor- and dissociation is independent of volume and hence pressure.
K_c for PCl₅ has volume in the denominator and hence dissociation of PCl₅ depends upon the volume and consequently pressure.

Question 8.
The reaction between ethyl acetate and water attains a state of equilibrium in an open vessel, but not the decomposition of CaCO3. Explain.

Answer:
CH₂ COOC₂H₅ (l) + H₂O (l) ⇌ CH₃COOH (l) + C₂H₅OH (l)
Here both reactants and products are liquids and they will not escape from the vessel even if it is open. Therefore equilibrium is attained.
CaSO₃ (s) ⇌ CaO (s) + CO₂ (g)
Here CO₂ is a gas. It will escape from the vessel if it is open and so backward reaction cannot take place. Therefore equilibrium is attained.

Question 9.
The degree of dissociation of N₂O₄ is α according to the reaction
N₂O₄ (g) ⇌ 2NO₂ (g) at temperature T and total pressure P.
Find the expression for the equilibrium constant of this reaction.

Answer:

Total moles at eqbm. = 1 – α + 2α = 1 + α
If P is total pressure

Question 10.
Why NH₄Cl is added in precipitating III group hydroxides before the addition of NH4OH?

Answer:
To prevent precipitation of IV group hydroxides (especially Mn) along with III group hydroxides. NH₄Cl decreases dissociation of NH₄OH and thus limited OH ions are present in solution to precipitate III group cations only
NH₄OH ⇌ NH₄ (aq) + OH (aq) to a small extent
NH₄Cl (aq) ⇌ NH₄ (aq) + Cl (aq) to a large extent
Due to common ion affect the degree of dissociation of NH₄OH decreases leaving only a small no. of OH̅ ions.

Question 11.
If concentrations are expressed in moles L^-1 and pressures in atmospheres. What is the ratio of K_p to K_c for the
2SO₂ + O₂ (g) ⇌ 2SO₃ (g)
Answer:

Question 12.
The equilibrium constants for the reactions
N₂ + O₂ ⇌ 2NOand
2NO + O₂ ⇌ 2NO₂ are K₁ and K₂ respectively, then what would be the equilibrium constant for the reactions.
N₂ + 2O₂ ⇌ 2NO₂?
Answer:

Question 13.
What qualitative information can be obtained from the magnitude of the equilibrium constant?

Answer:

1. Large values of equilibrium constant (> 10³) show that the forward direction is favored i.e. concentration of products is much larger than that of the reactants of equilibrium.
2. Intermediate values of K (10^-3 to 10³) show that the concentrations of the reactants and products are comparable.
3. The low value of K (< 10^-3) shows that the backward reaction is favored, i.e., the concentration of reactants is much large than that of. products.

Question 14.
The following reaction has attained equilibrium
CO (g) + 2H₂ (g) ⇌ CH₃OH (g); ΔH° = – 92.0 kJ mol^-1.
What will happen if

(i) the volume of the vessel is suddenly reduced to half?
Answer:
K_c = [CH₃OH]/[CO][H₂]², K_p = PCH₃OH/PCO × PH₂
When the volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus
Q_c = 2[CH₃OH]/2[CO] × {2[H₂]}² = i K..
As Q_c < K_p, equilibrium will shift in the forward direction.

(ii) The partial pressure of hydrogen is suddenly doubled (ii) an inert gas is added to the system.
Answer:
As volume remains constant, molar concentration will not change. Hence there is no effect on the state of equilibrium.

Question 15.
How does the degree of ionization of a weak electrolyte vary with concentration? Give exact relationship. What is this law called?

Answer:
α = Ki/c−−−−√ . It is called Ostwald’s dilution law.
(K_i is ionization constant and c is the molar concentration).

In This Post we are  providing Chapter-6 THERMODYNAMICS NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON THERMODYNAMICS

1.With the help of first law of thermodynamics and H = U + pv, prove = q_p

Ans.The enthalpy is defined as

H = U + pv

For a change in the stales of system,

= 

= 

= …………(i)

The first law of thermodynamics states that –

= …………………….(ii)

From (i) and (ii),

= 

When the pressure is constant,

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2.Why is the difference between and not significant for solids or liquids?

Ans. The difference between and is not usually significant for systems consisting of only solids and / or liquids because they do not suffer any significant volume changes upon heating.

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3.What is an extensive and intensive property?

Ans. Extensive property is a property whose value depends on the quantity or size of matter present in the system.

Intensive property is a property which do not depend upon the quantity or size of matter present.

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4.Calculate the heat of combustion of ethylene (gas) to from CO₂ (gas) and H₂O (gas) at 298k and 1 atmospheric pressure. The heats of formation of CO_2, H₂O and C₂H₄ are – 393.7, – 241.8, + 52.3 kJ per mole respectively.

Ans. C₂H₄ (g) + 30₂(g) 2CO₂(g) + 2H₂O (g)

reactants

= [2 x (CO₂) + 2 x ] – 

= 2 x[(-393.7)m+2x (-241.8)] – [(523.0) + 0)]

= [-787.4 – 483.6 ] -53.3

= – 1323.3 kJ.

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5.Give two examples of reactions which are driven by enthalpy change.

Ans. Examples of reactions driven by enthalpy change:

The process which is highly exothermic, i.e. enthalpy change is negative and has large value but entropy change is negative is said to be driven by enthalpy change, eg.

(i)

(ii)

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6.Will the heat released in the following two reactions be equal? Give reasons in support of your answer.

(i)H₂ (g) + 

Ans. No, the heats released in the two reactions are not equal. The heat released in any reaction depends upon the reactants, products and their physical states. Here in reaction (i), the water produced is in the gaseous state whereas in reaction (ii) liquid is formed. As we know, that when water vapors condensed to from water, heat equal to the latent heat of vaporization is released. Thus, more heat is released in reaction (ii).

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7.What is the relation between the enthalpy of reaction and bond enthalpy?

Ans .A chemical reaction involves the breaking of bonds in reactants and formation of new bonds in products. The heat of reaction (enthalpy change) depends on the values of the heat needed to break the bond formation .Thus

(Heat of reaction = (Heat needed to break the bonds in reactants – Heat liberated to from bonds in products).

= Bond energy in (to break the bonds) – Bond energy out (to form the bonds)

= Bond energy of reactants – Bond energy of products.

8.The are given + 61.17kJ mol^-1 and + 132 Jk^-1mol^-1 respectively. Above what temperature will the reaction be spontaneous?

Ans. The reaction

2Ag₂O (s) 

Will be spontaneous when is negative.

Since 

Shows that would be –ve when,

Or T > 

The process will be spontaneous above a temperature of .

9.Give the relationship between for gases.

Ans.For gases the volume change is appreciable.

let V_A be the total volume of gaseous reactants, and

V_B be the total volume of gaseous product.

n_A be the number of moles of the reactant and

n_B be the number of moles of the product,

Then at constant pressure and temperature,

p V_A = n_A RT

p V_B = n_B RT

or p V_B – pV_A = (n_B – n_A) RT

or p 

where and is equal to the difference between the number of moles of gaseous products and gaseous reactants.

Substituting the value of p awe get.

(heat change under constant pressure)

(heat change under constant volume)

for gaseous system.

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10.It has been found that 221.4J is needed to heat 30g of ethanol from 15⁰C to 18⁰C. calculate (a) specific heat capacity, and (b) molar heat capacity of ethanol.

Ans.(a) Specific heat capacity

C = 

= 

Since 1⁰C is equal to 1k, the specific heat capacity of ethanol = 2.46Jg^-1 0c^-1.

(b) Molar heat capacity, Cm = specific heat x molar mass.

Therefore, Cm (ethanol) = 2.46 x 46

= 113.2 Jmol^-1 0c^-1

The molar heat capacity of ethanol is 113.2 J mol^-1 0c^-1.

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In This Post we are  providing Chapter-5 STATES OF MATTER NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON STATES OF MATTER

Question 1.
Ammonia and Sulphur dioxide gases are prepared in two comers of a laboratory. Which gas will be detected first by a student working in the middle of the laboratory and why?

Answer:
Molecular mass of NH₃ = 17 Molecular mass of SO₂ = 64
NH₃ is a lighter gas and diffuses at a faster speed than SO₂
∴ NH₃ gas will be detected first.
[∵ Acc. to Graham’s law of diffusion r1r2=d2d1−−√

Question 2.
What is the effect of hydrogen bonding on the viscosity of a liquid?

Answer:
Hydrogen bonding leads to an increase in the effective size of the moving unit in the liquid. Due to an increase in the size and mass of the molecule, there is greater interval resistance of the flow of the liquid. As a result, the viscosity of the liquid rises.

Question 3.
Which are the two faulty assumptions in the kinetic theory of gases.

Answer:

1. There is no force of attraction between the molecules of the gas.
2. The volume of the molecules of the gas is negligibly small as compared to the total space (volume) occupied by the gas.

Question 4.
What is the relationship between the density and molar mass of a gaseous substance? Derive it.

Answer:
Ideal gas equation is PV = nRT
or nV=pRT
Replacing n by mM, we get

Question 5.
What is meant by the term: Non-ideal or real gas?

Answer:
The gas which does not obey the Cas law:
Boyle’s law, Charles’ law, Avogadro^ law at all temperatures and pressures is a called-non-ideal or real gas. Most of the real gases show ideal behaviour at low pressure and high temperature.

Question 6.
Derive the ideal gas equation PV = nRT.

Answer:
According to Boyle’s law V ∝ 1P if n and T are constant.
According to Charles’ law V ∝ T at constant P.and n
According to Avogadro’s law V ∝ n at constant T and P
Combining the three laws
In
V ∝ TnP
or
PV ∝ nT
or
PV = nRT where R is a constant of proportionality called universal gas constant.

Question 7.
Why liquids have a definite volume, but no definite shape?

Answer:
It is due to the fact that in liquids intermolecular forces are strong enough to hold the molecules together, but these are strong enough to hold the molecules together, but these forces are not strong enough to fix them into definite or concrete positions as in solids. Hence they possess fluidity but no definite shape.

Question 8.
How do the real gases deviate from ideality above and below the Boyle point?

Answer:
Above their Boyle point, real gases show positive deviations from ideality and the values of Z are greater than one. The forces of attraction between the molecules are very feeble. Below the Boyle point, real gases first show a decrease, in Z value with increasing pressure, the value Of Z increases continuously.

Question 9.
Write down the van der Waals, equation for n moles of a real gas. What do the constants ‘a’ and ‘b’ stand for?

Answer:
(P + an2V2) (V – nb) = nRT
where p, Vindicate gas pressure and its volume. T is the Kelvin temperature of the gas. R is gas constant. Value of ‘a’ is a measure of the magnitude of intermolecular attractive forces within the molecule and is independent of temperature and pressure, ‘rib’ is approximately the total volume occupied by the molecules themselves. ‘a’ and ‘b’ are called van der Waals constants and their value depends upon the nature of the gas.

Question 10.
One mole of S02 gas occupied a volume of 350 mL at 27°C and 50 atm pressure. Calculate the compressibility factor of the gas. Comment on the type of deviation shown by the gas from ideal behaviour.

Answer:
Compressibility factor, Z =
Here n = 1, P = 50 atm, V = 350 × 10^-3 L = 0.35 L
R = 0.0821 L atm K^-1 mol^-1; T = 27 + 273 = 300 K
∴ Z = 50×0.351×0.0821×300 = 0.711
For an ideal gas Z = 1
As for the given gas Z < 1, it shows a negative deviation, i.e., it is more compressible than expected from ideal behaviour.

Question 11.
A mixture of CO and CO₂ is found to have a density of 1.5 g L^-1 at 20°C and 740 mm pressure. Calculate the composition of the mixture.

Answer:
1. Calculation of average molecular mass of the mixture
M = dRTP=1.50×0.0821×293740760 = 37.06

2. Calculation of percentage composition
Let mol% of CO in the mixture = x
Then mol% of CO? in the mixture = 100 – x

Average molecular mass

Question 12.
A 5-L vessel contains 14 g of nitrogen. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K.

Answer:
N₂ ⇌ 2N

Initial moles = 1.428 = 0.05
Moles left = 0.05 – 30100 × 0.05 = 2 × 0.015 = 0.03

∴ Total no. of moles = 0.035 + 0.030 = 0.065
i.e. n = 0.065 mol, V = 5 L, T = 1800 K; P =?
P = nRTV=0.065×0.0821×18005
= 1.92 atm.

Question 13.
A given mass of a gas occupies 919.0 mL in a dry state at STP. The same mass when collected over water at 15°C and 750 mm pressure occupies on litre volume. Calculate the vapor pressure of water at 15°C.

Answer:
If p is the vapor pressure of water at 15°C, then P₂ = 750 – p
From the gas equation P1V1T1=P2V2T2, we get
760×919273=(750−p)×1000288
or
p = 13.3 mm

∴ Vapour pr. of water = 13.3 mm.

Question 14.
A steel tank containing air at 15 atm pressure at 15°C ¡s provided with a safety valve that will yield at a pressure of 30 atm. To what minimum temperature must the air be heated to below the safety valve?

Answer:
P1P2=T1T2
i.e., 1530=288 T2
or
T₂ = 576 K
or
t₂°C = 576 – 273 = 303°C

Question 15.
Calculate the pressure exerted by 110 g of CO₂ in a vessel of 2 L capacity at 37°C. Given that the van der Waals constants are a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1. Compare the value with the calculated value if the gas were considered ideal.

Answer:
According to van der Waals equation

110
Here, n = 11044 = 2.5 moles. Putting the given values, we get
P = 2.5×0.0821×310(2−2.5×0.0427)−3.59×2.52
= 33.61 atm – 5.61 atm = 28.0 atm

If the gas were considered an ideal gas. applying ideal gas equation
PV = nRT
or
P = nRTV
∴ P = 2.5×0.0821×3102 = 31.8 atm

In This Post we are  providing Chapter-4 CHEMICAL BONDING AND MOLECULAR STRUCTURE NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON CHEMICAL BONDING AND MOLECULAR STRUCTURE

1.Give the main feature of Kossel’s explanation of chemical bonding.

Ans. Kossel in relation to chemical bonding drew attention to the following facts –

(i) In the periodic table, the highly electronegative halogens and the highly electropositive alkali metals are separated by the noble gases.

(ii) In the formation of a negative ion from a halogen atom and a positive ion from an alkali metal, atom is associated with a gain and loss of an electron by the respective atoms.

(iii) The negative and positive ions so formed attain stable noble gas electronic configurations. The noble gases have particularly eight electrons, ns² np⁶.

The –ve and +ve ions are stabilized by electrostatic attraction.

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2.How can you explain the formation of NaCl according to kossel concept?

Ans. The formation of NaCl from sodium and chlorine can be explained as

Na ® Na⁺ + e^–

[Ne] 3s¹ ® [Ne]

Cl + e^– ® Cl^–

[Ne] 3s² 3p⁵ . [Ne] 3s² 3p⁶ or [Ar]

Na⁺ + Cl^– ® Na⁺ Cl^– or NaCl.

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3.Write the significance of octet rule.

Ans. Octet rule signifies –

(i) It is useful for understanding the structures of most of the organic compounds.

It mainly applies to the second period elements of the periodic table.

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4.Write the Lewis structure for CO molecule

Ans. (i) The outer (valence) shell configurations of carbon and oxygen atoms are

Carbon : (6) – 1s² 2s² 2p²

Oxygen : (8) – 1s² 2s² 2p⁴.

The valence electrons (4 + 6 = 10)

But it does not complete octet, thus multiple bond is exhibited.

Thus,

(ii) N (2s² 2p³), O (2s² 2p⁴)

5 + (2 x 6) + 1 = 18 electrons.

Thus,

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5.Give the Lewis dot structure of HNO₃

Ans. HNO₃ ®

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6.What changes are observed in atoms undergoing ionic bonding?

Ans. Due to the electron transfer the following changes occurs –

(i) Both the atoms acquire stable noble gas configuration.

(ii) The atom that loses electrons becomes +vely charged called cation whereas that gains electrons becomes –vely charged called anion.

(iii) Cation and anion are held together by the coulombic forces of attraction to form an ionic bond.

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7.Mention the factors that influence the formation of an Ionic bond.

Ans.Ionic bond formation mainly depends upon three factors –

(i) Low ionization energy – elements with low ionization enthalpy have greater tendency to form an ionic bonds.

(ii) High electron gain enthalpy – high negative value of electron gain enthalpy favours ionic bond.

(iii) Lattice energy – high lattice energy value favours ionic bond formation.

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8.Give reason why H₂⁺ ions are more stable than H₂^– though they have the same bond order.

Ans.In H₂^– ion, one electron is present in anti bonding orbital due to which destabilizing effect is more and thus the stability is less than that of H₂⁺ ion.

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9.How would the bond lengths vary in the following species? C₂, C₂^– C₂^2-.

Ans.The order of bond lengths in C₂ , C₂^– and C₂^2- is C₂ > C₂^– > C₂^2-.

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10.Out of covalent and hydrogen bonds, which is stronger.

Ans. Covalent bond.

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11.Define covalent radius.

Ans. The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation.

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12.Why NH₃ has high dipole moment than NF₃ though both are pyramidal?

Ans. In case of NH₃ the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N-H bonds, whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of the three N-F

bonds. The orbital dipole become of lone pair decreases, which results in the low dipole moment.

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13.Draw the resonating structure of NO₃^–

Ans.

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14.On which factor does dipole moment depend in case of polyatomic molecules.

Ans.The dipole moment of the polyatomic molecule depends on individual dipole moments of bonds and also on the spatial arrangement of various bonds in the molecule.

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15.Dipole moment of Be F₂ is zero. Give reason.

Ans. In BeF₂ the dipole moment is zero because the two equal bond dipoles point in opposite directions and cancel the effect of each other.

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In This Post we are  providing Chapter-3 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Question 1.
Calculate the energy required to convert all atoms of magnesium to magnesium ions present in 24 mg of magnesium vapors. IE₁ and IE₂ of Mg are 737.76 and 1450.73 kJ mol-1 respectively.

Answer:
Mg (g) + IE₁ → Mg+ (g) + e^– (g)
IE₁ = 737.76 kJ mol^-1

Mg+ (g) + IE₂ → Mg²⁺ + e^– (g)
IE₂ = 1450.73 kJ mol^-1

Total amount of energy needed to convert Mg (g) into ,
Mg²⁺ ion = IE₁ + IE₂

Now 24 mg of Mg = 241000 g = 241000×24 mol [1 Mole of Mg = 24 g]
= 10^-3 mole.

Question 2.
The IE₁ and IE₂ of Mg. (g) are 740 and 1450 kJ mol^-1. Calculate the percentage of Mg⁺ (g) and Mg²⁺ (g) if 1 g of Mg (g)
absorbs 50 kJ of energy.

Answer:
No. of moies of Mg vapours present in I g = 124 = 0.0147

Energy absorbed to convèrt Mg (g) to Mg⁺ (g) = 0.0417 × 740
=30.83 kJ

Energy left unused = 50 – 30.83
= 19.17 kJ
Now 19.17 kJ will be used to e Mg⁺ (g) to Mg²⁺ (g) .
∴ No. of moles of Mg+ (g) converted to Mg2+ (g) = 19.171450 = 0.132

%age of Mg⁺ (g) = 0.02850.0417 × 100 = 68.35
and %age of Mg²⁺ (g) = 100 – 68.35 = 31.65

Question 3.
Which of the elements Na, Mg, Si, P would have the greatest difference between the first and the second ionization enthalpies. Briefly explain your answer

Answer:
Among Na, Mg, Si, P, Na is an alkali metal. It has only one electron in the valence shell. Therefore, its IE₁ is low: However, after the removal of the first electron, it acquires a Neon gas configuration i.e., Na+ (1s², 2s² 2p⁶). Therefore its IE₂ is expected to be very high. Consequently, the difference in IE₁ and IE₂ comes to be greatest in the case of Na.

Question 4.
The IE₂ of Mg is higher than that of Na. On the other hand, the IE₂ of Na is much higher than that of Mg. Explain,

Answer:
The first electron in both cases has to be removed from the 3s orbital, but the nuclear charge of Na is less than that of Mg. After the removal of the first electron from Na, the electronic configuration of Na⁺ is 1s², 2s² 2p⁶, i.e., that of noble gas which is very stable and the removal of the 2nd electron is very difficult. In the case of Mg after the removal of the first electron, the electronic configuration of Mg⁺ is 1s², 2s² 2p⁶ 3s. The 2nd electron to be removed is again from 3s orbital which is easier.

Question 5.
The amount of energy released when 1 × 10¹⁰ atoms of chlorine in vapor state are converted to Cl^– ions according to the equation.
Cl (g) + e^– → Cl^– (g) is 57.86 × 10^-10 J
Calculate the electron gain enthalpy of the chlorine atom in terms of kJ mol^-1 and eV pet atom.
Answer:
The electron gain enthalpy of chlorine

Question 6.
Electronic configuration of the four elements are given below:
Arrange these elements in increasing order of their metallic character. Give reasons for your answer.

(i) [Ar]4s²
Answer:
[Ar]4s² is Calcium metal with At. no. = 20.

(ii) [Ar]3d¹⁰ 4s²
Answer:
[Ar]3d¹⁰ 4s² is Zinc metal with At. no. = 30.

(iii) [Ar]3d¹⁰ 4s² 4p⁶ 5s²
Answer:
[Ar]3d¹⁰ 4s² 4p⁶ 5s² is Strontium metal with At. no. = 38.

(iv) [Arl 3d¹⁰ 4s² 4p⁶ 5s¹
Answer:
[Ar] 3d¹⁰ 4s² 4p⁶, 5s¹ is*Rubidium metal with At. no. = 37.

Alkali metals are the most metallic, followed by alkaline earth metals and transition metals. Among alkali metals – Rubidium (37) is the most metallic. Among alkaline earth metals (Ca, Sr) Sr (Strontium) is more metallic than Calcium (Ca) as the metallic character increases from top to bottom in a group. Zinc – the transition metal is the least metallic. Thus metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv)

Question 7.
The formulation of F (g) from F (g) is exothermic whereas that of O^2- (g) from O (g) is endothermic. Explain.

Answer:
F (g) + e^– (g) → F (g); ΔH = Negative
Energy is released when an extra electron from outside is added to a neutral isolated gaseous atom of an element.
To convert, O (g) to O^2- (g) two steps are required
(i) O (g) + e^– (g) → O^– (g); ΔH₁ = – 141 kJ mol^-1
(ii) O^– (g) + e^– (g) → O^2- (g); ΔH₂ = + 780 kJ mol^-1

Hence the over all processes endothermic (+ 780 – 141 = + 639 kJ mol^-1) whereas F (g) to F^– (g) is exothermic.

Question 8.
Explain the important general characteristics of groups in the modem periodic table in brief.

Answer:
The elements of a group show the following important similar characteristics.
(0 Electronic configuration. All elements in a particular group have similar outer electronic configuration e.g., all elements of group I’, i.e., alkali metals have ns1 configuration in their valency shell. Similarly, group 2 elements (alkaline Earths) haye ns² outer configuration and halogens (group 17) have ns² np⁵ configuration (where n is the outermost shell).

(it) Valency. The valency of an element depends upon the number of electrons in the outermost shell. So elements of a group show the same valency, e.g., elements of group 1 show + 1 valency and group 2 show + 2 valencies i.e. valency i.e., NaCl > MgCl₂ etc.

(iii) Chemical properties. The chemical properties of the elements are related to the number of electrons in the outermost shell of their atoms. Hence all elements belonging to the same group show similar chemical properties. But the degree of reactivity varies gradually from top to bottom in a group. For example, in group 1 all the elements are highly reactive metals but the degree of reactivity increases from Li to Cs. Similarly, elements of group 17, i.e., halogens: F, Cl, Br, I are all non-metals and they’re- reactivity goes on decreasing from top to bottom.

Question 9.
Explain the electronic configuration in periods in the periodic table.

Answer:
Each successive period in the periodic table is associated with the filling Up of the next higher principal energy level (n – 1, n – 2, etc.). It can be readily seen that the number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. The first period starts with the filling of the lowest level (1s) and has thus the two elements – hydrogen (1s¹) and helium (1s²) when the first shell (K) is completed. The second period starts with lithium and the third electron enters the 2s orbital.

The next element, beryllium has four electrons and has the electronic configuration 1s² 2s². Starting from the next element boron, the 2p orbitals are filled with electrons when the L shell is completed’ at neon (2s² 2p⁶). Thus there are 8 elements in the second period. The third period (n = 3) being at sodium, and the added electron enters a 3s orbital. Successive filling of 3s and 3p orbitals give rise to the third period of 8 elements from sodium to argon.

The fourth period (n = 4) starts at potassium with the filling up of 4p of 4s orbital. Before the 4p orbital is filled, the filling up of 3d orbitals becomes energetically favorable and we come across the so-called 3d transition series of elements. The fourth period ends at krypton with the filling up of the 4p orbitals. Altogether we have 18 elements in the fourth period. The fifth period (n = 5) beginning with rubidium is similar to the fourth period and contains the 4d transition series starting at yttrium (Z = 39).

This period ends at xenon with the filling up of the 5p orbitals. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4/, 5d, and 6p orbitals, in that order. Filling up of the 4/ orbitals being with cerium, (Z = 58) and ends at lutetium (Z = 71) to give the 4/-inner transition series which is called the lanthanide series. The seventh period (n = 7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d, and 7p orbitals and includes most of the man-made radioactive elements.

This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z = 89) gives the 5f-inner transition series known as the actinide series. The 4f and 5f transition series of elements are placed separately in the periodic table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column.

Question 10.
Explain the variation of valence in the periodic table.

Answer:
Variation of valence in a group as well as across a period in the periodic table occurs as follows:
1. In a group: All elements in a group show the same valency. For example, all alkali metals (group 1) show a valency of 1+. Alkaline earth metals (group 2) show a valency of 2+.

However, the heavier elements of p-block elements (except noble gases) show two valences: one equal to the number of valence electrons or 8-No. of valence electron# and the other two less. For example, thallium (Tl) belongs to group 13. It shows valence of 3+ and 1+.

Lead (Pb) belongs to group 14. If shows valance of 4+ and 2+.
Antimony (Sb) and Bismuth (Bi) belong to group 15. They show valence of 5+ and 3+ being more stable.

This happens due to the non-participation of tie two s-electrons present in the valence shell of these elements. This non-participation of one pair of s-electrons in bonding is called the inert-pair effect.

2. In a period: The number of the valence electrons increases – in going from left to right in a period of the periodic table. Therefore the valency of the elements in a period first increases, and then decreases.

In This Post we are  providing Chapter-2 STRUCTURE OF ATOM NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON STRUCTURE OF ATOM

1.Which experiment led to the discovery of electrons and how?  

Ans:The cathode ray discharge tube experiment performed by J.J. Thomson led to the discovery of negatively charged particles called electron.

A cathode ray tube consists of two thin pieces of metals called electrodes sealed inside a glass tube with sealed ends. The glass tube is attached to a vacuum pump and the pressure inside the tube is reduced to 0.01mm. When fairly high voltage (10, 000V) is applied across the electrodes, invisible rays are emitted from the cathode called cathode rays. Analysis of this rays led to the discovery electrons.

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2.Give the main properties of canal ray experiment.

Ans:The canal ray experiment led to the discovery of –

(i)The anode rays, travel in straight line

(ii)They are positively charged as they get deflected towards the –ve end when subjected to an electric and magnetic field.

(iii)They depend upon the nature of gas present in the cathode tube.

(iv)The charge to mass ration (e/m) of the particle is found to depend on the gas from which they originate.

(v)They are also material particles

The analysis of these proportions led to the discovery of positively charged proton.

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3.Find out atomic number, mass number, number of electron and neutron in an element?

Ans: The mass no. of 

The atomic no. of  

No. of proton is = Z – A = 40 – 20 = 20

No. of electron its (A) = 20

No. of proton is (A) = 20

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4.Give the main features of Thomson’s Model for an atom.

Ans: J.J. Thomson proposed that an atom consists of a spherical sphere (radius of about 10^-10m)in which the positive charges are uniformly distributed the electrons are embedded into it in such a manner so as to give stable electrostatic arrangement.

This model is also called raisin pudding model.

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5.What did Rutherford conclude from the observations of   scattering experiment?

Ans: Rutherford proposed the nuclear model of an atom as

(i) The positive charge and most of the mass of an atom was concentrated in an extremely small region. He called it nucleus.

(ii) The nucleus is surrounded by electrons that move around the nucleus with a very high speed in orbits.

(iii) Electron and nucleus are held together by electrostatic forces of attraction.

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6.What is the relation between kinetic energy and frequency of the  photoelectrons?

Ans: Kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation.

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7.What transition in the hydrogen spectrum would have the same  wavelength as the Balmer transition, n = 4 to n = 2 of He⁺ spectrum?

Ans: For the Balmer transition, n = 4, to n = 2 in a He⁺ ion, we can write.

For a hydrogen atom

Equating equation (ii) and (i), we get

This equation gives n₁ = 1 and n = 2. Thus the transition n = 2 to n = 1 in hydrogen atom will have same wavelength as transition, n = 4 to n = 2 in He⁺

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8.Spectral lines are regarded as the finger prints of the elements. Why?

Ans: Spectral lines are regarded as the finger prints of the elements because the elements can be identified from these lines. Just like finger prints, the spectral lines of no two elements resemble each other.

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9.Why cannot the motion of an electron around the nucleus be determined

accurately?

Ans: Because there is an uncertainty in the velocity of moving electron around the nucleus (Heisenberg’s Uncertainty Principle).

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10.Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length

Ans: According to uncertainty Principle

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11.Give the mathematical expression of uncertainty principle.    

Ans:Mathematically, it can be given as 

Where is the uncertainty in position and  is the uncertainty in momentum (or velocity) of the particle.

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12.Which quantum number determines

(i)  energy of electron      

(ii) Orientation of orbitals.

Ans. (i) Principal quantum number (n), and

(ii) Magnetic quantum number (m).

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13.Arrange the electrons represented by the following sets of quantum number in decreasing order of energy. 

1.   n = 4,  l = 0, m = 0, s = +1/2

2.   n = 3,  l = 1, m = 1,  s = -1/2

3.   n = 3,  l = 2, m = 0, s = +1/2

Ans.(i)Represents 4s orbital

(ii) Represents 3p orbital

(iii)Represents 3d orbital

(iv)Represents 3s orbital

The decreasing order of energy   3d > 4s > 3p > 3s

n = 3,  l = 0, m = 0, s = -1/2

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14.What designations are given to the orbitals having

(i)    n = 2, l = 1     (ii)   n = 2, l = 0     (iii) n = 4, l = 3

(iv)   n = 4, l = 2      (v)    n = 4, l = 1? 

Ans.    (i)           Here, n = 2, and l = 1

Since l = 1 it means a p-orbital, hence the given orbital is designated as 2p.

(ii) Here, n = 2 and l = 0

Since l = 0 means s – orbital, hence the given orbital is 2s.

(iii) Here, n = 4 and  l = 3

Since, l = 3 represents f – orbital, hence the given orbital is a 4f orbital.

(iv) Here, n = 4 and l = 2

Since, l = 2 represents d – orbital, hence the given orbital is a 4d – orbital.

(v) n = 4  and l = 1

 since, l = 1 means it is a p – orbital, hence the given orbital can be designated as – 4p orbital.

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15.Write the electronic configuration of (i) Mn⁴⁺, (ii) Fe³⁺ (iii)  Cr²⁺ and Zn²⁺ Mention the number of unpaired electrons in each case.

Ans.(i)       Mn  (z = 25), Mn⁴⁺  (z = 21)

The electronic configuration of Mn⁴⁺ to Given by

1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

As the outermost shell 3d has 3 electrons, thus the number of unpaired

electrons is 3.

 (ii) Fe   (z = 26), Fe³⁺ (z = 23)

The electronic configuration of Fe³⁺ is given lay

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

The number of unpaired electron is 5.

(iii) Cr (z = 24),  Cr²⁺ (z = 22)

The electronic configuration of Cr²⁺ is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴

The number of unpaired electron is 4.

(iv) Zn (z = 30), Zn²⁺  (z = 28)

The electronic configuration of Zn²⁺ is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

The number of unpaired electron is 0.

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