P-Block Elements Class 11 Notes

Class 11 Chemistry Some P Block Elements Long Questions And Answers

Question 1. Unlike diamond, graphite is a good conductor of electricity —explain.
Answer:

In a diamond crystal, each carbon atom is sp³-hybridized, meaning that all electrons in the valence shell of the carbon atom engage in the formation of sigma bonds with four adjacent carbon atoms.

• Consequently, there are no free electrons remaining in the lattice, rendering the diamond incapable of conducting electricity.
• Conversely, each carbon atom in graphite, being sp²-hybridized, utilizes three of its valence electrons to establish three σ-bonds with three adjacent carbon atoms.
• The fourth electron in the 2pz orbital, oriented perpendicularly to the carbon layer, establishes an n-bond through lateral overlap with the 2pz orbital of any of the three neighboring carbon atoms.
• The electrons of the π-bonds are delocalized throughout the entire crystal via resonance. Graphite exhibits excellent electrical conductivity owing to the existence of mobile electrons.

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Question 2. Diamond is extremely hard but graphite is soft and slippery —explain with reason.
Answer:

In diamond crystal, each sp³-hybridized carbon atom is tetrahedrally bonded to four other carbon atoms via covalent connections.

• A multitude of tetrahedral units interconnected to create a three-dimensional extensive array of molecules with robust bonds extending in all directions. Due to its three-dimensional network of robust covalent connections, diamond exhibits exceptional hardness.
• Conversely, in graphite crystals, each sp²-hybridized carbon atom is bonded to three other carbon atoms, creating a network of planar hexagons.
• These two-dimensional layers are arranged in parallel planes, positioned horizontally above one another at a distance of 3.35 Å.
• Due to the weak van der Waals forces binding these layers, one layer can effortlessly glide over another with minimal pressure applied. This elucidates the reasons behind graphite’s softness and lubricity.

P-Block Elements Class 11 Notes

Question 3. Carbon monoxide possesses both oxidising and reducing properties—why?
Answer:

The oxidation slate of carbon in carbon monoxide is +2. It stands in between its highest oxidation state (+4) and lowest oxidation state (-4). As a result, in chemical reactions, the carbon atom in CO may increase its oxidation number from +2 to +4 or it may decrease its oxidation number from +2 to -4. When the oxidation number increases, CO is oxidised and plays the role of a reducing agent. For example, at high temperatures, CO reduces black CuO to red metallic Cu.

On the other hand, in case of a decrease in the oxidation number of carbon, carbon monoxide undergoes reduction, i.e., it acts as an oxidising agent. For instance, in the following reaction, carbon monoxide oxidises hydrogen-producing water and itself gets reduced to methane

Question 4. How will you convert a mixture of CO and CO₂ completely into

1. CO₂ and
2. CO?

Answer:

When a mixture of CO₂ and CO is passed over strongly heated CuO and kept in a combustion tube, CO₂ remains unchanged while CO gets oxidised to CO₂. As a consequence, only CO₂ comes out of the combustion tube. In this way, the mixture is completely converted into CO₂

On the other hand, when a mixture containing CO and CO₂ is passed over white-hot coke, CO remains unchanged but CO₂ gets reduced to CO. In this way, the mixture is completely converted into CO.

Mpbse Class 11 Chemistry P-Block Solutions

Question 5. Mention 3 similarities like B and Si.
Answer:

Three similarities between boron and silicon are

1. Both boron and silicon react with caustic soda to form H₂ gas.

2B + 6NaOH → + 3H₂

Si + 2NaOH + H₂O →  Na₂SiO₃ + 2H₂

2. Halides of both boron and silicon undergo hydrolysis to form weak acids

BCl₃ + 3H₂O→H₃BO₃+ 3HCl

SiCl₄ + 4H₂O→ H₄SiO₄ + 4HCl

3. Halides of both boron and silicon form complex compounds

BF₃ +HF →  HBF₄; SiF₄ + 2HF → H₂SiF₆

Question 6. When boron trichloride reacts with water, it only B3 forms [B(OH)₄]^– whereas aluminium trichloride forms [AI(H₂O)₄]³⁺ in acidified aqueous solution. State the hybridisation of boron and aluminium in these species and explain your answer
Answer:

Boron trichloride (BCl₃ ) undergoes hydrolysis to form orthoboric acid [B(OH)₃] at first As the atomic size of B is small and its electronegativity is high, B(OH)₃ polarises HO molecule by accepting an OH^– ion thereby forming [B(OH)₄]^– and releasing a proton.

BCl₃ + 3H₂O→B(OH)₃ + 3HCl

B(OH)₃ + H₂O→[B(OH)₄]^–+ H⁺

There are no vacant d -orbitals in B as it lies in the second period and has only one s – and three p – orbitals. So, B can have four pairs of electrons in its valence shell, i.e., its maximum coordination number is 4.

For this reason, B(OH)₃ accepts one OH” ion to form [B(OH)₄]- in which B-atom is sp³ -hybridised. On the other hand, AlCl₃ undergoes hydrolysis in acidic medium to form [Al(H₂O)g]³⁺

AlCl₃ + 6H₂O →[Al(H₂O)₆]₃+ + 3Cl(aq)^–

In an acidic medium, the concentration of OH^– ions is lower than H+ ions. Thus, Al³⁺ ions coordinate with H₂O molecules and not with OH^– ions. Due to the availability of vacant d -d-orbitals in Al³⁺ ions, it can expand its coordination number from 4 to 6. Thus, it can form the complex ion, [Al(H₂O)₆]³⁺ in which hybridisation state of Al is sp³d²

Chemical Properties Of P-Block Elements Class 11

Question 7. Give reasons for the following:

1. Diamond-tipped tools are used for drilling and cutting purposes.
2. Graphite is used as a lubricant.
3. Silicones are water-repelling in nature.
4. CO gets absorbed by ammoniacal cuprous chloride to form a complex but CO₂ does not

Answer:

1. Diamond has the highest thermal conductivity among all known substances. Because of its high thermal conductivity, diamond-tipped tools do not overheat. Diamond is extremely hard as well. For these reasons, diamond-tipped tools are extensively used for drilling and cutting purposes.

2. Graphite has a layered structure in which any two successive layers are held together by weak van der Waals forces of attraction and thus one layer can smoothly slide over the other. For this reason, graphite acts as a lubricant.

3. Silicones are water-repelling in nature because they are surrounded by non-polar alkyl groups.

4. Carbon monoxide acts as a Lewis base due to the presence of a lone pair of electrons on carbon :C=0: and forms a soluble complex with ammoniacal cuprous chloride (CuCl).

CuCl + NH₃ + CO→ [Cu(CO)NH₃]⁺Cl^–

On the other hand, carbon dioxide  does not have a lone pair ofelectrons on carbon and cannot form complexes.

Question 8. What happens when

1. A mixture of sand and sodium carbonate is melted on heating. 0 At 200°C and under high pressure, carbon monoxide is passed through caustic soda solution and the product is heated to 300°C.
2. At high temperatures, metallic calcium is made to react with carbon and the product obtained is treated with water.
3. Potassium ferrocyanide is heated in the presence of concentrated H₂SO₄ and the gas thus obtained is passed over finely divided nickel powder at 50°C.
4. Silicon is heated with methyl chloride at high temperatures in the presence of copper.
5. SiO₂ is treated with HF.

Answer:

1. When a mixture of sand and sodium carbonate is melted by tremendous heat, sodium silicate and carbon dioxide are obtained.

SiO₂ + Na₂ CO₃ →  Na₂SiO₃ + CO₂↑

2. When CO is passed through NaOH solution at 200°C under high pressure, sodium formate is produced.

NaOH + CO→ HCOONa

When sodium formate is heated at 300°C, sodium oxalate and hydrogen gas are formed.

3. Metallic calcium reacts with carbon at high temperatures to form calcium carbide. Calcium carbide on treatment with water produces acetylene and calcium hydroxide

4. When potassium ferrocyanide is heated with concentrated sulphuric acid, potassium sulphate, ammonium sulphate, ferrous sulphate and carbon monoxide are formed. When the resulting CO gas is passed over finely divided nickel at 50°C, nickel tetracarbonyl (a coordinated complex) is formed

Ni + 4CO →(50°C) Ni(CO)₄

5. When silicon is heated with methyl chloride at high temperature in the presence of copper, a mixture of mono-, di- and trimethylchlorosilane along with a small amount of tetramethylsilane is obtained

SiO₂ reacts with HF to form silicon tetrafluoride. The initially formed SiF₄ dissolves in HF to form hydrofluorosilicic acid.

SiO₂ + 4HF→SiF₄ + 2H₂O ; SiF₄ + 2HF → H₂SiF₆

Electronic Configuration Of P-Block Elements

Question 9. Starting from boric acid how can you prepare—

1. Boric anhydride
2. Boron trichloride
3. Boron trifluoride
4. Meta and tetraboric acid
5. Boron hydride
6. Ethyl borate

Answer:

1. Boric anhydride is prepared by heating boric acid at high temperatures

2. Boric acid is first converted to boric anhydride, which on heating with Cl₂ forms boron trichloride.

B₂O₃ + 3C + 3Cl₂→2BCl₃ + 3CO

3. Boric acid, on heating with CaF₂ and cone. H2S04 forms boron trifluoride.

CaF₂ + H₂SO₄→CaSO₄ + H₂F₂

2H₃BO₃ + 3H₂F₂ → 2BF₃ + 6H₂O

4. Orthoboric acid on heating at 100°C and 160°C forms metaboric acid and tetraboric acid respectively.

Metaboric Acid:

Tetraboric Acid:

5. Boric acid is first converted to boric anhydride which on heating with Mg-powder forms magnesium boride. On reacting dilute HCl with magnesium boride, boron hydrides are formed.

Ethyl borate is formed by heating a mixture of boric acid and ethanol in the presence of concentrated H₂SO₄.

3C₂H₅OH + H₃BO₃ → (C₂H₅)₃BO₃ + 3H₂O

Electronic Configuration Of P-Block Elements

Question 10. Consider the compounds, BCI₃ and CCl₄. How will they behave with water? Justify.
Answer:

BCl₃ is an electron-deficient molecule because the core boron atom possesses only six electrons in its valence shell. Consequently, it can take a pair of electrons provided by water and undergo hydrolysis to produce boric acid (H₃BO₃) and HCl.

The central atom of the molecule, designated as atom B, possesses six valence electrons. Shell

Conversely, the carbon atom in CCl₄ possesses 8 electrons in its valence shell and lacks unoccupied d-orbitals to expand its octet. CCl₄ cannot take a pair of electrons from H2O; therefore, it does not undergo hydrolysis.

Equilibrium Class 11 Chemistry Notes

Question 1. HPO₄^2- can act both as a Bronsted base and as a Bronsted acid. Write the equation of equilibrium established by HPO₄^2- as an acid and a base in an aqueous solution. Also, write the expressions of K_a & K_b in two cases.  What are the conjugate acid and base of HS^–?
Answer:

HPO₄^2-  can denate and accept protons in aqueous solution. Thus it can serve both as an acid and base.

⇒ \(\begin{aligned}
& \mathrm{HPO}_4^{2-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{PO}_4^{3-}(a q)+\mathrm{H}_3 \mathrm{O}_2^{+}(a q) \\
& \mathrm{HPO}_4^{2-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

⇒ \(K_a=\frac{\left[\mathrm{PO}_4^{3-}\right] \times\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HPO}_4^{2-}\right]} ; K_b=\frac{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]}\)

HS^– (aq) + H₂O(l) ⇌  H₂S(aq) + OH^– (aq)

Hence, the conjugate acid of HS^–  is H₂S.

HS^–(aq) + H₂O(l) ⇌  S^2-(aq) + H₃O+(aq)

Therefore, the conjugate base of HS^– is S^2--.

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Question 2. An aqueous solution of sodium bisulfate is acidic, whereas an aqueous solution of sodium bicarbonate is basic—Explain.
Answer:

Since sodium bisulfate (NaHSO₄) is a salt of strong acid (H₂SO₄) and strong base (NaOH), it is not hydrolyzed in an aqueous solution. NaHS04 in its solution dissociates completely to form Na⁺ and HSO₄ ions and HSO₄ ions so formed get ionized to form H₃O+ and SO₂^-4 ions. As a result, the aqueous solution of NaHSO₄ becomes acidic.

⇒ \(\begin{gathered}
\mathrm{NaHSO}_4(a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{HSO}_4^{-}(a q) \\
\mathrm{HSO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)
\end{gathered}\)

NaHCO₃ in its solution dissociates completely to form Na+ and HCO₃ ions [NaHCO₃(aq)-Na+(aq) + HCO₃ (aq) ]. HCO₃ can act both as an acid and a base in aqueous solution.

⇒ \(\begin{aligned}
& \mathrm{HCO}_3^{-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{CO}_3^{2-}(a q) \\
& \mathrm{HCO}_3^{-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

Equilibrium Class 11 Chemistry Notes

Question 3. Calculate the formation constant of [Ag(NH₃)₂]⁺
Answer:

⇒  \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; K_1=3.5 \times 10^3\)

⇒ \(\begin{aligned}
{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; } \\
K_1=1.7 \times 10^3 \ldots
\end{aligned}\)

Equation (3) represents the formation reaction of [Ag(NH₃)₂]2+. Therefore, the equilibrium constant for the reaction represented by equation (3) will be the formation constant for [Ag(NH₃)₂]2+. As equation (3) is obtained by adding equations (1) and (2), the formation constant (fcy) for [Ag(NH₃)₂]2+ equals ky x k2.

Thus, k_y = k_y X k₂ = (3.5 × 10³) × (1.7 ×  10³) = 5.95 ×  10⁶

Question 4. The first and second dissociation constants of an acid H₂A are 1 × 10^-5 and 5 × 10^-10  respectively. Calculate the value of the overall dissociation constant.
Answer:

⇒ \(\begin{aligned}
\mathrm{H}_2 \mathrm{~A}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{HA}^{-}(a q) ; \\
K_1=1 \times 10^{-5} \ldots(1)
\end{aligned}\)

⇒ \(\begin{array}{r}
\mathrm{HA}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}^{2-}(a q) ; \\
K_2=5 \times 10^{-10} \ldots
\end{array}\)

The overall dissociation reaction is the sum of the. reactions (1) and (2). H₂A(aq) + 2H₂O(Z) 2H30+(aq) + A3-(aq) Thus, the overall disputation constant K = K₁ × K₂ =  1 × 10^-5 and 5 × 10^-10= 5 ×10^-15

Question 5. a -D-glucose Beta -D glucose, the equilibrium constant for this is 1.8. Calculate the percentage of a -D glucose at equilibrium.
Answer:

Suppose, the initial concentration of a -D glucose is a mol L^-1 and its degree of conversion to p -D glucose at equilibrium is x mol-L^-1

Chemical Equilibrium Class 11 Questions

Therefore, the concentration of a -D glucose and D glucose at equilibrium will be as follows—

⇒ \(\begin{array}{ccc}
\begin{array}{c}
\text { Initial concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a & 0 \\
\begin{array}{c}
\text { Equilibrium concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a-a x & a x
\end{array}\)

Equilibrium constant for this process, \(K=\frac{[\beta-\mathrm{D}-\text { glucose }]}{[\alpha-\mathrm{D}-\text { glucose }]}\)

⇒ \(1.8=\frac{a x}{a(1-x)}=\frac{x}{1-x}\) or, \(x=\frac{1.8}{2.8}=0.6428\)

Thus, the percentage of a -D-glucose at equilibrium is \(\frac{a(1-0.6428)}{a} \times 100=35.72 \%\)

Question 6. Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 x 10-4 M Na₂CO₃ solution. At what concentration of Ba2+ will a precipitate begin to form? ( Ksp for BaCO₃ = 5.1 × 10^-9 )
Answer:

Ba(NO₃)₂ reacts with Na₂CO₃ to form BaCO₃. BaCO₃ is a sparingly soluble compound that forms the following equilibrium in its saturated solution.

⇒ \(\mathrm{BaCO}_3(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{CO}^{2-}(a q)\)

For \(\mathrm{BaCO}_3, K_{s p}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]\)

In the solution \(\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M}\)

BaCO₃ will precipitate when [Ba²⁺][CO₂^–] → Ksp, i.e., when

⇒ \(\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]>5.1 \times 10^{-9}\left(\text { as } K_{s p}\left[\mathrm{BaCO}_3\right]=5.1 \times 10^{-9}\right)\) \(\text { If }\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M} \text {, then }\left[\mathrm{Ba}^{2+}\right]>\frac{5.1 \times 10^{-9}}{1 \times 10^{-4}}\)

Or, [Ba²⁺] > 5.1 × 10^-5M

Thus, the precipitation of BaCO₃ will start when [Ba2+] in the solution is grater than 5.1 × 10^-5 M.

Mpbse Class 11 Chemistry Solutions Pdf

Question 7. 2.5 m l of \(\frac{2}{5} \mathrm{M}\) weak monoacdic base Kb = 1 x 10-125at 25°C ) is titrated with \(\frac{2}{15}\) in water at 25°C. Calculate the concentration of H + at the equivalence point. (kw = 1 × 10-14)
Answer:

Ana. 2.5 mL of \(\frac{2}{15}\) M monobasic acid \(\equiv \frac{2}{5} \times 2.5 \equiv 1\) mmol of the base.

Suppose, VmL of \(\frac{2}{15} \mathrm{M}\) HC1 is required for the neutralisation. \(V \mathrm{~mL} \text { of } \frac{2}{15} \mathrm{M} \mathrm{HCl} \equiv \frac{2}{15} \times V \mathrm{mmol} \mathrm{HCl} .\)

Therefore \(\) mmol HCL will neutralise 1mmol of the base Hence \(\frac{2 \times V}{15}=1 \text { or, } V=7.5 \mathrm{~mL}\)

The total volume of the solution after neutralization = (2.5 + 7.5) mL = 10 mL.

The number of mmol of the salt formed in the neutralization =1 mmol.

So, the concentration of the salt in the final solution \((C)=\frac{1}{10}=0.1 \mathrm{M}\) As the resulting salt is formed from a weak base and a strong acid, it undergoes hydrolysis. The pH of the solution of such a salt is given by the pH
\(=7-\frac{1}{2} p K_b-\frac{1}{2} \log C\) \(\text { As } K_b=10^{-12}, p K_b=-\log _{10}\left(10^{-12}\right)=12\)

Therefore \(p H=7-\frac{1}{2} \times 12-\frac{1}{2} \log (0.1)=7-6+0.5=1.5\) and the concentration of H+(aq) ions at the equivalence point is [H⁺]

= 10-PH M = 10-1-5 M = 0.316 M.

Chemical Thermodynamics Class 11 Notes

Question 1. In which of the following two processes, the change in entropy of the system will be negative?

• Fusion of Ice.
• Condensation of water vapor

Answer: The change in entropy of the system is negative during the condensation of water vapor. In this process \(\Delta S_{\text {sys }}=S_{\text {water }}-S_{\text {water vapour }}<0\)

As the molecules in water vapor have more freedom of motion than they have in the water, the molecular randomness is higher in water vapor than in water. Thus, \(S_{\text {water }}<S_{\text {water vapour }}\) Consequently, the value of AS becomes negative.

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Question 2. The change in internal energy in different steps of the process A→ B → C→ D are given: A→Bx.kJ-mol-1; B→ C, -y kj-mol-1; C-D,z kj-mol-1. What will the value of AH be for the change A → D?
Answer:

⇒ \(A →{\Delta U_1} B→{\Delta U_2} C →{\Delta U_3} D\)

Δ Ux = UR- UA = x kj.mol-1 ;

ΔU2 = Hc.— UB = -ykjmol-1;

ΔU3 = UD-uc = zkj.mol-1

∴ For the change A D

A U = UD-UA =(UD-UC) + (UC-UB) + (UB-UA)

=(z-y + x) kj-mol-1.

Chemical Thermodynamics Class 11 Notes

Question 3. Change in enthalpy in different steps of the process A→B→C→A are given: A→B, x kj-mol-1; C→A, y kj-mol-1. Find the value of AH for step B→C.
Answer:

The initial and final states (A) are the same in the given process. So it is a cyclic process. Since H is a state function, AH = 0 for the process. Thus in this process \(\begin{aligned}
& \Delta H=\left(H_B-H_A\right)+\left(H_C-H_B\right) \Psi\left(H_A-H_C\right)=0 \\
& \left.x+\left(H_C-H_B\right)+y=0 \text { or }\left(H_C\right)_B H_B\right)=-(x+y) \mathrm{kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

So change in enthalpy for BC =- (x + y) kj-mol-1.

Question 4. Why are the standard reaction enthalpies of the following two reactions different?
Answer:

⇒ \(\begin{aligned}
& \mathrm{C}(\text { graphite, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-393.5 \mathrm{~kJ} \\
& \mathrm{C}(\text { diamond }, s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-395.4 \mathrm{~kJ}
\end{aligned}\)
Anszwer: Graphite and diamond are two allotropes ofsolid carbon. Different allotropic forms have different enthalpies. As the given U reactions involve different allotropes, the standard reaction enthalpies ofthese reactions are different.

Question 5. At 25°C, the standard reaction enthalpy for the reaction is -221.0 kj. Does this enthalpy change indicate the standard enthalpy of the formation of CO(g)? If not, then what would be the value of the enthalpy of j formation of CO(g) at 25°C?
Answer:

In the given reaction 2 mol CO(g) is formed from its stable constituent elements. Sd,’ definition, AH0 of this reaction does not represent standard enthalpy of formation of CO(g). 2C(graphite,s) + O₂(g) 2CO(g) ; AH0 = -221.0 kj

Dividing both sides by 2, we get

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g) ; \Delta H^0=-110.5 \mathrm{~kJ} \cdots[1]\)

In reaction [1], 1 mol of CO(g) is formed from its stable constituent elements. Thus, in this reaction AH0 = standard enthalpy of formation of CO(g).

So, at 25°C the standard enthalpy of formation of CO(g) =-110.5 kj-mol-1.

Chemical Thermodynamics Class 11 Notes

Question 6. Determine the standard reaction enthalpy for the reaction: \(\mathrm{A}_2 \mathrm{~B}_3(s)+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(s)+3 \mathrm{CB}_2(g)\) Given: \(2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow \mathrm{A}_2 \mathrm{~B}_3(s); \Delta H^0=-x \mathrm{~kJ}\)
Answer:  Reversing equation 1 we get \(\mathrm{A}_2 \mathrm{~B}_3(s) \rightarrow 2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(\mathrm{~g}) ; \quad \Delta H^0=+x \mathrm{~kJ}\)

Multiplying the equation by 3, we get

⇒  \(3 \mathrm{CB}(g)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow 3 \mathrm{CB}_2(g) ; \Delta H^0=-3 y \mathrm{~kJ}\)

Adding equations [3] and [4], we get

⇒ \(\mathrm{A}_2 \mathrm{~B}_3(\mathrm{~s})+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(\mathrm{~s})+3 \mathrm{CB}_2(g) ; \Delta H^0=(x-3 y) \mathrm{kJ}\)

Therefore, the standard reaction enthalpy for the given reaction =(x- 3y)kJ.

Question 7. In which ofthe following reactions does AH0 at 25 °C indicate the standard enthalpy of formation \(\left(\Delta \boldsymbol{H}_f^0\right)\) of the compound formed in each of the reactions?
Answer:

Here, 2 mol of NH₃(g) is produced from the stable constituent elements, H2(g) and N2(g). Hence according to the definition, at 25°C the standard reaction enthalpy of this reaction is not equal to the standard enthalpy of formation of NH3(g).

The standard state of oxygen at 25°C is O₂(g). So, the given equation does not represent the formation reaction of NO(g). Consequently, at 25°C the standard reaction enthalpy of this reaction is not equal to die standard enthalpy offormation of NO(g).

Here, 1 mol of solid NaCl is formed from its stable constituent elements. Hence at 25°C, the die standard reaction enthalpy of this reaction is equal to the die standard enthalpy of formation of NaCl(s).

Question 8. Discuss the change in die degree of randomness for the following cases— Combustion of kerosene, Sublimation of dry ice, Extraction of salt from seawater, Condensation of water vapour, Crystallisation of a solid from its aqueous solution,
Answer: Combustion of kerosene: Kerosene is a mixture of liquid hydrocarbons. The combustion of kerosene produces C02 and water vapour. As in the reaction, a liquid converts into a gaseous mixture, the molecular randomness ofthe system increases.

Sublimation of dry ice: In dry ice, (solid carbon dioxide), the molecules of C02 exist in an orderly state. When this solid sublimes, the gaseous molecules formed move randomly, i.e., the degree of randomness ofthe molecules increases.

Extraction (Crystallisation) of salt from seawater: In seawater, the attractive forces between Na+ and Cl- ions are weak as the distance between these ions is large. Hence, these ions are virtually free to move randomly. NaCl extracted from the seawater is in a solid state with a crystal structure in which Na+ and Cl- are arranged in a definite order. Hence, the extraction of salt from seawater is associated with a decrease in the degree of randomness.

Condensation of water vapor: In water vapor, the intermolecular forces of attraction between H2O molecules are weak, so the molecules remain in a state of randomness. However, water obtained by condensation of water vapor has less freedom of motion and hence less degree of randomness because of the stronger intermolecular forces of attraction compared to water vapour. Hence, in the case of condensation of water vapour, the degree of randomness decreases.

Crystallization of a solid from its Aqueous solutions:

In an aqueous solution, the solute molecules exist in a state of random motion. When the solute is crystallised from its solution, the solute molecules in the die crystal remain at fixed positions and become almost motionless. So, the crystallization of a solid from its solution causes a decrease in randomness.

⇒ \(\mathrm{Cr}^{3+}+6 \mathrm{H}_2 \mathrm{O}(a q) \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)\): In this process, one Cr3+ ion combines with six water molecules to form a single complex ion, [Cr(H20)6]3+. Consequently, the number of particles in the system reduces (from 7 to 1 for each combination), thereby decreasing the degree of randomness.

⇒ \(\mathrm{NH}_4 \mathrm{NO}_2(s) \xrightarrow{\Delta} \mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}):\) On decomposition, 1 formula unit of solid ammonium nitrite produces 1 molecule of N2(g) and 2 molecules of water vapour i.e., H20(g). In this process, a solid (in which the molecules are orderly arranged) converts into gaseous substances. Furthermore, the number of molecules also increases. Naturally, this process increases the randomness ofthe system.

Thermodynamics Important Questions Class 11

Question 9. Mention if the entropy of the system increases or decreases In each of the following cases: Bolling of water, Sublimation of solid iodine,

⇒ \(\begin{aligned}
& 2 \mathrm{O}_3(g) \rightarrow 3 \mathrm{O}_2(g) \\
& \mathrm{NH}_4 \mathrm{Cl}(s) \rightarrow \mathrm{NH}_3(g)+\mathrm{HCl}(g) \\
& \mathrm{Hg}(l) \rightarrow \mathrm{Hg}(g) \quad \text { (6) } \mathrm{I}_2(g) \rightarrow \mathrm{I}_2(s) \\
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \text { (8) } 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \\
& \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2(g) \\
& \mathrm{PCl}_5(g) \rightarrow \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \\
& \text { Haemoglobin }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \text { Oxyhaemoglobin } \\
& \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightarrow \mathrm{Ni}(\mathrm{CO})_4(g) \\
& 4 \mathrm{Fe}(s)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(s) \\
& \mathrm{C}(\text { diamond }) \rightarrow \mathrm{C} \text { (graphite) } \\
&
\end{aligned}\)

Answer: The change in entropy in a phase transition or a reaction depends on the following factors.

For a given amount of substance, the entropy of the substance in different physical states varies in the order: of S_gas > S_liquid > S_soild where S is the molar entropy. Because of this, the entropy of the system increases in the phase changes such as liquid → vapour, solid → vapour, while it decreases in the phase changes such as liquid → solid, vapour → liquid and vapour → solid.

In a reaction, indie reactants are solids or liquids or in a dissolved state in solution and they convert into gaseous products, then the entropy of the die system increases, (b) If the number of gaseous particles (atoms or molecules) in a reaction increases or decrease, then the entropy of the system increaser or decreases.

Increases (liquid — vapour transition), Increases (solid-gas transition), Increases (the number of gaseous particles increases) Increases (gaseous substances are formed from a solid reactant) Increases (liquid — gas transition) Decreases (gas-solid transition) Increases (the number of gaseous particles increases).

Decreases (the number of gaseous molecules decreases) Increases (gaseous substance is formed through the reaction of the reactants, one of which is solid and the other is in a dissolved state in solution) Increases (the number of gaseous molecules increases) Decreases (the number of gaseous molecules decreases)

Thermodynamics Important Questions Class 11

Decreases (the number of gaseous molecules decreases)

Decreases (the number of gaseous molecules decreases) (g) Increases (the crystal structure of diamond is more compact than that of graphite. As a result, the molecular randomness in graphite is more than that in diamond.

Question 10. At 25°C and 1 atm pressure, for the reaction 3O₂(g) → 20₂(g); H = 286kJ and AS = -137.2 J.K-1. Is this reaction spontaneous? Does the spontaneity of this reaction depend on temperature? Is the reverse reaction spontaneous? If so, then why? Does the spontaneity of the reverse reaction depend on temperature?
Answer: No. (2) For the given reaction ΔH > 0 and ΔS < 0. So, according to the equation ΔG = ΔH-TΔS, AG will be positive at any temperature. Hence, the spontaneity of this reaction is independent of temperature.

The reverse reaction is spontaneous.

In the reverse reaction [20₃(g)→ 3O₂(g)] ; AH = —286 kj and AS = +137.2 J K- . So, according to the equation, AG = AH- TAS, AG is negative.

The spontaneity of the reverse reaction is also independent of temperature as AG is negative.

H<0 and S> 0 at any temperature.

Hydrogen Class 11 Questions And Answers

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Question 1. Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer: In nature, there are three isotopes of hydrogen. These are protium \(\left({ }_1^1 \mathrm{H}\right) \text {, deuterium }\left[{ }_1^2 \mathrm{H} \text { or } \mathrm{D}\right] \text { and tritium }\left[{ }_1^3 \mathrm{H} \text { or } \mathrm{T}\right] \text {. }\)

The mass ratio of \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{H} \text { and }{ }_1^3 \mathrm{H} \text { is } 1: 2: 3 \text {. }\)

Question 2. Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer: A hydrogen atom has only one electron and thus, it has one electron less than the stable configuration of the nearest noble gas helium. Thus, to achieve a stable configuration it shares its single electron with the electron of another H -atom to form a stable diatomic molecule (H₂ ).

Read and Learn More Class 11 Chemistry

Question 3. Complete the following reactions:

1. \(\mathrm{H}_2(g)+\mathbf{M}_m \mathrm{O}_{\mathrm{o}}(s) \stackrel{\Delta}{\longrightarrow}\)
2. \(\mathrm{CO}(g)+\mathrm{H}_2(g) \frac{\Delta}{\text { catalyst }}\)
3. \(\mathrm{C}_3 \mathrm{H}_8(g)+3 \mathrm{H}_2 \mathrm{O}(g) \underset{\text { catalyst }}{\longrightarrow}\)
4. \(\mathrm{Zn}(s)+2 \mathrm{NaOH}(a q) \stackrel{\Delta}{\longrightarrow}\)

Answer:

1. \(o\mathrm{H}_2(g)+\mathrm{M}_m\mathrm{O}_o(s)\stackrel{\Delta}{\longrightarrow}m\mathrm{M}(s)+o\mathrm{H}_2\mathrm{O}(l)\)
2. \(\mathrm{CO}(g)+2 \mathrm{H}_2(g) \underset{\text { catalyst }}{\longrightarrow} \mathrm{CH}_3 \mathrm{OH}(l)\)
3. \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \stackrel{\mathrm{Ni}, 1270 \mathrm{~K}}{\longrightarrow} 3 \mathrm{CO}(g)+7 \mathrm{H}_2(g)\)
4. \(\mathrm{Zn}(s)+2 \mathrm{NaOH}(a q) \stackrel{\Delta}{\longrightarrow} \underset{\text { Sodium zincate }}{\mathrm{Na}_2 \mathrm{ZnO}_2(a q)+\mathrm{H}_2(g)}\)

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 4. Discuss the consequences of high enthalpy of the H —H bond in terms of the chemical reactivity of dihydrogen.
Answer: H —H bond length is very small (74 pm) because the H-atom has a very small atomic size. Consequently, H—H bond dissociation enthalpy is very high (435.9 kJ-mol^-1) which makes hydrogen completely inert at ordinary temperature. However, at higher temperatures, the H—H bond undergoes dissociation in the presence of a catalyst to form hydrides with metals and non-metals.

Question 5. What characteristics do you expect from an electron-deficient hydride concerning its structure and chemical reactions?
Answer:

1. There is an insufficient number of electrons in the valence shell of the central atom. So, in this type of hydride, the valence shell of the central atom does not have a complete octet configuration. For example, in BH₃, the valence shell of B has six electrons and the hydrides are trigonal planar shape.
2. Due to electron deficiency, this type of hydrides act as Lewis acids, i.e., they accept electron pairs. For example,\(\mathrm{H}_3 \mathrm{~B}+\mathrm{NH}_3 \rightarrow \mathrm{H}_3 \mathrm{~B} \leftarrow: \mathrm{NH}_3\)
3. To compensate for the electron deficiency, the hydrides form dimers, trimers, and polymers and attain stability. for example B₂H₆,B₄H₁₀,(AIH₃)n etc.
4. Electron-deficient hydrides are extremely reactive. They easily react with both metals and non-metals.

Question 6. Do you expect the carbon hydrides of the (C_nH_2n+2) to act as ‘Lewis’ acid or base? Justify your answer.
Answer: The hydrides of carbon of the type C_nH_2n+2 electron-precise hydrides, i.e., they have an exact number of electrons in the valence shell of the central atom so as to write their conventional Lewis structures. Therefore, they do not have a tendency to either gain or lose electrons and consequently, they do not act as Lewis acids or bases.

Question 7. What do you understand by the term “nonstoichiometric hydrides”? Do you expect this type of hydride to be formed by alkali metals? Justify your answer.
Answer:

Hydrides of d and f-block elements that are low in hydrogen, characterized by a fractional metal-to-hydrogen ratio, are termed non-stoichiometric hydrides.

• The alkali metals, due to their strong reducing properties, donate their lone pair of electrons to the hydrogen atom, resulting in the formation of hydride ions.
• As a hydride ion H- is generated through the complete transfer of an electron, the ratio of metal to hydrogen remains constant in these hydrides, resulting in compositions that reflect a straightforward whole-number ratio. Consequently, the alkali metals only produce stoichiometric hydrides.

Chemical Properties Of Hydrogen

Question 8. How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer:

• Certain transition metals, such as palladium (Pd) and platinum (Pt), adsorb significant quantities of hydrogen atoms on their surfaces, resulting in the formation of hydrides.
• The incorporation of hydrogen atoms causes the metal lattice to expand, resulting in decreased stability.
• Consequently, upon heating, these metal hydrides breakdown to liberate dihydrogen and revert to a finely split metallic state. The dihydrogen produced in this method can serve as a fuel.
• Consequently, transition metals or their alloys can be utilized for the storage and delivery of hydrogen as a fuel.

Question 9. Among NH₃, and H₂O midIIF, which would you expect to have the highest magnitude of hydrogen bonding & why?
Answer:

The strength of a hydrogen bond depends upon the atomic size mid the electronegativity of the atom to which the H-atom Is covalently linked. Smaller size and higher electronegativity of the other atom favor the formation of stronger H -bonding and that Is due to increase In magnitude of bond polarity. Among N, F, and O atoms, V has the lowest atomic size and the highest electronegativity. Hence the II — F bond is maximum polar and as a result, it will have the highest magnitude of H-bonding.

Question 10. Saline hydrides are known to react with water violently producing fire. Can CO₂, a well-known fire extinguisher, be used in this case? Explain.
Answer: The saline hydrides (For example, NaH, CaH₂, etc.), react with water violently to yield the corresponding metal hydroxides with the evolution of H₂ gas. The liberated H₂ gas undergoes spontaneous combustion causing fire and this is because of the highly exothermic nature of the combustion reaction.

⇒ \(\mathrm{NaH}(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{H}_2(g) ;\)

⇒ \(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

In this case, CO₂ cannot used as a fire extinguisher because its gels are reduced by the hot metal hydride to form formate ions.

⇒ \(\stackrel{-1}{\mathrm{NaH}}+\stackrel{+4}{\mathrm{CO}_2} \rightarrow \stackrel{+1+2}{\mathrm{H} C O O N a}\)

Question 11. Arrange the following:

1. CaH₂, BeH₂, and TiH₂ in order of increasing electrical conductance.
2. LiH, NaH, and CsH in order of increasing ionic character.
3. If —H, D—D, and F—F in order of increasing bond dissociation enthalpy.
4. NaH, MgH₂, and H₂O in order of Increasing and reducing properties.

Answer:

1. Being a covalent hydride BeH2 does not conduct electricity at all. Being an Ionic hydride CaH₂ conducts electricity in the fused state while TiH₂, being a metallic hydride, conducts electricity at room temperature, ‘t hus, the order of increasing electrical conductance is: BeH₂ < CaH₂ < TiH₂.
2. The electronegativity of the alkali metals decreases down the group from Li to Cs. Therefore, the ionic character of their hydrides also increases in the same order, l.e., LiH < NaH < CsH.
3. The bond dissociation enthalpy of the: F—F: bond is the lowest (242.6 kJ • mol^-1) and this is due to the high concentration of electron density around each F atom in the form of three unshared pairs which significant repulsive interactions. Again, because of the marginally smaller size of D as compared to H, the bond dissociation enthalpy of the D—D bond (443.35 kJ-mol^-1) is slightly higher than that of the H —H bond (435.88 kJ-mol-1). Hence, the bond dissociation enthalpy increases in the order: of F —F < H —H < D —D.
4. NaH, being an ionic hydride, is a more powerful reducing agent than the covalent hydrides MgH₂ and H₂O. MgH₂ is a stronger reducing agent than H20 because the bond dissociation enthalpy of the Mg—H bond is much lower than that of the O —H bond. Therefore, the reducing property increases in the order: of H₂O < MgH₂ < NaH.

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 12. Compare the structures of H20 and H202.
Answer: The oxygen atom in water is sp³ -sp³-hybridized. The two O —H bonds are sp3-s sigma bonds. The H —O —H bond angle is 104.5°. This value is a little less than the tetrahedral angle (109°28/) because of stronger lone pair-lone pair and lone pair-bond pair repulsions than bond pair-bond pair repulsion. Thus, water is a bent molecule. Each oxygen atom in H₂O₂ is also sp3 hybridised. The O —0 bond is a sp³-sp³ sigma bond and the two O —H bonds are sp3-s sigma bonds. The two O —H bonds are, however, present in different planes. In the gas phase, the dihedral angle between the two planes (i.e., the planes containing H —O —O system) is 111.5°. So, the molecule has an open-booklike structure.

Question 13. What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Answer: Self-ionization of water is called auto-protolysis. Self-ionization of water can be expressed by the given equation

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \\ & \text { Acid-1 Base-2 } \quad \text { Acid-2 } \quad \text { Base-1 } \\ & \end{aligned}\)

Water exhibits amphoteric properties because of auto-protolysis. Thus water reacts with both acids and bases. It usually acts as a base in the presence of an acid stronger than it and acts as an acid in the presence of a base stronger than it. For example,

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{NH}_3(a q) \longrightarrow \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q) \\ & \begin{array}{llll} \text { Acid-1 } & \text { Base-2 } & \text { Acid-2 } & \text { Base-1 } \end{array} \\ & \end{aligned}\)

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{HCl}(a q) \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q) \\ & \text { Base-1 } \quad \text { Acid-2 } \quad \text { Acid-1 } \quad \text { Base-2 } \\ & \end{aligned}\)

Question 14. Consider the reaction of water with F2 and suggest, In terms of oxidation and reduction, which species are oxidized/reduced.
Answer:

⇒ \( 2 \mathrm{~F}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{~F}^{-}(a q) Oxidant Reductant\)

⇒ \( 3 \mathrm{~F}_2(g)+3 \mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{O}_3(g)+6 \mathrm{H}^{+}(a q)+6 \mathrm{~F}^{-}(a q) Oxidant Reductant\)

In these reactions, water acts as a reductant and itself gets oxidized to oxygen or ozone. In this case, highly electronegative fluorine acts as an oxidant and gets reduced to F-.

Question 15. Complete the following chemical reactions.

1. \(\mathrm{PbS}(s)+\mathrm{H}_2 \mathrm{O}_2(a q) \rightarrow\)
2. \(\mathrm{MnO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}_2(a q) \rightarrow\)
3. \(\mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow\)
4. \(\mathrm{AlCl}_3(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow\)
5. \(\mathrm{Ca}_3 \mathrm{~N}_2(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow\)

Classify the above into [a] hydrolysis, [b] redox, and [c] hydration reactions.

Answer:

⇒ \(\mathrm{PbS}(s)+4 \mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow \mathrm{PbSO}_4(s)+4 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\begin{array}{r}
2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{H}_2 \mathrm{O}_2(l)+6 \mathrm{H}^{+}(a q) \longrightarrow \\
2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_2 \mathrm{O}(l)+5 \mathrm{O}_2(g)
\end{array}\)

⇒ \(\mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(g) \longrightarrow \mathrm{Ca}(\mathrm{OH})_2(a q)\)

⇒ \(\begin{array}{r}
\mathrm{AlCl}_3(g)+6 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \\
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)}
\end{array}\)

⇒ \(\begin{array}{r}
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow} \\
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{OH})\right]^{2+}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)}
\end{array}\)

⇒ \(\mathrm{Ca}_3 \mathrm{~N}_2(s)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2(a q)+2 \mathrm{NH}_3(a q)\)

Reactions 1 and 2 are redox reactions. Reactions 3 and 4 are hydrolysis reactions. Reaction 5 is the hydration reaction.

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 16. Is demineralized or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer: Although very pure, demineralized or distilled water is not useful for drinking purposes and this is because it does not contain even useful minerals. To make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralized or distilled water.

Question 17. Describe the usefulness of water in biosphere and biological systems.
Answer:

Water is essential for the survival of various life forms on Earth. It comprises around 65-70% of the body mass of flora and fauna.

• Due to its elevated specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant relative to other liquids, water is essential to the biosphere.
• The significant heat of vaporization of water enables it to regulate body temperature. Water also contributes indirectly to climate regulation.
• Water facilitates the transfer of minerals, nutrients, and enzymes throughout the body and influences metabolic processes.
• Water is an essential element in the process of photosynthesis. Consequently, rivers play a crucial function in the ecosystem.

Question 18. Knowing the properties of H20 and D20, do you think that D20 can be used for drinking purposes?
Answer: Drinking heavy water (D₂O) is harmful to the human beings. Heavy water being highly hygroscopic, absorbs water from different parts of the body. As a result, the body cells may get destroyed. Apart from this, heavy water slows down different biochemical reactions such as mitosis, cell division, etc.

Question 19. What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer: Hydrolysis refers to the reaction of salt with water to form acidic or basic solutions.

For example:

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { (basic solution) }}{2\left[\mathrm{Na}^{+}+\mathrm{OH}^{-}\right]+\mathrm{H}_2 \mathrm{CO}_3 ;}\)

⇒ \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { (acidic solution) }}{\left[\mathrm{H}^{+}+\mathrm{Cl}^{-}\right]+\mathrm{NH}_4 \mathrm{OH}}\)

Hydration, on the other hand, refers to the addition of H20 to ions or molecules to form hydrated ions or hydrated salts. For example:

⇒ \(\underset{\text { Salt }}{\mathrm{NaCl}(s)+\mathrm{H}_2 \mathrm{O}} \rightarrow \underbrace{\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)}_{\text {Hydrated ions }}\)

⇒ \(\begin{array}{cc} \mathrm{CuSO}_4(s)+5 \mathrm{H}_2 \mathrm{O}(l) & \rightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(s) \\ \text { Anhydrous salt } & \text { Hydrated salt } \\ \text { (colourless) } & \text { (blue) } \end{array}\)

Question 20. How can saline hydrides remove traces of water from organic compounds?
Answer: Saline hydrides (i.e., NaH, CaH₂, etc.) react with water forming their corresponding metal hydroxides with the liberation of dihydrogen. Thus, traces of water present in organic solvent can be easily removed by distilling them over saline hydrides when dihydrogen escapes into the atmosphere, metal hydroxides are left in the distillation flask, and the dry organic solvent is distilled over.

Class 11 Chemistry Chemical Properties Of Hydrogen

Question 21. What do you expect the nature of hydrides to be if formed by elements of atomic numbers 15, 19, 23, and 44 with dihydrogen? Compare their behavior towards water.
Answer:
The element with Z = 15 is non-metal phosphorus and hence it forms the covalent hydride PH₃.

The element with Z = 19 is the alkali metal potassium and hence it forms the saline or ionic hydride K+HT.

The element with Z = 23 is the transition metal vanadium of group-3 and hence it forms the interstitial hydride (VHj 6).

The element with Z = 44 is the transition metal ruthenium (Ru) belonging to group 8. It does not form anhydride (hydride gap). Only the ionic hydride, KH reacts with water evolving dihydrogen.

⇒ \(\mathrm{KH}(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{H}_2(g)\)

Question 22. Do you expect different products in solution when aluminum (III) chloride and potassium chloride are treated separately with

Normal water

Acidified water and

Alkaline water? Write equations wherever necessary.

Answer: In normal water: KC1, being a salt of a strong acid and strong base, does not undergo hydrolysis in normal water. It simply dissociates to form K+(aq) and Cl~(aq) ions.

⇒ \(\mathrm{KCl}(s) \stackrel{\text { water }}{\longrightarrow} \mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

On the other hand, A1C13, being a salt of a weak base Al(OH)3 and a strong acid HC1, undergoes hydrolysis giving acidic solution.

⇒ \(\mathrm{AlCl}_3(s)+3 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Al}(\mathrm{OH})_3(s)+3 \mathrm{H}^{+}(a q)+3 \mathrm{Cl}^{-}(a q)\)

In acidified water: The H+ ions react with Al(OH)3 to form Al3+(aq) ions and H20. Therefore, in acidic water, A1C13 exists as A13+(aq) and Cl~(aq) ions.

⇒ \(\mathrm{AlCl}_3(s) \stackrel{\text { acidified } \mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{Al}^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)\)

⇒ \(\mathrm{KCl}(s) \stackrel{\text { acidified water }}{\longrightarrow} \mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

In alkaline water: A1(0H)3 reacts with OH ions to form soluble tetrahydroxoaluminate complex or meta-aluminateion (A10ÿ2).

KC1 does not react and dissociate to give K+(aq) and Cl~(aq) ions.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Question 1. What is ‘hydrogenite’? Mention its use.

Answer: A mixture of silicon, caustic soda, and slaked lime is called hydrogenite. Dihydrogen can be prepared by heating hydrogenate with water.

⇒ \(\mathrm{Si}+2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Na}_2 \mathrm{SiO}_3+\mathrm{H}_2 \uparrow\)

∴ \(\mathrm{Si}+\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CaSiO}_3+2 \mathrm{H}_2 \uparrow\)

Question 2. What is denoted by [Hg04]+ ion? Explain

Answer: High charge density and high hydration energy of the H⁺ ion (proton) cause the H⁺ ion to remain solvated to form a hydroxonium ion or hydronium ion (H30+). Hydronium ion again forms hydrogen bonds with three water molecules and remains solvated. Therefore, in water, a proton forms [H₃O(H₂O)₃]⁺ or [H₉O₄]⁺ ion.

Question 3. Pure para-hydrogen is available but not pure ortho hydrogen. Explain.

Answer: At ordinary temperature, ordinary hydrogen is a mixture of 75% of ortho and 25% of para-isomer. With a decrease in temperature, the amount of ortho-hydrogen decreases while that of para-hydrogen increases. At 20K, pure para-hydrogen is obtained. As para-hydrogen is more stable, it is found in the pure form. However, if the temperature is increased, the amount of ortho-isomer of dihydrogen increases but at 400K or above, the ratio of ortho-and para-isomer is fixed (3: 1). Therefore, pure para-hydrogen is available but not pure ortho-hydrogen.

Question 4. How many hydrogen-bonded water molecule(s) are associated with CuS0₄-5H₂0?

Answer: Only one molecule of water, which remains outside the brackets (coordination sphere) is linked by a hydrogen bond to SO as shown below. The remaining four water molecules are associated to Cu²⁺ ion by coordination bonds.

Question 5. Write down the reaction between H₂OZ and hydrazine (NH₂NH₂) in the presence of Cu(II) catalyst. Mention the use of this reaction.

Answer: A highly concentrated solution (about 40%) of H₂0₂ (called high test peroxide) oxidizes hydrazine (N₂H₄) in the presence of Cu(II) into nitrogen gas, itself being oxidized to water (steam).

⇒ \(\stackrel{-2}{\mathrm{~N}_2} \mathrm{H}_4(l)+2 \mathrm{H}_2{ }_{-1}^{-1}(l) \underset{\text { catalyst }}{\stackrel{\mathrm{Cu}(\mathrm{II})}{\longrightarrow}} \stackrel{0}{\mathrm{~N}}(\mathrm{~g})+4 \mathrm{H}_2 \stackrel{-2}{\mathrm{O}}(g)+\text { heat }\)

The reaction is highly exothermic and is accompanied by a large increase in volume which in turn generates high pressure. Due to this, the reaction is employed for propelling rockets.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Hydrogen Multiple Choice Questions And Answers

Question 1. The normality of’30 volume’ of H₂O₂ is—

1. 2.678 (N)
2. 5.336 (N)
3. 8.034 (N)
4. 6.685 (N)

Answer: 2. 5.336 (N)

Volume strength = 5.6 x normality

or, 30= 5.6 x normality

⇒ \(\text { or, normality }=\frac{30}{5.6} \mathrm{~N}=5.357 \mathrm{~N}\)

The normality of 30 volumes of H₂0₂ is 5.357N.

Question 2. A commercial sample of H₂0₂ is labeled as 10V. Its % strength is nearly

1. 3
2. 6
3. 9
4. 12

Answer: 1. 3

10V (10 volume) H₂0₂ means that lmL of H₂0₂ solution will produce lOmL of 02 at STP. Now, the decomposition of H₂O₂ is as below:

⇒ \(\begin{array}{lr} 2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow & 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ (2 \times 34) \mathrm{g} & 22400 \mathrm{~mL} \\ =68 \mathrm{~g} & \text { (at STP) } \end{array}\)

At STP, lOmL of 0₂ is obtained from lmL of H₂0₂

⇒ \(22400 \mathrm{~mL} \text { of } \mathrm{O}_2 \text { is obtained from } \frac{22400}{10} \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2\)

= 2240 mL of H₂0₂

2240mL of H₂0₂ solution contains 68g of H₂0₂

∴ \(100 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { solution contains } \frac{68 \times 100}{2240}=3.03 \mathrm{~g}\) of H202 = 3g of H202

The percent strength of 10V H₂O₂ is 3

Class 11 Chemistry Chemical Properties Of Hydrogen Question 3. In aqueous alkaline solution, two-electron reduction of HO₂ gives

1. HO
2. H₂O
3. 0₂
4. O₂–

Answer: 1. Ho

Question 4. Which statement is not correct for ortho- and para-hydrogen

1. They have different boiling points
2. Ortho-form is more stable than para-form
3. They differ in their nuclear spin
4. The ratio of ortho to para-hydrogen changes with a change in temperature

Answer: 2. Ortho-form is more stable than para-form

At normal or high temperatures, ortho-hydrogen is more stable than para-hydrogen but at very low temperatures para-hydrogen is more stable than orthohydrogen.

Question 5. At room temperature, the reaction between water and fluorine produces—

1. HF and H₂0₂
2. HF,0₂ and F₂0₂
3. F©, 0₂ and H®
4. HOF and HF

Answer: 3. F©, 0₂ and H®

Question 6. Which one of the following statements about water is false

1. Water can act both as an acid and as a base
2. There is extensive intramolecular hydrogen bonding in the condensed phase
3. Ice formed by heavy water sinks in normal water
4. Water is oxidized to oxygen during photosynthesis

Answer: 2. There is extensive intramolecular hydrogen bonding in the condensed phase

Water molecules are associated with the formation of intermolecular hydrogen bonding.

Question 7. Hydrogen peroxide oxidises [Fe(CN)₆]₄– to [Fe(CN)₆]₃– in acidic medium but reduces [Fe(CN)₆]₃~ to [Fe(CN)₆]₄– in alkaline medium. The other products formed are, respectively

1. H₂0 and (H₂0 + 0₂)
2. H₂0 and (H₂0 + OH”)
3. (H₂0 + 0₂) and H₂0
4. (H₂0 + 0₂) and (H₂0 + OH-)

Answer: 1. H₂0 and (H₂0 + 0₂)

Acidic medium:

Question 8. Which of the following statements about hydrogen is incorrect

1. Hydrogen has three isotopes of which tritium is the most common
2. Hydrogen never acts as a cation in ionic salts
3. Hydronium ion, H₃0 exists freely in solution
4. Dihydrogen does not act as a reducing agent

Answer: 4. Dihydrogen does not act as a reducing agent

Among three isotopes of hydrogenprotlum (H1), deuterium H₂), and tritium (H₃), normal hydrogen is protium. Among these three, tritium is radioactive and unstable for which it is negligible in amount. In an aqueous solution, metal is reduced by dihydrogen. Again, metallic oxide is also reduced by dihydrogen to free metal.

Question 9. In ice, the oxygen atom is surrounded

1. Tetrahedrallyby 4 hydrogen atoms
2. Octahedrallyby 2 oxygen and 4hydrogen atoms
3. Tetrahedrallyby 2 hydrogen 2 oxygen atoms
4. Octahedrallyby 6 hydrogen atoms.

Answer: 1. X-ray studies have shown that in ice, four hydrogen atoms tetrahedrally surround each oxygen atom.

Class 11 Chemistry Chemical Properties Of Hydrogen MCQs Question 10. Predict the product of the reaction of I₂ with H₂0₂ in a basic medium.

1. I-
2. I_2O3
3. I0₃
4. I₃

Answer: 1. I-

Question 11. Strength of H₂0₂ is 15.18 g.L-1 , then it is equal to—

1. 1 volume
2. 10 volume
3. 5 volume
4. 7 volume

Answer: 3. 5 Volume

⇒ \(\text { Volume strength }=\frac{5.6 \times \text { strength in } \mathrm{g} \cdot \mathrm{L}^{-1}}{\text { equivalent mass of } \mathrm{H}_2 \mathrm{O}_2}\)

∴ \(=\frac{5.6 \times 15.18}{17}=5 \text { volumes }\)

Question 12. Which of the following reactions increases the production of dihydrogen from synthesis gas—

1. \(\mathrm{CH}_4(g)+\mathrm{H}_2 \mathrm{O}(g)\stackrel{\mathrm{1270K}}{\mathrm{Ni}} \mathrm{CO}(g)+\mathrm{H}_2(g)\)
2. \(\mathrm{C}(g)+\mathrm{H}_2 \mathrm{O}(g) \stackrel{1270 \mathrm{~K}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{H}_2(g)\)
3. \(\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \underset{\text { Catalyst }}{\stackrel{673\mathrm{~K}}{\longrightarrow}}\mathrm{CO}_{2(g)}+\mathrm{H}_{2(g)}\)
4. \(\mathrm{C}_2 \mathrm{H}_6(g)+2 \mathrm{H}_2 \mathrm{O}(g) \underset{\mathrm{Ni}}{\stackrel{1270 \mathrm{~K}}{\longrightarrow}} 2 \mathrm{CO}+5 \mathrm{H}_2\)

Answer: 3. \(\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \underset{\text { Catalyst }}{\stackrel{673\mathrm{~K}}{\longrightarrow}}\mathrm{CO}_{2(g)}+\mathrm{H}_{2(g)}\)

The production of dihydrogen can be increased by reacting carbon monoxide of syngas with steam in the presence ofiron chromate as catalyst.

⇒ \(\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2\mathrm{O}_{(\mathrm{g})} \underset{\text { Catalyst}}{\stackrel{673\mathrm{~K}}{\longrightarrow}}\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}\)

Question 13. Which of the following reactions produces hydrogen

1. Mg + H₂0
2. H₂S₂O₈ + H₂O
3. Ba0₂ + HCl
4. Na₂0₂ + 2HC1

Answer: 1. Mg + H₂0

Alkali and alkaline earth metals react with water to produce hydrogen gas and metal hydroxides. This occurs due to high electropositive character ofthe metals.

⇒ \(\mathrm{Mg}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow\mathrm{Mg}(\mathrm{OH})_2+\mathrm{H}_2\)

Question 14. H₂0₂ can be obtained when following reacts with H₂S0₄ except with

1. Ba0₂
2. Pb0₂
3. Na₂0₂
4. SrOz

Answer: 2. Pb0₂

H₂0₂ is prepared by the reaction of peroxide with H₂S0₄.Pb0₂ is a dioxide. Hence, it does not give H₂0₂ with dilute H₂S0₄.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen

Very Short-Type Questions

Question 1. Which is the lightest gas known?
Answer: Dihydrogen

Question 2. Which isotope of hydrogen is radioactive?
Answer: Tritium (jH)

Question 3. Give examples of an ionic, a covalent, and a metallic hydride.
Answer: CaH₂ (ionic); NH₃ (covalent) & CrH (metallic)

Question 4. What is hydrolith?
Answer: Calcium hydride (CaH₂)

Chemical Properties Of Hydrogen

Question 5. Name the two nuclear spin isomers of dihydrogen.
Answer: ortho- hydrogen and

Question 6. Give an example of an electron-deficient hydride in which three centre-two electron bonds are present.
Answer: para- hydrogen; 6. B₂Hg

Question 7. Which gaseous compound on treatment with dihydrogen produces methanol?
Answer: Carbon monoxide;

Question 8. How will you prove that a colorless liquid is water?
Answer: 1. White anhydrous CuS04 becomes blue in contact with water

Question 9. What is the unit for expressing the degree of hardness of water?
Answer: ppm (parts per million);

Question 10. Write the names of two chemical substances that are used for removing dissolved oxygen from water meant for the boiler.
Answer: Hydrazine (NH₂NH₂) and sodium sulphite (Na₂S0₃);

Question 11. Why is heavy water used in atomic reactors?
Answer: It is used as a moderator

Question 12. Name a solid and a liquid absorbent of water.
Answer: Cone. H₂S0₄ and P₂0₅;

Question 13. Which chemical is commercially known as ‘perhydrol’?
Answer: H₂0₂ solution;

Question 14. What is called ‘hyper lol or horizon’?
Answer: 2-ethylanthraquinol;

Chemical Properties Of Hydrogen Question 15. What is the die volume strength of a molar solution of H₂0₂?
Answer: A compound of hydrogen peroxide and urea is called hyper lol (NH₂C0NH₂-H₂0₂);

Question 16. Which organic reagent is used for the manufacture of H₂O₂?
Answer: 2-ethylanthraquinol;

Question 17. 10 volume of H₂0₂ = x(N)H₂0₂ .What is the value of x?
Answer: 17. X = 56; 18.

Question 18. What are how water molecules are bonded to the anhydrous salt to form hydrates?
Answer: Coordinate bond and H bond

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen Short-Type Questions

Question 1. How can dihydrogen be obtained from nitric add?
Answer: By the reaction of Mg with HN0₃(2%)

Question 2. Concentrated H₂S0₄ cannot be used for drying H₂ gas why?
Answer: Cone. H₂S0₄ on absorbing H₂0 from moist H₂ produces so much heat that hydrogen catches fire

Chemical Properties Of Hydrogen Question 3. What type of hydrides can be formed by each of the following elements: Li, Zr, P, Hf, N, Ca?
Answer: Ionic, metallic, covalent, metallic, covalent, and ionic respectively.

Question 4. What are the different types of bonds formed by hydrogen in its compounds?
Answer: Ionic, covalent and H-bond

Question 5. In the preparation of H₂0₂, MnOz or Pb0₂ cannot be used instead of Ba0₂ —why?
Answer: Ba0₂ contains peroxo linkage ( —O —O — ), but Mn0₂(0=Mn=0) or Pb0₂(0=Pb=0) does not contain peroxo linkage.

MPBSE Class 11 Chemistry Chemical Properties Of Hydrogen Numerical Problems

Question 1. 1L of a hard water sample contains 1 mg CaCl₂ and 1 mg MgCl₂. Estimate the degree of hardness of this sample of water.
Answer: Molecular mass of CaCl₂ =111

Nowlllg of CaCl₂= lOOg of CaC0₃

⇒ \(1 \mathrm{mg} \mathrm{CaCl}_2 \equiv \frac{100 \times 0.001}{111} \mathrm{~g}^2 \text { of } \mathrm{CaCO}_3\)

= 9 X 10^-4g of CaC0₂ = 0.9mg of CaC0₃

The molecular mass of MgCl₂ = 95

Now, 95g of MgCl₂=100g of CaC0₃

⇒ \(1 \mathrm{mg} \text { of } \mathrm{MgCl}_2 \equiv \frac{100}{95} \mathrm{mg} \text { of } \mathrm{CaCO}_3=1.05 \mathrm{mg} \text { of }\)CaC03

Equivalent amount of CaC03 corresponding to CaCl₂

and MgCl₂ presentin or 103g of hard water

= (0.90 + 1.05)mg = 1.95mg [1L water = 103g = 106mg water]

The mass of the equivalent amount of CaC03 corresponding to CaCl2 and MgCl2 present in 106mg of water is 1.95 mg. Hence, the degree of hardness of the given sample is 1.95 ppm.

Hence, the hardness of that sample of water is 75 ppm.

Chemical Properties Of Hydrogen

Question 2. Determine the strength of ’30 volume’ H₂0₂ in normality.
Answer: Volume strength = 5.6 x normality

or, 30 = 5.6 x normality

⇒ \(\text { or, normality }=\frac{30}{5.6} \mathrm{~N}=5.35 \mathrm{~N}\)

The normality of 30 volume of H₂0₂ is 5.35N

Question 3. Determine the volume (in liter) of 0₂ obtained at STP when 0.1 liter of 2(M)H₂O₂ solution is decomposed.
Answer:

Now, 1L1(M) 1I₂0₂ = 34g of H₂0₂

1L₂(M) H₂0₂ → (2 X 34)g of H₂0₂

0.1L 2(M) H₂0₂ = (2 X 34 X 0.1 )g of H₂0₂ = 6.8g of H₂O₂

Again at STP, 68g of M₂0₂ produces 22.4L of 0₂.

⇒ \(6.8 \mathrm{~g} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { produces } \frac{22.4}{68} \times 6.8 \mathrm{~L} \text { of } \mathrm{O}_2\)

= 21241, of 0₂

Question 4. When 100 ml of tube-well water is titrated using methyl orange as an indicator, it requires 15 ml of 0.01 (N) HCl. Estimate the hardness of that sample of water.
Answer: 15 mL 0.01 (N) HCI = 1 mL 0.15(N) HCI

⇒ \(1000 \mathrm{~mL} 1(\mathrm{~N}) \mathrm{HCl} \equiv \frac{100}{2} \mathrm{~g} \mathrm{CaCO}_3\)

⇒ \(1 \mathrm{~mL} 0.15(\mathrm{~N}) \mathrm{HCl}=50 \times \frac{1}{1000} \times 0.15 \mathrm{~g} \text { of } \mathrm{CaCO}_3\)

= 7.5x 10^-3g of CaCO₃

Therefore, 100ml, or 100g of that sample of water contains some hardness-producing substance which is equivalent to 7.5 x 10^-3g of CaC0₃. 106g of water contains the hardness producing substance equivalent to\(\frac{7.5 \times 10^{-3} \times 10^6}{100}=75 \mathrm{~g} \text { of } \mathrm{CaCO}_3\)

Chemical Properties Of Hydrogen

Question 5. Calculate the amount of H₂0₂ present in 600 mL of 10-volume H₂0₂ solution.
Answer: 10 volume H₂0₂ means that 1 mL of H₂O₂ solution will produce 10 mL of 0₂ at STP

⇒ \(\begin{gathered} 2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ \left(2^{\prime} 34\right) \mathrm{g}=68 \mathrm{~g} \quad 22400 \mathrm{~mL} \text { (at STP) } \end{gathered}\)

At STP, 10 mL 0₂ is obtained from 1 mL of lOvol H₂0₂ solution.

⇒ \(22400 \mathrm{~mL} \text { of } \mathrm{O}_2 \text { is obtained from } \frac{22400}{10} \mathrm{~mL} \text { of } 10 \mathrm{vol}\)

H₂0₂ solution =2240 mL of H₂0₂

2240 mL of H₂0₂ solution contains 68g of H₂0₂

⇒ \(100 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { solution contains } \frac{68 \times 100}{2240}=3.03 \mathrm{~g} \text { of }\)H2O2.

⇒ \(\text { So, } 600 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \text { solution contains }=\frac{3.03 \times 600}{100} \mathrm{~g} \text { of }\)

H₂0₂ = 18.18g of H₂0₂ =18.2g of H₂0₂

Chemical Properties Of Hydrogen

Question 6. An excess of acidic KI solution is added to 25 mL of a H₂0₂ solution when iodine is liberated. 20 mL of 0.1(N) sodium thio-sulfate solution is required to titrate the liberated iodine. Calculate the percentage strength, volume strength, and strength in normality of the H₂0₂ solution.
Answer: 25 x ;e(N) = 20 x 0.1(N)

⇒ \(x=\frac{20\times0.1}{25}=\frac{2}{25}=0.08(\mathrm{~N})\)

Amount of H₂0₂ in 25mL0.08(N) II₂0₂ solution

1000 mL 1(N) H₂0₂ solution = 17g H₂0₂

⇒ \(\begin{aligned} & 25 \mathrm{~mL} 0.08(\mathrm{~N}) \mathrm{H}_2 \mathrm{O}_2 \text { solution } \equiv \frac{17 \times 25 \times 0.08}{1000}=0.034 \mathrm{~g} \\ & \mathrm{H}_2 \mathrm{O}_2 \end{aligned}\)

⇒ \(100 \mathrm{ml} 0.08(\mathrm{~N}) \mathrm{H}_2 \mathrm{O}_2 \text { solution contains }=\frac{0.034 \times 100}{25}\)

= 0.136g H₂O₂

% strength of H₂0₂ solution = 0.136

68g H₂0₂ gives → 22.4 L 0₂ at STP

⇒ \(0.034 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2 \text { solution } \rightarrow \frac{22.4 \times 0.034}{68} \text { Litre } \mathrm{O}_2 \text { at STP }\)

= 0.0112 Litre 0₂ at STP

= 11.2 mL 0₂ at STP

25 mL H₂O₂ solution gives 11.2 mL 0₂ at STP.

⇒ \(\text { ImL } \mathrm{H}_2 \mathrm{O}_2 \text { solution gives } \frac{11.2}{25}=0.448 \mathrm{~mL} \mathrm{O}_2 \text { at STP. }\)

Volume strength of the given H₂0₂ solution = 0.448

structure of atom class 11 questions and answers

Question 1. Why is the charge of an electron considered the smallest measurable unit of electricity?
Answer:

American scientist, R. A. Millikan determined the charge of an electron by his famous classical oil drop experiment value was found to be 4.8 x 10_1° esu or 1.602 x 10-19 C. No other particles carry a lesser negative charge.

This is the smallest measurable quantity of charge. The electric charge carried by positively or negatively charged particles is integral multiples of this minimum possible quantity.

Hence, the charge of an electron is considered the smallest unit of electricity.

Question 2. Estate two differences between cathode & anode rays.
Answer:

Cathode rays are composed of negatively charged particles whereas positively charged particles constitute anode rays.

The ratio of charge and mass (e/m ) of the constituent of cathode rays is always constant and it does not depend on the nature of the gas in the discharge tube and the cathode used.

On the other hand, the ratio of charge and mass (e/m ) of the particle constituting the anode rays is not definite. It assumes different values if different gases are used in the discharge tube.

Read and Learn More Class 11 Chemistry

Question 3. Calculate the mass and charge of 1 mol electrons.
Answer:

Charge carried by 1 electron = 1.602 x 10-19 coulomb i.e., the charge associated with 1 mol electron.

=1.602 ×10^-19 × 6.022 × 10²³ coulomb

= 9.647 × 10⁴ coulomb

Mass of an electron = 9.108 ×  10^-28 g

Mass of1mol electrons = 9:108 × 10^-28 ×  6.022×10²³

= 5.485 × 10²⁴ g

Question 4. Compare the values of e/m of electron and proton.
Answer:

If the mass and charge of an electron are mx and ex respectively and the mass and charge of a proton are m2 and e2 respectively, then.

⇒ \(\frac{(\mathrm{e} / \mathrm{m}) \text { of electron }}{(\mathrm{e} / \mathrm{m}) \text { of proton }}-\frac{e_1 / m_1}{e_2 / m_2}=\frac{e_1}{\mathrm{e}_2} \times \frac{m_2}{m_1}=\frac{m_2}{m_1}=1837\)

[As both electron and proton carry a unit charge and a proton is 1837 times heavier than an electron.]

e/m ofelectron = 1837 x e/m of proton.

Structure Of Atom Class 11 Questions And Answers

Question 5. If an electron is promoted from the first orbit to the the third orbit of a hydrogen atom, by how many times will the radius of the orbit be increased?
Answer:

The radius of the orbit of H -atom, rn \(=\frac{n^2 h^2}{4 \pi^2 m e^2}\)

Radius of the first orbit (r1) \(=\frac{1^2 \times h^2}{4 \pi^2 m e^2}\)

and radius of the third orbit (r3) \(=\frac{3^2 \times h^2}{4 \pi^2 m e^2}\)

∴ \(\frac{r_3}{r_1}=\frac{3^2 \times h^2}{4 \pi^2 m e^2} \times \frac{4 \pi^2 m e^2}{h^2}=9 \text { i.e., } r_3=9 \times r_1\)

∴ If the electron moves from the 1st orbit to the 3rd orbit, the radius of the orbit will be increased 9 times

Question 6. The ionization energy for the H-atom in the ground state is x1-atom-1. Show that the energy required for the process, He+(g) — He2- + e, is 4x J-atom-1.
Answer:

The energy of the electron in the n-th orbit of an H-like particle is

⇒ \(\text { [where } \left.K=\frac{2 \pi^2 m e^4}{h^2}=\text { constant }\right]\)

⇒ \(\text { Now, } \begin{aligned}
I . E_{\mathrm{H}} & =E_{\propto}-E_1=0-\left(-K \times \frac{1^2}{1^2}\right)=K \\
& =x \mathrm{~J} \cdot \text { atom }^{-1}
\end{aligned}\)

The energy required for the given process is the ionization energy of He+.

Since He+ is a hydrogen-like species, so

⇒ \(\begin{aligned}
L . E_{\mathrm{He}^{+}} & =E_\alpha-E_1=0-\left(-K \times \frac{2^2}{1^2}\right) \quad\left[\text { for } \mathrm{He}^{+}, z=2\right] \\
& =4 K=4 x \mathrm{~J} \cdot \mathrm{atom}^{-1}
\end{aligned}\)

Question 7. What does the statement “electronic orbits at stationary state” mean? Does an electron remain stationary in a stationary orbit? Explain.
Answer: According to Bohr’s theory of the hydrogen atom, electrons revolve around the nucleus in some fixed orbits and during its motion, an electron does not lose energy.

For this reason, these orbits are known as “electronic orbits at stationary states”.

When an electron stays in such an orbit, it does not remain stationary at all. Had it been so, the electron, being attracted by the nucleus would have fallen on the nucleus.

The electron always remains in motion so as to overcome the influence of the nuclear attractive force.

Class 11 Chemistry Structure Of Atom Important Questions

Question 8. If the radius of the first Bohr orbit is x, then find the de Broglie wavelength of the electron in the third orbit.
Answer: The radius of the the n -th orbit \(r_n \propto n^2\)

∴ r3 = 9x r1 = 9x

Since [r1 =x]

Now, the circumference of the n-th orbit is an integral multiple of the wavelength associated with the motion of the electron, i.e.

2nrn = n x A

∴ 2nr3 = 3 x A

Or, \(\lambda=\frac{2 \pi r_3}{3}=\frac{2 \pi}{3} \times 9 x=6 \pi x\)

Question 9. A photon of wavelength A collides with an electron. After the collision, the wavelength of the photon changes to A’. Calculate the energy of the scattered electron
Answer: Energy ofthe photon before collision \(=h v=\frac{h c}{\lambda}\)

Energy ofthe photon after collision \(=h v^{\prime}=\frac{h c}{\lambda^{\prime}}\)

The difference in energy is imparted to the electron, which suffers collision. Hence energy ofthe scattered electron.

⇒ \(=\frac{h c}{\lambda}-\frac{h c}{\lambda^{\prime}}=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)

Question 10. Find the condition under which the de Broglie wave¬ length of a moving electron becomes twice that of a moving proton. Given that a proton is 1836 times heavier than an electron.
Answer: \(\lambda_e=\frac{h}{m_e v_e} \text { and } \lambda_p=\frac{h}{m_p v_p}\)

⇒ \(\text { Since } \lambda_e=2 \times \lambda_p\)

∴ \(\frac{h}{m_e v_e}=2 \times \frac{h}{m_p v_p}\)

or, \(2 \times m_e v_e=m_p v_p\)

Or, \(v_e=\frac{m_p v_p}{2 m_e}=\frac{v_p}{2} \times \frac{m_p}{m_e}=\frac{v_p}{2} \times 1836=918 v_p\)

i.e., vg = 9lSvp, this is the necessary condition.

Structure Of Atom Class 11 Questions And Answers

Question 11. If the kinetic energy of an electron increases by nine times, the wavelength of the de Broglie wave associated with it will increase by how many times
Answer: The de Broglie wavelength associated with a moving electron of mass m and kinetic energy E is given by

⇒ \(\lambda=\frac{h}{\sqrt{2 E m}}\)

⇒ \(\text { Now } \lambda_1=\frac{h}{\sqrt{2 E m}}, \lambda_2=\frac{h}{\sqrt{2 \times 9 E \times m}}=\frac{h}{\sqrt{2 E m}} \times \frac{1}{3}\)

∴ \(\frac{\lambda_2}{\lambda_1}=\frac{h}{\sqrt{2 E m}} \times \frac{1}{3}+\frac{h}{\sqrt{2 E m}}\)

or, \(\lambda_2=\frac{1}{3} \lambda_1\)

So, de Broglie wave associated with the electron will increase by \(\frac{1}{3}\) times.

Question 12. Mention three differences between wave and partide.
Answer: Differences between wave and partied

Question 13. Discuss the limitation of Bohr’s theory based on Heisenberg’s uncertainty principle.
Answer:

• Bohr’s theory posits that negatively charged particles (electrons) within an atom orbit the nucleus in precisely defined paths with predetermined radii. To counterbalance the nuclear attraction force, electrons must exhibit a specific velocity.
• According to Heisenberg’s uncertainty principle, it is impossible to simultaneously ascertain the precise position and momentum (or velocity) of a minuscule particle such as an electron. Consequently, Bohr’s model is in conflict with Heisenberg’s uncertainty principle.

Question 14. What is the difference between the notations L & Z
Answer:

Represents the second Bohr orbit for which n = 2. On the other hand, l denotes azimuthal quantum number which may have values 0, 1, 2, etc.

Shapes of various subshells present within the same principal shell, the number of sub-shells in a ‘certain shell, and their relative energies, etc. depend on the values of’ l ’.

Structure Of Atom Class 11 Notes Pdf

For example, in the case of s, p, d, and /-subshells, the values of l are 0, 1, 2, and 3 respectively.

Question 15. How many quantum numbers are required to identify an orbital? Explain it with the help of a specific example.
Answer:

To identify an orbital, three quantum numbers need to be mentioned.

These are the principal quantum number (n), azimuthal quantum number (Z), and magnetic quantum number (m).

For example, in order to denote 2px, 2py and 2Pz > the respective values of the said quantum numbers are—

1. 2px: n = 2, l = 1, m = -1
2. 2py: n = 2, l = 1, m = 0
3. 2pz: n = 2, l = 1, m = +1

Incidentally, it may be noted that the values of all four quantum numbers need to be mentioned to specify a particular electron in any atom.

Question 16. Give the electronic configuration of 24Cr3+. Find the no. of unpaired electrons present in its ion.
Answer: 24Cr: ls22s22p63s23p63d54s1

Question 17. Write down the electronic configurations of Ni and Ni2+ (atomic number of Ni = 28). How many odd electrons do each of them contain?
Answer: 28Ni: ls22s22p63s23p63d84s2

∴ Number of odd electrons = 2;

Electronic configuration of Ni2+: ls22s22p63s23p63d84

∴ A number of odd electrons present in Ni2+ ion 2.

Structure Of Atom Class 11 Questions And Answers

Question 18. Which orbit of the Be3+ ion has the same radius as that of the ground state of hydrogen atoms?
Answer: For hydrogen-like atoms or ions, the radius of the n-th orbit is given by, rn \(r_n=0.529 \times \frac{n^2}{Z}\)

In the ground state of H-atom, n =1 (1st orbit) and Z(at. no.) =1

∴ Radius of 1st orbit of H-atom \(r_1=0.529 \times \frac{1^2}{1}=0.529 \)

Let the n -th orbit of Be3+ ion has the same radius as that of the ground state of H-atom.

Now, rn \(r_n\left(\mathrm{Be}^{3+} \text { ion }\right)=0.529 \times \frac{n^2}{4}\) \(=0.529 \times \frac{n^2}{4}\)

Thus, \(0.529 \times \frac{n^2}{4}=0.529\)

or, n2 = 4 or, n = 2

Hence, second orbit of the Be3+ ion has the same radius as that of the ground state of the H-atom.

Question 19. Explain why Fe3+ is more stable than Fe2+ ion.
Answer: Atomic number of Fe-atom =26 and its electronic configuration: ls22s22p63s23p63rf64s2

Electronic config. of Fe2+: lsa2s22p63s23p63d6

and electronic config. of Fe3+ : ls22s22p63s23p63rf5

We know that the electronic configurations of halt-filled or fully-filled subshells are more stable compared to others.

Now, the 3d -subshell of Fe3+ ion is exactly half-filled, and hence Fe3+ is more stable than Fe2.

Ncert Solutions For Structure Of Atom Class 11

Question 20. The electronic configuration of the electrons in the outer shells of Cu & Cr are 3d104s1 & 3ds4s1 respectively instead of being 3d94s2 and 3d44s2 explain why.
Answer: Half-filled and fully-filled subshells have greater stability. In the case of Cu, the 3d104s1 configuration is more stable than 3d94s2 because in 3d104s1, the 3d -subshell is completely filled, and the 4s -subshell is half-filled.

Similarly, in the case of Cr, the 3d54sx configuration contains half-filled 3d -and 4s -subshells, and hence this will be more stable than the 3d44s2 configuration.

Question 21. All atoms having an even number of electrons always contain paired electrons—is the statement true? If so, then by what principle or rule the above statement can be explained?
Answer:

All the electrons, present in atoms with an even number of electrons, do not necessarily get paired. This can be seen from the electronic configuration of carbon (atomic number 6): ls²2s²2p

In spite of having an even number of electrons (total 6 electrons) the 2p -orbitals of carbon atoms have 2 unpaired or odd electrons. This electronic configuration can be explained by Hund’s rule.

According to this rule—the pairing of electrons in the orbital of a particular subshell (degenerate orbitals) does not occur until all the orbitals of that subshell are singly filled up and the singly occupied orbitals must have electrons with parallel spins.

Question 22. Find the total number of orbitals present in the principal quantum number ‘3’ of an atom. Write the symbols of different types of orbitals and also indicate the number of orbitals in each type
Answer:

S. Total number of orbitals present for the principal quantum number 3 is given by, 32 = 9.

Now n – 3 may have the values 0, 1, and 2 so the types of orbitals present in the given shell tire s. p and d.

Again the no. of -orbitals =(2/ + l) =(2×0) + l =1

Number of p -orbitals = 2/ +1 =(2xl) +1=3

Number of d -orbitals = 2/+l=(2×2)+l=5

MPBSE Class 11 Chemistry Notes For Amorphous Carbon

1. Charcoal

Vegetable charcoal:

1. Wood charcoal: When wood is subjected to destructive distillation in an iron retort, the volatile organic compounds present escape and the residue left in the retort is called wood charcoal.
2. Sugar charcoal: It can be prepared by heating pure sugar in a closed vessel or by eliminating water from sugar by reacting it with concentrated sulphuric acid

Animal charcoal:

• Bone charcoal:
  • Small pieces of animal bones are treated with superheated steam to remove the adhering fat and marrow.
  • The dried bones are then subjected to destructive distillation in an iron retort when volatile substances are distilled out and a black residue containing carbon and about 90% impurities like calcium phosphate and calcium carbonate is left behind.
  • This is known as bone black or bone charcoal. These impurities are removed by dissolving the black material in dilute HCl.
  • The insoluble deep black powder thus obtained is almost pure charcoal and is called ivory black.
• Blood charcoal:  Charcoal obtained by destructive distillation of blood is known as blood charcoal.

MPBSE Class 11 Chemistry For Carbon nanotubes Physical properties

Charcoal is black, soft and porous.It is a bad conductor of heat and electricity. Its specific gravity lies between 1.4 and 1.9. But because of its porosity, air enters in its pores.

• As a result, its specific gravity gets reduced to 0.2 and hence charcoal floats on water. When porous charcoal b freed from entrapped air, it can retain any other ga* In ib pores, This phenomenon is known as adsorption.
• The adsorbed gas escapes on heating and is much more reactive than the ordinary gas.

Amorphous Carbon Class 11 Notes

2. Activated charcoal

Activated charcoal has high adsorption power as compared to ordinary charcoal.

Activated charcoal Preparation

• When the charcoal obtained by destructive distillation of coconut shell is heated to about 800-900°C  a limited supply of air or steam, activated charcoal is obtained.0
• Besides this, activated charcoal may also be obtained by the destructive distillation of sawdust soaked In aqueous solution of ZnCl₂ or MgCl

Activated charcoal Properties

• Activated charcoal is not only a good adsorbent for gases but it also has the power of decolourising a coloured substance and absorbing the taste of a substance
• Moreover, as a catalyst accelerates the rates of many chemical reactions.

MPBSE Class 11 Chemistry For Activated charcoal Chemical properties

1. At higher temperatures, charcoal burns in air or oxygen to form CO₂ gas. However, in a limited supply of oxygen, its combustion produces carbon monoxide.

C + O₂→ CO₂; 2C + O₂→ 2CO

2. Charcoal forms different compounds with sulphur, nitrogen and hydrogen at high temperatures.

Amorphous Carbon Class 11 Notes

2C + N₂→ (CN)₂ (cyanogen)

2C + H₂→ C₂H₂ (acetylene)

In the first reaction, two solids (C and S) react together to form a liquid (carbon disulphide

3. Charcoal combines with heated Ca, Al, Fe etc. to form their corresponding carbides.

Ca + 2C→ CaC₂; 3Fe + C→ Fe₃C; 4Al + 3C→Al₄C₃

4. When a mixture of silica (SiO₂) and coke dust is heated at 1500’2000°Cin an electric furnace, silicon carbide (SiC) is formed. It is black, bright and very hard solid. It is known as carborundum which is used for polishing metals

5. When steam is passed over white-hot charcoal or coke, a mixture containing equal volumes of CO and H₂, called water gas is produced

C+H₂ O → CO + H₂ (water gas)

6. Reducingproperty: Charcoal is a good reducing agent.

Amorphous Carbon Chemistry Class 11

1. At higher temperatures, various metal oxides are reduced by charcoal to their corresponding metals.

CuO + C→ Cu + CO; PbO + C→ Pb + CO

Fe₂O₃ + 3C→  2Fe + 3CO

2. At higher temperatures, charcoal reduces carbon dioxide to CO and sodium sulphate to sodium sulphide.

3. Charcoal in burning condition is oxidised by concentrated nitric acid or sulphuric acid to CO₂.

MPBSE Class 11 Chemistry For Activated Charcoal Uses

Wood charcoal serves as a fuel and a reducing agent in metal extraction.

• It is employed in the decolorization of sugar syrup and the refinement of oils, fats, and glycerin.
• It is utilized in the treatment of potable water as it adsorbs surplus chlorine following chlorination.
• It is utilized in gas masks due to its ability to adsorb toxic gases.
• It is additionally utilized in the formulation of gunpowder and black paint (Black Japan).

MPBSE Class 11 Chemistry For Lamp Black Or Carbon Black

When organic liquids rich in carbon such as kerosene, petrol, turpentine oil, benzene etc. are subjected to bum in controlled air, a black sooty smoke is produced. This smoke on condensation in a cold container forms soot.

• This soot is called lamp black or carbon black.
• It may also be obtained when natural gas (methane) is subjected to albumin-controlled air.
• It is the purest form of all the amorphous allotropes of carbon.

Mpbse Class 11 Chemistry Notes Pdf

Lamp Black Or Carbon Black Properties:

It is amorphous, black and non-conductor of heat and electricity.

Lamp Black Or Carbon Black Uses:

It is used in the preparation of printing ink, shoe polish and blackpaints

MPBSE Class 11 Chemistry For Coke And Gas Carbon

When anthracite coal (96% carbon) is subjected to destructive distillation, the solid residue left in the iron retort is called coke. At higher temperatures (1000 -1200°C), hard coke is called coke. At higher temperatures (1000 -1200°C), hard coke is obtained whereas at 600-650°C, we get soft coke.

Properties Of Amorphous Carbon Class 11

The black hard dense residue deposited on the relatively cooler upper part of the retort is as gas carbon. it possesses thermal water gas is produced. and electrical conductivity and electrical conductivity.

MPBSE Class 11 Chemistry For Coke and gas carbon Uses

• Hard coke serves as a fuel and a reducing agent for residential applications.
• Soft coke serves as a residential fuel source.
• Carbon gas is extensively utilized as electrodes in batteries, arc lamps, and during electrolysis.

All allotropic forms of carbon comprise the same element: When a certain mass of a pure allotropic variant of carbon is heated in an elongated, robust combustion tube in the presence of pure oxygen, carbon dioxide (CO₂) and carbon monoxide (CO) are generated.

• CuO, contained within the tube, transforms CO into CO₂. The collected CO₂ is absorbed in a pre-weighed potash bulb connected to the exit end of the tube.
• An rise in the bulb’s weight indicates the quantity of CO2 produced. When the experiment is conducted independently with the same constant weight of any other allotrope, an equivalent quantity of CO₂ is produced in each instance.
• This experiment demonstrates that the various allotropes comprise the same element, namely carbon.

Periodicity In Properties Class 11 Questions

Question 1. Would you expect the second electron-gain enthalpy of 0 as positive, more negative, or less negative than the first? Justify your answer.
Answer:

There are several valence electrons in oxygen and it requires two more electrons to complete its octet. So, the O-atom accepts one electron to convert into an O-ion and in the process liberates energy. Thus, the first electron-gain enthalpy of oxygen is negative.

⇒ \(\mathrm{O}(g)+e \rightarrow \mathrm{O}^{-}(g)+141 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\left(\Delta_i H_1=-v e\right)\)

However, when another electron is added to O- to form 02-ion, energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged O ion and the second incoming electron. Thus, the second electron-gain enthalpy of oxygen is positive.

⇒ \(\mathrm{O}^{-}(\mathrm{g})+e \rightarrow \mathrm{O}^{2-}(\mathrm{g})-\left(780 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right)\left(\Delta_i H_2=+v e\right)\)

Read and Learn More Class 11 Chemistry

Question 2. How would you react to the statement that the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds?
Answer:

The electronegativity of any element depends on the hybridization state and oxidation state of that element in a particular compound, i.e., the electronegativity of an element varies from compound to compound. For example, the electronegativity of Natom varies as sp3 —N < sp2—N < sp—N. So, the, given statement is not correct.

Question 3. Would you expect the first ionisation enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer:

Isotopes of an element have the same number of electrons and similar electronic configurations. So their nuclear charge and atomic radii are identical. Consequently, two isotopes of the same element are expected to have the same ionisation enthalpy.

Question 4. What are the major differences between metals and
non-metals?
Answer:

Metals exhibit a pronounced propensity to lose electrons, resulting in the formation of cations.

• They serve as potent reducing agents, possess low ionization enthalpies, exhibit less negative electron gain enthalpies, have low electronegativity, and generate basic oxides and ionic compounds.
• Non-metals exhibit a pronounced inclination to acquire electrons, resulting in the formation of anions.
• They are potent oxidizing agents, possess elevated ionization enthalpies, exhibit substantial negative electron-gain enthalpies, demonstrate high electronegativity, and generate acidic oxides and covalent compounds.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 5. Use the periodic table to answer the given questions. Identify an element with 5 electrons in an outer subshell. Identify an element that would tend to lose 2 electrons. Identify an element that would tend to gain 2 electrons. Identify the group having metal, non-metal, liquid, and gas at room temperature
Answer:

Fluorine. Its configuration is ls²2s²2p⁵

Magnesium. Its configuration is ls²2s²2p⁶3s². So, Mg loses 2 electrons from its outermost shell to form Mg2+ and attains a stable configuration.

Oxygen. Its configuration is ls²2s²2p⁴ So, O gains 2 electrons to form O₂– and attains stable configuration.

Group-17. The metallic character of astatine (At) is much greater than its non-metallic character and its melting point is very high (3O₂°C).

So, astatine is considered as a metal. So in group-17 there is a metal (At), non-metals (F2, Cl₂, Br₂, I₂), liquid (Br₂) and gas (F₂, Cl₂).

Question 6. The order of reactivity of group-1 LI < Na < K < Rb < Cs whereas that of group-17 elements Is F > Cl > Br >I. Explain.
Answer:

There is only one electron in the valence shell of the elements of group 1. Thus, they have a strong tendency to lose this single electron.

The tendency to lose electrons depends on the ionization enthalpy. As ionization enthalpy decreases down the group, the correct order of increasing reactivity of group-1 elements is Li < Na < K < Rb < Cs. On the other hand, there are 7 electrons in the valence shell of the elements of group-17. Thus, they have a strong tendency to gain a single electron.
The tendency to gain electrons depends on the electrode potentials of the elements. As the electrode potential of elements decreases down the group, the correct order of activity is F > Cl > Br >I.

Class 11 Chemistry Important Questions

Alternate explanation: In the case of halogens, their reactivity increases with the increase in electron-gain enthalpy.

Order of electron-gain enthalpy: F < Cl > Br >I. As electron gain enthalpy decreases from Cl toI, the order of activity also follows this sequence. However, fluorine is the most reactive halogen as its bond dissociation energy is very low.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 7. Assign the position of the element having outer electronic configuration: ns2np4 for n = 3, 0 (n-l)d2ns2 for n = 4 & (n-2)/7(n-l)d1ns2 for tt = 6, in the periodic table.
Answer: (T) As n= 3, the element belongs to the period. Since the last electron enters the p-orbital, the given element is a p-block element. For p-block elements, group no. of the element = 10+no. of electrons in the valence shell.

The element is in the (10+6) = 16th period

As n = 4, the element belongs to the fourth period.

Since is present in the element, it is a block element. For d-block elements, group no. of the element = no. of ns electrons + no. of(n-l) f electrons = 2+2 = 4. Therefore, the element is in the 4th period.

As n – 6, the element belongs to the sixth period. Since, the last electron enters the f-orbital, the element is a f-block element. All f-block elements are situated in the third group of the periodic table.

The first (A1H1) and second (A1H2) ionisation enthalpies (klmol-1) and the (AegH) electrongain enthalpy (in kj.mol-1 ) of a few elements are given below: Which of the above elements is likely to be the P least reactive clement. the most reactive metal.

the most reactive non-metal. the least reactive non-metal. the metal which can form a stable binary halide of the formula MX2(X = halogen).

Elements And Periodicity In Properties Pdf

Question 8. The metal that can form a predominantly stable covalent halide of the formula MX (X = halogen)? Element V is the least reactive metal as it has the highest first ionisation enthalpy & positive electron-gain enthalpy.|
Answer

The element is the most reactive metal due to its lowest initial ionization enthalpy and low negative electron gain enthalpy.

• Element 3 is the most reactive non-metal due to its exceptionally high initial ionization enthalpy and significantly high negative electron uptake enthalpy.
• Element 4 is the least reactive non-metal due to its large negative electron-gain enthalpy and very moderate initial ionization enthalpy.
• Element 4 possesses low first and second ionization enthalpies. The initial ionisation enthalpy of this element exceeds that of the alkali metals. Consequently, the specified element is an alkaline earth metal capable of forming a stable binary halide with the formula MX2.
• The initial ionization enthalpy of elements is low, whereas the subsequent ionization enthalpy is elevated. It is an alkali metal capable of forming stable covalent halides (MX).
• Predict the formulas of the stable binary compounds generated by the following pairs of elements: OLi and O, Mg and N, Al and Si, P and F, and the element with atomic number 71 and F.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 9. In the modern periodic table, period indicates the value of atomic number atomic mass principal quantum number azimuthal quantum number
Answer: Each period in the modern periodic table begins with the filling of a new shell. So, the period indicates the value of a principal quantum number.

Question 10. Which of the following statements related to the modern periodic table is incorrect?

p -block has 6 columns because a maximum of 6 electrons can occupy all the orbitals in a p-shell.

d -block has 8 columns, as a maximum of 8 electrons can occupy all the orbitals in a d -subshell.

Each block contains some columns equal to the number of electrons that can occupy that subshell.

Question 11. Block indicates the value of azimuthal quantum number (l) for the last subshell that received electrons in building up electronic configuration
Answer: The statement is incorrect because d -the block has 10 columns because a maximum of ten electrons can occupy all the orbitals in a d -subshell.

Class 11 Chemistry Periodic Table Notes

Question 12. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

1. Valence principal quantum number (n)
2. Nuclear charge (Z)
3. Nuclear mass
4. Number of core electrons.
5. Nuclear mass does not affect the valence shell electrons (such as, and H have similar chemical properties.
6. The size of isoelectronic species F-, Ne, Na+ is affected by: nuclear charge (Z)
7. valence principal quantum number (n)
8. electron-electron interaction in the outer orbitals is one of the factors because their size is the same.
9. The size of isoelectronic species depends on the nuclear charge. As nuclear charge increases, the size of species decreases.
10. electron-electron interaction in the outer orbitals
11. none of the factors because their size is the same.
12. The size of isoelectronic species depends on the nuclear charge. As nuclear charge increases, the size of species decreases.

Question 13. Which one ofthe following statements is incorrect about ionization enthalpy?

1. Ionization enthalpy increases for each successive electron.
2. The greatest increase in ionization enthalpy is experienced in the removal of electrons from the core noble gas configuration.
3. The end of valence electrons is marked by a big jump in ionization enthalpy.
4. Removal of electrons from orbitals with lower n values is easier than from orbitals with higher values.

Answer: This is incorrect as the removal of electrons from orbitals with lower values is more difficult than the removal of electrons from orbitals with higher values. Because in the former case, the electron remains tightly bound as it more closer to the nucleus.

Question 14. Considering the elements B, Al, Mg, and K, the correct order of their metallic character is:

• B > Al > Mg > K
• Mg > Al > K > B
• Al > Mg > B > K
• K > Mg > Al > B

Answer: The statement is correct because across a period from left to right, the metallic character of the elements decreases but down a group, the metallic character increases. Thus, the metallic character of K is the highest, and that of B is the lowest.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 15. Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is:

• B>C>Si>N>F
• F>N>C>B>Si
• Si>C>B>N>F
• F>N>C>Si>B

Answer: On moving across a period from left to right, the nonmetallic character of the elements increases. Thus, the order of
non-metallic character ofthe given elements: F > N > C > B. However, on moving down a group, the non-metallic character
of elements decreases. Thus, S is more non-metallic than C. Thus, the non-metallic character of the elements follows the
sequence F>N>C>B>Si.

Question 16. For the elements F, Cl, O, and N correct order of their chemical reactivity in terms of oxidizing property is:

• F > Cl > O > N
• Cl > F > O > N
• F > O > Cl > N
• O > F > N > Cl

Answer: Oxidising power increases across a period from left to right. Thus, the order of oxidizing power decreases in the order F > O > N. On moving down a group, oxidizing power decreases. So, F is a stronger oxidizing agent than Cl. Again, O being more electronegative than Cl, O is a stronger oxidizing agent than Cl. Therefore, the overall decreasing order of oxidizing powers F > 0 > Cl > N.

MPBSE Class 11 Chemistry  Nascent Hydrogen

Hydrogen at the moment of its formation is called nascent hydrogen. The reducing property of nascent hydrogen is much greater than that of ordinary molecular hydrogen.

Nascent Hydrogen Example:

Ordinary’ molecular hydrogen cannot reduce acidic potassium permanganate or ferric chloride solution. However, when zinc dust is added to these solutions, the solutions become colorless and this is because, nascent hydrogen liberated in situ by the action of zinc on dilute sulphuric acid brings about the reduction of these compounds, discharging the colors of these solutions.

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+2[\mathrm{H}](\text { nascent hydrogen })\)

Nascent hydrogen is a hydrogen atom released during a reaction, and it reacts with something else very, very rapidly. Now, there isn’t any physical difference, both are atoms, but based on the surrounding conditions, if the atom is stable, it is called atomic. If not, it is called nascent.

MPBSE Class 11 Chemistry  Identification And Uses Of Dihydrogen

Identification of dihydrogen:

1. The gas burns in the air with a pale blue flame and produces water. That the product is water is identified by the fact that it turns white anhydrous CuSO, blue.
2. The gas is completely adsorbed by spongy palladium metal at ordinary temperature and die adsorbed gas is set free on heating the hydrogenised palladium to dull-red.

Uses of dihydrogen:

1. Dihydrogen is largely used in the synthesis of ammonia which is a starting material for the manufacture of nitric acid and various fertilizers such as urea, ammonium sulfate, etc.
2. Large quantities of dihydrogen are consumed in the vanaspati industry for the conversion of vegetable oil into vanaspati ghee (hardening of oils).
3. It is used in the manufacture of bulk chemicals such as methanol.\(\mathrm{CO}(g)+2 \mathrm{H}_2(g) \underset{\mathrm{Co}+\text { catalyst }}{\stackrel{700 \mathrm{~K} / 200 \mathrm{~atm}}{\longrightarrow}} \mathrm{CH}_3 \mathrm{OH}(l)\)
4. It is used in the manufacture of metal hydrides.
5. In metallurgy, dihydrogen is used to reduce heavy metal oxides to metals.
6. It is used in the preparation of many useful chemicals such as hydrochloric acid.
7. It is used in tire atomic hydrogen torch (produces temperature ~ 4000°C ) and oxy-hydrogen torch (produces temperature between 2270-2770°C ) for cutting and welding purposes.
8. Liquid dihydrogen mixed with liquid oxygen is used as a rocket fuel in space research. It is also used in fuel cells for generating electrical energy.
9. It is used in the manufacture of synthetic petrol.
10. With helium, it is used for filling balloons employed for atmospheric study.

MPBSE Class 11 Chemistry  Dihydrogen As A Fuel Hydrogen Economy

The production of energy in our present-day life involves mainly the use of fossil fuels such as coal, petroleum, natural gas, etc.

• In order to meet the growing demand for energy, the world’s fossil fuel reserves are getting depleted at an alarming rate and possibly they may get exhausted by the middle of the 21st century.
• In fact, the rate of consumption of fossil fuels is very much faster than the rate of their renewal in nature and it is the basic reason for their shortage. Therefore, new sources of energy are to be found.
• Dihydrogen has been considered as an alternative source of energy. The basic principle of a hydrogen economy is the storage and transportation of energy in the form of liquid or gaseous dihydrogen.

MPBSE Class 11 Chemistry  Advantages of using dihydrogen as a fuel:

1. The major advantage of using dihydrogen as a fuel is that it provides a pollution-free atmosphere because its combustion product is only water.
2. The heat of combustion per gram of dihydrogen is higher than any other fuel. For example, the amount of energy obtained from dihydrogen is 142 kJ. g-1 while those obtained from gasoline, coal, paper, and wood are 48, 29.3, 20, and 15 kJ . g-1 respectively.
3. It is abundant in a combined state as water.
4. An automobile engine using dihydrogen is about 25- 50% more efficient than an automobile engine using gasoline.
5. The time required for the regeneration of dihydrogen from its combustion product water is much shorter than that of fossil fuel from its combustion product CO₂.
6. Although dihydrogen appears to be a very good future fuel, there are some typical problems that are to be solved before we adopt the hydrogen economy.

MPBSE Class 11 Chemistry  Hydrides

Dlhydrogcn, under appropriate reaction conditions, combines with almost all elements, except noble gases, to form binary compounds called hydrides. The hydrides may be represented by the general formula EH_x (Example, MgH₂) or E_mH_n (ExampleB₂H₆ ), where E is a symbol of the element.

Nascent Hydrogen Class 11 Notes

All the main group elements except noble gases and probably indium and thallium, all lanthanides, all actinides, and transition metals such as Sc, Y, La, Ac, Tc, Zr, and Hf (to a lesser extent V, Nb, Tb, Cr, Cu, and Zn) form hydrides.

Depending upon the behavior and the nature of bonding, the hydrides can be classified into three main categories:

1. Ionic or saline or salt-like hydrides;
2. Covalent or molecular hydrides and
3. Metallic or interstitial hydrides

MPBSE Class 11 Chemistry  Ionic Or Salt-Like Hydrides

These hydrides are formed when hydrogen combines with electropositive alkali metals or alkaline earth metals of s -s-block having electronegativity less than hydrogen (2.1). The formation of ionic hydrides involves the transfer of one electron from the metal atom to the H-atom. Some common examples are LiH, NaH, CaH₂, SrH_2, etc. The hydride ion (H ) with [He] configuration is soft enough to be strongly polarised by the small Be²⁺ and Mg²⁺ ions and for this reason, BeH₂ and MgH₂ are significantly covalent. In fact, they exist in polymeric form.

Ionic Or Salt-Like Hydrides Preparation: These hydrides are prepared by heating the metal in an atmosphere of hydrogen under pressure over the temperature range of 150-600°C. For example:

⇒ \(2 \mathrm{Li}+\mathrm{H}_2 \stackrel{600^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{LiH}\)

⇒ \(2 \mathrm{M}+\mathrm{H}_2 \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{MH}[\mathrm{M}=\mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}]\)

⇒ \(\mathrm{M}^{\prime}+\mathrm{H}_2 \stackrel{150-300^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{M}^{\prime} \mathrm{H}_2\left[\mathrm{M}^{\prime}=\mathrm{Ca}, \mathrm{Ba}, \mathrm{Sr}\right]\)

Nascent Hydrogen Definition Class 11

Ionic Or Salt-Like Hydrides Physical properties:

1. Ionic hydrides are non-volatile, non-conducting white crystalline solids.
2. They have high melting and boiling points.
3. They- have high density, generally higher than that of the metals from which they are formed. This is due to the fact that H“ ions occupy holes in the lattice of the metal, without distorting the metal lattice.
4. They have high values of heat of formation\(\left(\Delta H_f^0\right)\)
5. Their thermal stability decreases with increasing the cationic size of the metal down the group in the periodic table: LiH > NaH > KH > RbH > CsH
6. In the molten state, they conduct electricity with the liberation of dihydrogen at the anode. This confirms the presence of hydride ions (H^–) in them. For example:

⇒ \(\mathrm{CaH}_2(\text { molten }) \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{H}^{-}\)

Cathode: \(\mathrm{Ca}^{2+}+2 e \rightarrow \mathrm{Ca}\)

Anode: \(2 \mathrm{H}^{-} \rightarrow \mathrm{H}_2+2 e\)

MPBSE Class 11 Chemistry  Ionic Or Salt-Like Hydrides  Chemical Properties

1. They react vigorously with water and other protic solvents such as ethanol, ammonia, etc., to liberate dihydrogen.

⇒ \(\mathrm{NaH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NaOH}+\mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{LiH}+\mathrm{CH}_3 \mathrm{OH} \rightarrow \mathrm{LiOCH}_3+\mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{NaH}+\mathrm{NH}_3 \rightarrow \mathrm{NaNH}_2+\mathrm{H}_2 \uparrow\)

2. Except LiH, they bum in the air on strong heating (400°-500°C) due to their decomposition into hydrogen which is inflammable.

3. There are strong reducing agents (especially under heated conditions). For example:

⇒ \(\mathrm{LiH}+\mathrm{CO}_2 \rightarrow \mathrm{HCOOLi} \text { [lithium formate] }\)

⇒ \(\mathrm{NaH}+2 \mathrm{CO} \rightarrow \mathrm{HCOONa} \text { [sodium formate] }+\mathrm{C}\)

Mpbse Class 11 Chemistry Notes Pdf

⇒ \(\mathrm{PbSO}_4+2 \mathrm{CaH}_2 \rightarrow \mathrm{PbS}+2 \mathrm{Ca}(\mathrm{OH})_2\)

⇒ \(6 \mathrm{LiH}+\mathrm{N}_2 \rightarrow 2 \mathrm{Li}_3 \mathrm{~N}+3 \mathrm{H}_2 \uparrow ; 3 \mathrm{CaH}_2+\mathrm{N}_2 \rightarrow \mathrm{Ca}_3 \mathrm{~N}_2+3 \mathrm{H}_2 \uparrow\)

MPBSE Class 11 Chemistry  Jonic hydrides Uses:

Jonic hydrides are used as

1. A ready source of di¬ hydrogen,
2. Solid fuel,
3. Reducing agents and
4. To prepare the very important reducing agents like lithium aluminum hydride (liajh4) & sodium borohydride (nabh4).

Covalent Or Molecular Hydrides

• When hydrogen combines with elements of comparatively high electronegativity such as p -p-block elements, covalent or molecular hydrides are formed. The common elements forming molecular hydrides are B, C, N, O, F, Si, P, S, Cl, Ga, Ge, As, Se, Br, In, Sn, Pb, etc.
• The two s -s-block elements namely Be and Mg also form this type of hydride. The formation of covalent hydride is primarily due to the fact that the electronegativity difference between these elements and hydrogen is not much higher.
• Their general formula is XH_n (for s -block elements) or XH_8-n (for p -block elements) where n is the number of valence electrons of the element. These hydrides usually exist as discrete covalent molecules which are held together by weak van der Waals forces of attraction and hence are called covalent or molecular hydrides.

Molecular Hydrides Classification: Covalent or molecular hydrides can be further classified into three categories according to their relative number of electrons and bonds present in their Lewis structures.

1. Electron-deficient hydrides: These hydrides have less number of electrons in the valence shell of the central atom (usually, less than eight electrons in the valance shell). So their monomers do not satisfy the usual octet rule.

The elements of group-13 form this type of hydride, which gains stability through the formation of dimers (Example., B₂H₆, Ga₂H₆, etc.) and polymers [Example (AlH₃)w, etc.]. Due to a deficiency of electrons, these hydrides act as Lewis acids and form complex entities with Lewis bases such as \(\ddot{\mathrm{N}}_3\) H^– ion, etc.

Hydrogen Chapter Class 11 Short Notes

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{LiH} \rightarrow 2 \mathrm{Li}^{+}\left[\mathrm{BH}_4\right]^{-} \text {[Lithium borohydride] }\)

2. Electron-precise hydrides: These hydrides have an exact number of electrons in the valence shell of the central atom so as to write their conventional Lewis structures. Examples are hydrides of the elements of the group- 14 such as CH₄, SiH₄, GeH_4, etc. They are tetrahedral in structure.

3. Electron-rich hydrides: These hydrides have more electrons in the valence shell of the central atom so as to write their conventional Lewis structures. The excess electrons are present in the form of lone pairs. Examples are hydrides of elements of group-15 to 17 such as \(\ddot{\mathrm{N}_H}{ }_3, \mathrm{H}_2 \ddot{\mathrm{O}}:, \mathrm{H} \ddot{\mathrm{i}} \mathrm{i}:, \mathrm{H}_2 \ddot{\mathrm{s}}:\) etc. Due to the presence of lone pairs of electrons, these hydrides act as Lewis bases and form complex entities with Lewis acids such as BF₃, B₂H₆, Ga₂ H,_6, etc.

⇒ \(\mathrm{BF}_3+\ddot{\mathrm{NH}}_3 \rightarrow\left[\mathrm{BF}_3 \leftarrow: \mathrm{NH}_3\right]\)

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \ddot{\mathrm{NH}}_3 \rightarrow\left[\mathrm{BH}_2\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4^{-}\right]\)

Nomenclature : The systematic names of the molecular hydrides are usually derived from the name of the element by attaching the suffix ‘ane! For example chlorane for HCl, phosphane for PH₃, oxidane for H₂O, azane for NH₃, sulphane for H₂S, etc. However, common names like hydrogen chloride, phosphine, water, ammonia, and hydrogen sulfide are more commonly used.

MPBSE Class 11 Chemistry  Nomenclature  Preparation:

1. These hydrides can be obtained by direct combination of the elements at high temperatures.

⇒ \(\mathrm{N}_2+3 \mathrm{H}_2 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NH}_3 ; \mathrm{O}_2+2 \mathrm{H}_2 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{H}_2 \mathrm{O}\)

2. These may be obtained by the hydrolysis of compounds such as borides, silicides, phosphides, sulfides, carbides, etc. For example:

⇒ \(\mathrm{Mg}_3 \mathrm{~B}_2+6 \mathrm{HCl} \rightarrow 3 \mathrm{MgCl}_2+\mathrm{B}_2 \mathrm{H}_6\)

⇒ \(\mathrm{Mg}_2 \mathrm{Si}+4 \mathrm{HCl} \rightarrow 2\mathrm{MgCl}_2+\mathrm{SiH}_4\)

⇒ \(\mathrm{Al}_4 \mathrm{C}_3+12 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{Al}(\mathrm{OH})_3+3 \mathrm{CH}_4\)

⇒ \(\mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{NH}_3\)

Hydrogen Chapter Class 11 Short Notes

⇒ \(\mathrm{FeS}+2 \mathrm{HCl} \rightarrow \mathrm{FeCl}_2+\mathrm{H}_2 \mathrm{~S}\)

⇒ \(\mathrm{Ca}_3 \mathrm{P}_2+6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{PH}_3\)

3. These may also be prepared by the reduction of the appropriate anhydrous chloride by LiAlH₄. For example:

⇒ \(4 \mathrm{BCl}_3+3 \mathrm{LiAlH}_4 \rightarrow 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{LiCl}+3 \mathrm{AlCl}_3\)

⇒ \(\mathrm{MCl}_4+\mathrm{LiAlH}_4 \rightarrow \mathrm{LiCl}+\mathrm{AlCl}_3+\mathrm{MH}_4[\mathrm{M}=\mathrm{Si}, \mathrm{Ge}, \mathrm{Sn}]\)

MPBSE Class 11 Chemistry  Nomenclature  Physical Properties:

1. These hydrides consist of discrete covalent molecules held together by relatively weaker van der Waals forces and dipole-dipole interaction and in some cases by hydrogen bonding. Hence, these axe gases, liquids, or solids of low melting and boiling points.
2. The hydrides of the first element of groups 15, 16, and 17 (i.e.„ NH₃, H₂O, and HF ) have abnormally high boiling points as compared to the hydrides of other elements of each group. The highly electronegative central atoms of these compounds strongly polarise the covalent bond with H and as a result, they remain associated through intermolecular hydrogen bonding. Due to such associations, they have abnormally high boiling points.
3. They are bad conductors of electricity.
4. Being covalent in nature, they are much soluble in organic solvents.

MPBSE Class 11 Chemistry  Nomenclature  Chemical Properties

1. Since the electronegativity of the central atoms increases from left to right along a period of the periodic table, covalent hydrides become progressively more acidic. For example, NH₃ (basic), H₂O (amphoteric), and HF (acidic).

2. The electron-rich hydrides behave as Lewis bases (electron donors), while the electron-deficient hydrides behave as Lewis acids (electron acceptors). They react with each other to form complex compounds.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \ddot{\mathrm{NH}}_3 \longrightarrow 2\left[\mathrm{BH}_3 \leftarrow: \mathrm{NH}_3\right]\)

Nascent Hydrogen Class 11 Notes

3. They undergo decomposition to their respective elements when heated.

⇒ \(\mathrm{SiH}_4 \stackrel{>300^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Si}+2 \mathrm{H}_2\)

The thermal stability in a group decreases as the electronegativity decreases with an increase in the size of the central atom down the group. For example, the thermal stability of group-15 hydrides follows the order:

NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

MPBSE Class 11 Chemistry  Metallic Or Interstitial Hydrides

• These hydrides are binary compounds of hydrogen and transition elements and are generally formed when d -d-block elements of group-3, 4, 5 (Sc, Ti, V, Zr, Nb, La, Hf, Ta, Ac, etc.), 6 (only Cr), 10, 11, 12, (Pb, Cu, Zn, etc.) and /-block elements (Ce, Eu, Yb, Th, U, etc.) are subjected to react with dihydrogen at elevated temperatures.
• The metals of group-7, 8, and 9 do not form hydrides, and the region of the periodic table from group- 7 to 9 is called hydride gap. These hydrides exhibit properties similar to those of the parent metal and hence are called metallic hydrides.
• In these hydrides, hydrogen atoms being small in size occupy the interstitial space of the metal lattice producing distortion without any change of its type. That’s why, these hydrides are called interstitial hydrides. This distortion of the lattice makes the hydrides brittle.
• However, it has been known from recent studies that except the hydrides of Ni, Pb, Ce, and Ac, other hydrides of this class have lattices different from those of the parent metal. These hydrides may also be regarded either as alloys or as solid solutions of hydrogen in metal.
• The composition of hydrides of some lanthanides and actinides belonging to f-block may not correspond to a simple whole number ratio and therefore, they are called non-stoichiometric hydrides. Their composition has been found to vary with temperature and pressure. LaH_2.87’ YbH_2.55’ ZrH_1.3-1.75’ VH_0.56’  NiH_0.6-0.7’ PdH_0.6-0.8 etc.

MPBSE Class 11 Chemistry  Metallic Or Interstitial Hydrides Properties

1. These hydrides are hard and conduct heat and electricity.
2. They have a metallic luster and magnetic properties.
3. They generally undergo reversible decor position into dihydrogen and metal and therefore, they are strong-reducing agents.

Occlusion of dihydrogen:

• Some transition metals can adsorb large volumes of dihydrogen on their surface due to the formation of interstitial hydrides. For example, when red-hot Pd is cooled in the atmosphere of dihydrogen, it adsorbs 935 times its own volume of dihydrogen.
• This property of adsorption of a gas by a metal is known as occlusion. The amount of dihydrogen occluded depends upon the nature and the physical state of the metal involved.
• For example, the occlusion power decreases in the order of colloidal Pd > Pt > Au >Ni. Due to the occlusion of H₂, the metal lattice becomes expanded and unstable. For this reason, the occluded dihydrogen is liberated on strong heating.

Chemistry Nascent Hydrogen Notes

MPBSE Class 11 Chemistry  Occlusion of dihydrogen Uses

1. Metal hydrides formed as a result of the occlusion of H₂ gas can be used as a hydrogen storage media (i.e., as a source of energy).
2. Occlusion can be used to remove impurities like N₂ present in H₂.
3. Occlusion can also be used to separate H₂ from He.
4. Metals like Ni, Pd, Pt, etc. which can adsorb large volumes of dihydrogen are widely used in catalytic hydrogenation (reduction) reactions for preparing a large number of compounds.

1. 1. Besides the three categories of hydrides mentioned above, elements having electronegativity between 1.4 to 2.0 form some polymeric hydrides in which the monomer molecules are held together in two or three dimensions by hydrogen bridges. Some common examples are (BeH₂)_n< (AlH₃)_n, (InH₃)_n, (SiH₄)_n, etc.
   2. Some complex hydrides are also formed. In these hydrides, the hydride ion H^– becomes involved in forming a coordinate bond with the central metal atom. B, A1, Ga, etc., of group IIIA form this type of hydride. Some common examples are lithium aluminum hydride (LiAlH₄), sodium borohydride (NaBH₄), etc. These are extensively used in organic and inorganic chemistry as a versatile reducing agent.