Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight: Overview, Questions, Preparation

65.  (i) - (c); (ii)- (d); (iii)- (b); (iv)- (e); (v)- (a)

1. CuCO₃ CuO  +  CO₂

  (D) 

Ca (OH)₂ +  CO₂ → CaCO₃ + H₂O

  (E) 

CaCO₃ +  CO₂ + H₂O → Ca (HCO₃)₂

                                        clear sol. 

CuO   +  CuS 3Cu  +  SO2 

  (A)  

 Cu + 4HNO3 → Cu (NO3)2 + 2NO2 +  2H2O

           (conc.)           (B) 

Cu (NO₃)₂ +  4NH₃ →  [Cu (NH₃)₄] (NO₃)₂

                                    (C) blue solution 

2.  (A)  =  FeCr₂O₄

(B)  =  Na₂CrO₄

(C)  =  Na₂Cr₂O₇

(D)  = K₂Cr₂O₇ 

4FeCrO₄ +  8NaCO₃ +  7O₂ → 8Na₂CrO₄ +  2Fe₂O₃ +  8CO₂

  (A)                                                     (B)

2NaCrO₄ +  2H +  → Na₂Cr₂O₇ +  2Na +  + H₂O

Na₂Cr₂O₇ +  KCl→ K₂Cr₂O₇ +  2NaCl 

  (C)                            (D)  

3. (A)  =  MnO₂

(B)  = K₂MnO₄

(C)  =  KMnO₄ 

(D)  =  KIO₃ 

2MnO₂ +  4KOH  + O₂ → 2K₂MnO₄ +  2H₂O

  (A)                                              (B)

3MnO₄^2-  +  4H⁺  → 2MnO₄^-  +  MnO₂ +  2H₂O

  (C)

2MnO₄^2-  +  2H₂O  +  KI→ 2MnO₂ +  2SO₄^-  +  KIO₃

  (A)                         (D)  

4. (a) As per fajan's rule, smaller the size of ion, greater is its tendency to make covalent bonds as the positively charged cation strongly attract the negatively charged electron cloud.

On moving from La to Lu in the lanthanoid series, the atomic size decreases so the covalent character increases. Hence, La₂O₃ is ionic and Lu₂O₃ is covalent.

(b) The stability of Oxo salts is directly proportional to the size of atom, as we move from La to Lu, the size of the atom decreases and hence the stability of oxo salts also decreases.

(c) As we move along the lanthanide series, the atomic size decreases. As a result, the charge/size ratio increases and the stability of the complexes also increases.

(d) As an effect of lanthanide contraction, zirconium and hafnium have similar radius of 160pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. Lanthanoid contraction: Because the elements in Row 3 have 4f electrons.

These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii.

(e) Acidic or basic character of oxides is related to the nature of oxides. As we move along the lanthanide series from La to Lu, the covalent character increases. 

Therefore, their basic character decreases or acidic character increases. 

5. (A) (i) Out of all the elements of the first transition series copper has the highest second ionisation enthalpy.

Electronic configuration of Copper is: 3d¹⁰4s¹

After the Loss of first electron from the 4s copper acquires 3d¹⁰ configuration which is stable. Therefore, removal of second electron from the field 3-D orbital is very difficult and requires high amount of energy.

(ii) Among the elements of first transition series zinc has the highest third ionisation enthalpy. Electronic configuration of zinc is: 3d¹⁰4s²

After the loss of two electrons from 4s orbital, Z and +2 Ion acquires 3d¹⁰ fully filled configuration which is highly stable therefore removal of third electron from 3d 10orbital will be more difficult and requires a large amount of energy.

(iii) Zinc is the element of the first transition series that has the lowest enthalpy of atomisation. The electronic configuration of zinc is: 3d¹⁰4s².This is because it has filled 3d-subshell and filled 4s-subshell so no unpaid electron is available for metallic bonding.

(B) (i) CarbonylmM (CO)₅

Ans: Fe (CO)₅ as per EAN Rule. 

EAN Rule: Effective atomic number (EAN), number that represents the total number of electrons surrounding the nucleus of a metal atom in a metal complex. 

(ii) MO₃F

Ans: The M in the given complex is in +7 O.S.

6. Interstitial compounds are formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals.

Characteristics:

They are hard and rigid in nature having high melting points. 

Like pure metals only, they are conductors. They can gain chemical inertness. 

7. (A) 2I^-  + S₂O₈^2-  →Fe (III) I₂ +  2SO₄^2- 

 role of fe ions:

2Fe³⁺  +  2I^- → 2Fe²⁺  + I₂

2Fe²⁺  + S₂O₈^2- → 2Fe³⁺  +  2SO₄^2-  

(B) 1. Fine powdered state of Nickel is used in the hydrogenation of oils into fats. 

2. Iron is used in the formation of ammonia in Haber’s process. 

3. In contact process in manufacture of H₂SO₄, vanadium is used in its oxide form- V₂O₅.

8. KMnO₄ K₂MnO₄ +    MnO₂ + O₂

  (A)                           (B)          (C)

MnO₂ +  KOH → 2K₂MnO₄ +  2H₂O

MnO₂ +  4NaCl  +  4H₂SO₄→ MnCl₂ +  2NaHSO₄+2H₂O  +  Cl₂

  (D) 

(A)  =  KMnO₄

(B)  = K₂MnO₄

(C)  =  MnO₂

(D)  =  MnCl₂ 

9. Copper lies below hydrogen in electrochemical series which means it has positive standard reduction potential i.e.   + 0.34 V. and hence cannot liberate hydrogen from acids.

10. Mn has half-filled 3d⁵ electrons and Zn has fully-filled 3d¹⁰ electrons which give them extra stability and they both resist to lose electrons and get reduced. 

The E? value also depends on the hydration enthalpy. More negative is the enthalpy of hydration, high is the E? value. This is the reason behind Mn, Ni and Zn are having higher E? value than expected

11. Cr (24)= [Ar]3d⁵4s²

Zn  (30) =  [Ar] 3d¹⁰4s² 

The stability of orbitals according to the electrons filled is in order: 

Fully filled > half-filled > partially filled 

As 3d¹⁰ is completely filled orbital of Zn  is more stable than the 3d⁵ half-filled orbital of Cr . It is comparatively easy to remove electron from 3d⁵ than from 3d¹⁰. 

Therefore, the first ionization enthalpy of Cr is lower than that of Zn. 

12. Transition elements have the ability to involve their d-electrons along with their s-electrons in the process of bond formation. This results in formation of strong metal-metal bonds which makes their melting point high. 

13. 2Cu²⁺  +    4I^-  → Cu₂I₂ + I₂

When Cu²⁺  reacts with potassium iodide white precipitate of Cu₂I₂ is formed.

14. Higher enthalpy of hydration provides extra stability to Cu²⁺  oxidation state than  +1 oxidation state. 

Enthalpy of hydration is described as the amount of energy released on dilution of one mole of gaseous ions.

15. The reactions are explained as:

MnO₂ +  4HCl → MnCl₂ +  Cl₂ +  2H₂O

  (A)                                       (B)

3Cl₂ +     2NH₃ → NCl₃ +  3HCl

(excess)                          (C) 

16. Fluorine - 1s²2s²2p⁵

Oxygen  -  1s² 2s² 2p⁶ 3s² 3p⁶ 

As we can infer from the electronic configuration, fluorine only has one unpaired electron in its 2p-orbital which can be involved in formation of only one bond and fluorine lacks d-orbital. 

While oxygen has d-orbitals in its configuration which allow the oxygen to form multiple bonds. Therefore, oxygen has greater stability to stabilise its oxidation states than fluorine. 

17. Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.

Cr³⁺  =  [Ar] 3d³ =  3 unpaired electrons

Co²⁺  =  [Ar] 3d⁷ =  3 unpaired electrons 

But as both the ions have the same no. of unpaired electrons, there’s one other factor called orbital contribution which also influences the magnetic moment. As, the three unpaired electronic configuration is symmetrical in Cr³⁺  this contributes zero orbital contribution but this is quite appreciable in case of Co²⁺  and hence it has larger value of Magnetic moment than Cr³⁺ 

18. Ce, Pr and Nd belong to the lanthanide series whereas Th, Pa and U belong to the actinides family. 

When electrons start accommodating the 4f and 5f orbitals, the 5f electrons penetrate less into the inner core. They are more effectively shielded nuclei in comparison to 4f-electrons in lanthanides. This leads to the fact that 5f-electrons experience reduced nuclear force of attraction and hence they have Lower ionisation enthalpies than lanthanoids. 

19. As an effect of lanthanoid contraction, zirconium and hafnium have similar radius of 160 pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii. 

20. Ce  =  4f¹ 5d¹ 6s²

Ce^{3 +}  =  4f¹ 5d⁰ 6s⁰ 

As in Ce³⁺ ,

4f has only single electron. Cerium holds the capacity to lose this electron also and attain 4f⁰ configuration which is more stable. 

Ce⁴⁺  =  4f⁰ 5d⁰ 6s⁰

Hence it shows 4⁺ oxidation states.

21. 2MnO₄^-  +  16H⁺ +  5C₂O₄^2-  → 2Mn₂ +  +  8H₂O  +  10CO₂

KMnO₄ oxidises the oxalic acid to CO₂ and reduces itself to Mn²⁺  state. Mn²⁺ is colourless that’s why it seems that KMnO₄ has been disappeared. 

22. This is due to the inter conversion of dichromate (orange) to chromate ion (yellow). 

Cr₂O₇^2-   CrO₄^2-  

(orange)     (yellow) 

23. KMnO₄ act as an oxidising agent however it’s activity as oxidising agent is influenced by the pH of the solution i.e. acidic, basic or neutral solution. 

K₂Cr₂O₇ + 2KOH → 2K₂CrO₄ + H₂O

(Orange)                      (yellow)

 2K₂CrO₄ + H₂SO₄ → K₂Cr₂O₇ + K₂SO₄ + H₂O

(yellow)                                       (orange) 

24. As an effect of lanthanide contraction, the second and third rows of transition elements resemble each other. For example, zirconium and hafnium have similar radius i.e. 160pm and 159pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii.

25. The sum of sublimation energy and ionisation enthalpy to oxidise cu (s) to Cu²⁺ is so highly that it is not compensated by the hydration enthalpy of Cu. Due to this, the Eof Cu is positive. 

While in case if Zn, the E value is negative or more negative than the expected value because when the electrons are removed from the 4s-orbital. Zn acquires a stable 3d¹⁰ configuration state.

26. As the positive charge of the ion increases or we can say that oxidation state of a transition element increases, its size decreases and as per Fajan's rule, more the charge on the metal ion, more is its tendency to form covalent compounds because positively charged cation attracts the electron cloud strongly towards itself. 

27. As per (n + l) rule, 4s has lower energy than 3d-orbital. 

3d−n+l = 3+2 = 5

4s  -  n  +  l  =  4 + 0  =  4 

So, 4s are filled first.

After filling of electrons, 4s-orbital moves beyond 3d-orbital and 4s electrons are loosely held by the nucleus. Hence, electrons are removed first during the process of ionisation.

28. Ionisation enthalpies are the main factor that influence the reactivity of transition elements. Higher the ionisation enthalpy, lesser is the reacting of the transition element. 

When we move along the period from Sc to Cu, a regular increase in the ionisation enthalpy is observed which results in the almost regular decrease in the reactivity of elements.

29.  (B) 26

Positive oxidation states indicate the loss of electrons from the atom. If X is in +3 oxidation state, then three electrons have been removed from it. Find the atomic number of the parent atom, we will add three in the given electronic number. i.e.

X³⁺  (Z = 23)  =  [Ar] 3d⁵

X  =  23 + 3  = 26  

30. (A) Cu (II) is more stable .

Electronic configuration of Cuis [Ar] 3d¹⁰ 4s¹

Cu (I) - [Ar] 3d¹⁰ 4s0

Cu (II)- [Ar] 3d⁹ 4s0

Despite the fact that Cu (I)  has fully filled 3d-orbital but Cu (II) is more stable than Cu (I) due to the greater effective nuclear charge of Cu (II) as nucleus has to hold 17 electrons rather than 18 electrons like in Cu (I).

31.  (D) Cu

Density is mass by volume. As we move from left to right for a long period, the atomic radii decrease. Hence, volume decreases. Also, an increase in atomic masses is observed. 

So overall the density increases out of the above option from iron to copper, copper will have the highest density.

32.  (B) CuF₂

(A) Ag₂SO₄  (Ag⁺ )→ 5d¹⁰6s⁰

(B) CuF₂  (Cu²⁺ )→ 3d⁹4s⁰

(C) ZnF₂  (Zn²⁺)→ 3d¹⁰4s⁰

(D) Cu₂Cl₂  (Cu⁺)→ 3d¹⁰6s⁰

Unpaired electrons present in any compound impart colour to the salt of transition metal. Only CuF₂ has unpaired electrons in its 3d orbital that’s why it is white coloured in its solid-state while rest of the salts are colourless.

33.  (A) Mn₂O₇

2KMnO₄ +  2H₂SO₄ (conc.) → Mn₂O₇ +    2KHSO₄ + H₂O

Manganese heptoxide is an acid anhydride of permanganic acid ( HMnO₄). It is a dangerous oxidiser, volatile and highly reactive liquid.

34. (B) 3d⁵

Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.

3d⁷ =  3 unpaired electrons 

3d⁵ =  5 unpaired electrons

3d⁸ =  2 unpaired electrons 

3d² =   2 unpaired electrons 

Out of all, 3d⁵ has five unpaired electrons that is maximum and hence it has the highest magnetic moment.

35. (B)+3

Out of all +3 is the most common oxidation state for all the lanthanide elements.

36.  (A)i, ii

Disproportionation/redox reaction is a reaction in which one compound is oxidised as well as reduced at the same time.

(A) Cu⁺  ?  Cu²⁺  +  Cu 

  ( + 1)       ( + 2)        (0) 

(B) 3MnO₄^- +  4H⁺  ?  2MnO₄^- +  MnO₂ +  2H₂O 

  ( + 6)                          ( + 7)           ( + 4) 

37. (D) Mn²⁺ acts as auto catalyst

Mn²⁺  formed in the reaction acts as an autocatalyst. Auto catalyst is a compound that is one of the products that are formed in the reaction itself catalyses the reaction.

Reaction: 2MnO₄^-  +  16H⁺  +  5C₂O₄^2- → 2Mn₂ +  +  10CO₂ +  8H₂O

38. (C) Tm

In the periodic table, actinides lie from atomic number 90 to 103. 

Thulium is not an actinide; it is a lanthanide. 

39.  (A) 25

2MnO₄^-  +  5S^2-  + 16H⁺  → 2Mn²⁺  +  5S  + H₂O

From the reaction,

For 5 moles of S, two moles of KMnO₄ were required.

Therefore, for one  mole of S, the number of moles of KMnO₄ required will be = 25.

40.  (A) V₂O₅, Cr₂O₃

Oxides in the lower oxidation state are basic and in the higher oxidation state they are acidic. They are amphoteric in the intermediate oxidation state. Therefore, V₂O₅, Cr₂O₃ are amphoteric in nature.

41.  (A) [Xe] 4f⁷ 5d¹ 6s²

Gadolinium belongs to the 4f series; it has atomic no.= 64. It has extra stability due to a half-filled 4f subshell.

Gd  (Z=64) ?   [Xe] 4f⁶ 5d² 6s² 

42.  (D) They are chemically very reactive.

Interstitial compounds/alloys are substances that are formed when a small atom like carbon, hydrogen, boron, nitrogen can occupy space in their lattices. All the properties mentioned above are true for interstitial compounds except (D), they are chemically inert.

43.  (B) 3.87 B.M. 

Cr →  (Z = 24)

Cr3 +  →  (Z = 21)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s⁰

n  =  3 ( 3 unpaired electrons )

μ = $\sqrt[]{n(n+2)}$

μ = $\sqrt[]{3(3+2)}$

μ = $\sqrt[]{3\left(5\right)}$

μ = $\sqrt[]{15}$

=  3.87 B.M. 

44.  (C) IO₃^- 

Iodide is oxidised to iodate.  

2KMnO₄ +  KI  + H₂O → 2KOH  +  2MnO₂ +  KIO₃

               iodide                                               iodate 

45.  (A) is incorrect.

Copper lies below hydrogen in electrochemical series and hence cannot liberate hydrogen from acids. 

Cu  +  2H₂SO₄ → CuSO₄ +  SO₂ +  2H₂O

 46.  (C) Sn⁴⁺

Cr₂O₇^2-  +  14H⁺  +  3Sn²⁺  → 2Cr₃ +  +  3Sn⁴⁺  +  7H₂O

In the above redox reaction, oxidation of Sn²⁺  is taking place. It gets converted to Sn⁴⁺  whereas, reduction of chromium is occurring from +6 oxidation state to +3. 

47.  (D) in covalent compounds fluorine can form single bond only while oxygen forms double bond

Oxygen has the capacity to form multiple bonds which enables it to form a variety of covalent compounds. 

In (Mn₂O₇) also, 6 oxygen are doubly bonded to two manganese atoms and one oxygen is forming bridge between two. 

While in (MnF₄), four fluorine atoms are singly bonded to manganese atom giving it a +4 oxidation state.

Therefore, due to capability of oxygen to have multiple bonds in covalent compounds, manganese is having higher oxidation state of +7 in (Mn₂O₇) . 

48.  (C)  both have similar atomic radius

As an effect of lanthanide contraction, zirconium and hafnium have similar radii of 160 pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

Lanthanide contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield well, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and results in the decrease in their size and atomic radii. 

49.  (B) KMnO₄ oxidises HCl into Cl₂  which is also an oxidizing agent.

HCl is not used in titrations involving KMnO₄  as it produces unsatisfactory results. When HCl  is added to KMnO₄, it itself acts as reducing agent and coverts HCl to Cl₂. 

50.  (A) & (B)

Due to charge transfer phenomenon, which can be understood, for an instant, an electron is transferred from oxygen to the central metal atom, where oxidation state of metal is reduced by one and oxygen gets converted from O^- to O^2- . 

This transfer imparts Colour to the compound. 

51.  (A) & (D) 

The magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of the magnetic moment.

Co²⁺  and Cr³⁺ have same unpaired electrons i.e. 3 unpaired electrons. Hence they have the same spin only magnetic moment. 

52.  (B) and (C) 

For the heavier metals of d-block, higher oxidation state is more favourably stable than the lower oxidation state. For example: Mo (VI) and W (VI) are more stable than Cr (VI). For lighter d-block elements like chromium, +3 oxidation state is more stable. So, Cr (VI) in dichromate act as a strong oxidising agent by reducing itself from +6 to +3 oxidation states While Mo (VI) and W (VI) does not. 

53.  (B) & (D) 

Actinides show a variety of oxidation states like +3, +5, +6 and +7. 

From the given option, plutonium and neptunium show all oxidation states up to +7

54.  (A) and (B)

Electronic configuration:

92U -  5f³ 6d¹ 7s²

93Np -  5f⁴ 6d¹ 7s² 

55.  (B) and (C) 

Ce (Z = 57)  =  [Xe] 4f⁵ 5d⁰ 6s² , O.S.  =   + 3, + 4

Eu (Z = 63)  =  [Xe] 4f⁷ 5d⁰ 6s², O.S.  =   + 2, + 3

Yb (Z = 70)  =  [Xe] 4f¹⁴ 5d⁰ 6s², O.S.  =   + 2, + 3

Ho (Z = 67)  =  [Xe] 4f¹¹ 5d⁰ 6s², O.S.  =   + 3 

56.  (B) and (C) 

Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.

Electronic configuration:

Ti³⁺  =  3d¹ =  1 unpaired electron

Mn²⁺  =  3d⁵ =  5 unpaired electrons

Fe²⁺  =  3d⁶ =  4 unpaired electrons

Co²⁺  =  3d⁷ =  3 unpaired electrons 

Out of all given options, Mn²⁺ and Fe²⁺  have unpaired electrons in number of five and four respectively. Hence, have higher spin only magnetic moment. 

57.  (A) and (B) 

Cr and Co  form tri halide compounds.

58.  (B) and (C) 

A compound act as oxidising agent when it's central atom is reduced to its lower oxidation state. This occurs only when the lower oxidation state of metal is more stable than the higher oxidation states. 

In metal WO₃ and CrO₄^2- , both W and Cr are stable in their higher oxidation state and won't get reduced to their lower oxidation state. Therefore, will not act as oxidising agents. 

59.  (B) and (C) 

Electronic configuration of cerium:

Ce  -  4f¹ 5d¹ 6s²

Ce⁴⁺?  4f⁰ 5d⁰ 6s⁰ 

Lead has the tendency to lose its four outer electrons to attain Noble gas configuration and this removal of electrons corresponds to the f? configuration and cause it to be stable in +4 oxidation state

60. All halogens combine with copper to form copper halides except iodine. The reason behind this is that Cu²⁺  oxidises iodide (-1)  to iodine (0). Therefore, Cu²⁺ iodide does not exist .Both the statements assertion and reason are correct and reason is the correct explanation of assertion. 

61. Although Zr and Hf lie in same group but As an effect of lanthanide contraction, zirconium and hafnium have similar radius of 160pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. As a result, Zr and Hf are difficult to separate. Both the statements assertion and reason are correct but reason is not the correct explanation of assertion. 

62. Actinoids are more reactive than lanthanoids as they involve their 5d-orbital in the bond formation process while due to effectively shielded and greater nuclear charge 4f-orbitals are not available for bonding. Thus, lanthanoids comparatively form less complexes than actinoids. 

Assertion is not true but reason is true.

63. Cu lies below hydrogen in electrochemical series and has positive electrode potential. hence cannot liberate hydrogen from acids. 

Both the statements assertion and reason are correct and reason is the correct explanation of assertion. 

64. Osmium has the capacity to expands its octet by utilising all the electrons from its 6s and 5d orbitals this results it to attain an oxidation state of +8.

Both the statements assertion and reason are correct but reason is not the correct explanation of assertion. The correct option is B.

66.  (i) - (b); (ii)- (a); (iii)- (d); (iv)- (e); (v)- (c)

67.  (i) - (c); (ii)- (a); (iii)- (b)

68.  (i) - (c); (ii)- (a); (iii)- (e); (iv)- (b)

69.  (i) - (d); (ii)- (a); (iii)- (b); (iv)- (e); (v)- (f)

70.  (i) - (b); (ii)- (d); (iii)- (a); (iv)- (e); (v)- (c)

71.  (i) - (c); (ii)- (d); (iii)- (b); (iv)- (e)

Structure PCl₅ is trigonal bipyramidal,

Hybridization of P is sp³d

Equatorial bonds lie in a plane

Axial bonds are longer than equatorial bonds so axial bonds are weaker than equatorial bonds.

$N{H}_{4}C\mathcal{l}\left(aq\right)+NaN{O}_2\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(\mathcal{l}\right)+NaC\mathcal{l}\left(aq\right)$  

Hence, N₂ gas is produced.

A. ${\left[Fe{F}_6\right]}^{3-}$           

Fe³⁺ - 3d⁵4s⁰

F^- is weak field ligand, so no pairing of electrons :

                      

Number of unpaired electrons, n = 5

B. ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$          

$F{e}^{3+}-3d^{5}4s^0$                              

CN^- is strong field ligand, so pairing of electrons takes place.

                            

Number of unpaired electrons, n = 1

C. ${\left[MnC{\mathcal{l}}_6\right]}^{3-}$            

$M{n}^{3+}-3d^{3}4s^0$  

                    

Number of unpaired electrons, n = 3

D. ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$           

$M{n}^{3+}-3d^{3}4s^0$                              

                            

Number of unpaired electrons, n = 3

Higher the number of unpaired electrons. Then higher be the magnetic moment as;

Magnetic moment, $\mu =\sqrt[]{n\left(n+2\right)}B.M$

Hence, order of magnetic moment is ;

B < D = C < A

Moles of chlorine in the given compound = Moles of chlorine in AgCl

= moles of AgCl

$n_{C\mathcal{l}}=\frac{0.4}{143.5}mo\mathcal{l}$                

Mass of chlorine = $\frac{0.4}{143.5}\times 35.5g$  

= 0.098 g

$\mathrm{\%}ofC\mathcal{l}=\frac{0.098}{0.25}\times 100=39.6\mathrm{\%}$        

For ideal gas using ;

PM = dRT

$P=\left(\frac{RT}{M}\right)d$

Comparing with y = mx + C

Slope, m =  $\frac{RT}{M}$

Slope $\propto$  T

So; higher the slope higher the T

Hence, T₃ > T₂ > T₁

Aromatic species are most stable, here

Leaching involves the given reaction,

$4Au\left(s\right)+8C{N}^{-}\left(aq\right)+2H_{2}O\left(aq\right)+O_2\left(g\right)\to 4{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+4O{H}^{-}\left(aq\right)$              

Here, O₂ is required for formation of Au (l) cyanide complex but no complex in absence of O₂.

$2{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+Zn\left(s\right)\to {\left[Zn{\left(CN\right)}_4\right]}^{2-}\left(aq\right)+2Au\left(s\right)$                

In above displacement reaction, Zn is oxidized.

Higher the E.N. difference between hydrogen and other atom then higher be the strength of intermolecular H-bond

Here, order of difference in E.N is

O - H > N – H > C - H

Hence, correct order of H bond strength is,

CH₄ < HCN < NH₃

For isoelectronic species

$Ionicradii\propto \frac{Numberofe}{Numberofp}$              

$\begin{array}{l}N{a}^{+}M{g}^{2+}{N}^{3-}{O}^{2-}{F}^{-}\\ e=1010101010\\ p=1112789\end{array}$                

Hence, correct order of ionic radii is;

$M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$                

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

In 3d-series, all metals except Cu have negative value of $E_{M^{2+}/M}^0$ due to low hydration enthalpy and high I.E and sublimation enthalpy.

Electronic configuration of the given ions is :

$E{u}^{2+}-4f^{7}6s^0$                 

$\begin{array}{l}T{m}^{2+}-4f^{13}6s^0\\ S{m}^{2+}-4f^{6}6s^0\\ T{m}^{3+}-4f^{12}6s^0\end{array}$

$\begin{array}{l}T{b}^{4+}-4f^{7}6s^0\\ Y{b}^{2+}-4f^{14}6s^0\\ D{y}^{3+}-4f^{9}6s^0\\ y{b}^{3+}-4f^{13}6s^0\end{array}$              

Hence, Eu²⁺ and Tb⁴⁺ have half | filled f-orbitals and Yb²⁺ have completely filled f-orbitals.

Chlorine nitrate as hydrolysis gives hypohalous acid and nitric acid as,

$C\mathcal{l}ON{O}_2+H_{2}O\to \underset{A}{HOC\mathcal{l}}+\underset{B}{HN{O}_3}$                

Chlorine nitrate on reaction with HCl produces Cl₂ and HNO₃ as,

$C\mathcal{l}ON{O}_2+HC\mathcal{l}\to \underset{C}{C{\mathcal{l}}_2}+\underset{B}{HN{O}_3}$            

Above are the six possible forms of diaminobenzoic acid.

On decarboxylation we get;

Buna-N is obtained by copolymerization of 1, 3-butadiene and acrylonitrile,

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \alpha -D-glucose+\alpha -D-glucose$

In maltose, two units of a-D-glucose are linked by glycosidic linkage at C₁ and C₄ as shown.

    

a - D – glucose                              a - D - glucose

Maltose has hemiacetal link so it is reducing sugar.

A. Antipyretic reduces fever.

B. Analgesic reduces pain.

C. Tranquilizer reduces stress.   

D. Antacid reduces acidity of stomach.

A. $\underset{Salt}{N{a}_{2}C{O}_3}+\underset{di\mathcal{l}}{H_{2}S{O}_4}\to NaS{O}_4+H_{2}O+C{O}_2\uparrow$            

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C{O}_2+Ca{\left(OH\right)}_2\to CaC{O}_3\downarrow +H_{2}O$                              

B. $\underset{Salt}{N{a}_{2}S}+\underset{di\mathcal{l}}{H_{2}S{O}_4}\to N{a}_{2}S{O}_4+H_{2}S\uparrow$            

H₂S + (CH₃COO)₂Pb → $\underset{Black}{PbS}+2C{H}_{3}COOH$  

C. $\underset{Salt}{N{a}_{2}S{O}_3}+\underset{di\mathcal{l}}{H_{2}S{O}_4}\to N{a}_{2}S{O}_4+H_{2}O+S{O}_3\uparrow$            

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S{O}_3+KMn{O}_4+H^{+}\to \underset{Green}{MnS{O}_4}+S{O}_4^{2-}+H_{2}O$                             

D. $\underset{Salt}{2NaN{O}_2}+\underset{di\mathcal{l}}{H_{2}S{O}_4}\to N{a}_{2}S{O}_4+2HN{O}_2$          

$3HN{O}_2\to NH{O}_3+2NO+H_{2}O$                              

$2NO+O_2\to \underset{Browngas}{2N{O}_2}$                              

$2HN{O}_2+2Kl+2C{H}_{3}COOH\to 2C{H}_{3}COOK+2H_{2}O+2NO+I_2$                             

I₂ + starch → Blue colour complex.

Determining empirical formula of the given compound.

$\mathrm{\%}ofC=\frac{48.5}{116}\times 100\mathrm{\%}=41.8\mathrm{\%}$               

$\mathrm{\%}ofH=\frac{7.5}{116}\times 100\mathrm{\%}=6.5\mathrm{\%}$               

$\mathrm{\%}ofO=\frac{60}{116}\times 100\mathrm{\%}=51.7\mathrm{\%}$               

              Mass         Moles                  Simplest ratio

C            41.8g        $\frac{41.8}{12}=3.48$         $\frac{3.48}{3.23}\approx 1$                 

H            6.5 g        $\frac{6.5}{1}=6.5$             $\frac{6.5}{3.23}\approx 2$                      

O           51.7 g        $\frac{51.7}{16}=3.23$         $\frac{3.23}{3.23}=1$                  

Ratio of moles of C, H and O is 1 : 2 : 1

A. CH₃COOH ⇒ C : H : O = 2 : 4 : 2 = 1 : 2 : 1

B. HCHO ⇒ C : H : O = 1 : 2 : 1

C. CH₃OOCH₃ ⇒ C : H : O = 2 : 6 : 2 = 1 : 3 : 1

D. CH₃CHO ⇒ C : H : O = 2 : 4 : 1

So, CH₃COOH and HCHO formulae agree with the given data

n = 1, 2, 3 …….;   $\mathcal{l}=0$ ……. to n – 1, $m_{\mathcal{l}}=\mathcal{l}.......0......+\mathcal{l}$  

A. n = 3   $\mathcal{l}=3$ $m_{\mathcal{l}}=-3$  is incorrect as $\mathcal{l}$  can not be equal to n.

B. n = 3   $\mathcal{l}=2$   $m_{\mathcal{l}}$ = -2 is correct set.

C. n = 2   $\mathcal{l}=1$ $m_{\mathcal{l}}$ = +2 is incorrect set as  $\mathcal{l}$ can not be equal to n.

So; correct set of quantum numbers is B and C.

  $BeO+2N{H}_3+4HF\to {\left(N{H}_4\right)}_2\left[Be{F}_4\right]$  

${\left(N{H}_4\right)}_2\left[Be{F}_4\right]\underset{}{\overset{\Delta }{\to }}Be{F}_2+2N{H}_{4}F$                

here, oxidation state of be in A is +2.

Number of moles, n = 5 mol

Temperature, T= 300 K

Initial volume, V₁ = 10L

Final volume, V₂ = 20 L

Using;

Work done; w = -2.303 nRT log₁₀   $\frac{V_2}{V_1}$

$=-2.303\times 5\times 8\times 300lo{g}_{10}\frac{20}{10}J$                

= -8630 J

So, magnitude of work done is 8630 J.

Mass of solute ; w_B = 2.5 × 10^-3 kg

Mass of solvent, $w_A=75\times 10^{-3}kg$  

Boiling point of solution ; $T_b=373.535K$  

Boiling point of water ;  $T_b^0=373.15K$

So, elevation in boiling point, $\Delta T_b=T_b-T_b^0$  

$\Delta T_b=0.385K$                                                           

Using ; $\Delta T_b=K_b\times \frac{w_B}{M_B\times w_A}\times 1000$  

$0.385=0.52\times \frac{2.5\times 10^{-3}}{M_B\times 75\times 10^{-3}}\times 1000$               

Molar mass of solute ; M_B = 45 g/mol

0.001 M NaOH solution has [OH^-] = 0.001 M = 10^-3 M

Using ; pOH = $-lo{g}_{10}\left[O{H}^{-}\right]$  

=  $-lo{g}_{10}10^{-3}$

pOH = 3

pH = 14 – pOH

pH = 14 – 3

pH = 11

For the given cell, net redox reaction is as;

$H_2\left(g\right)+2A{g}^{+}\left(aq\right)\to 2H^{+}\left(aq\right)+2Ag\left(s\right)$               

n = 2

$E_{cell}^0=+0.5332V$               

Using;  $\Delta G^0=-nF{E}_{cell}^0$  

$=-2\times 96487\times 0.5332J/mo\mathcal{l}$               

$=-102.89kJ/mo\mathcal{l}$               

Now, for the reaction

$\frac{1}{2}{H}_2\left(g\right)+A{g}^{+}\left(aq\right)\text{?}\to H^{+}\left(aq\right)+Ag\left(s\right)$               

n = 1

So; $\Delta G^o=-1\times 96487\times 0.5332J/mo\mathcal{l}$  

=  $-51.44kJ/mo\mathcal{l}$

Initial temperature ; T₁ = 300 K

Final temperature ; T₂ = 309 K

Rate constant gets doubled i.e K₂ = 2K₁

$Usinglo{g}_{10}\frac{k_2}{k_1}=\frac{Ea}{2.3R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$log2=\frac{Ea}{2.3\times 8.3}\left[\frac{9}{300\times 309}\right]$

$E_a=\frac{2.3\times 8.3\times 300\times 309\times log2}{9}J/mo\mathcal{l}$

$E_a=58.98kJ/mol$    

Using, Freundlich adsorption isotherm ;

$\frac{x}{m}=k.p^{1/n}$       ………………. (i)

$log\frac{x}{m}=\frac{1}{n}logp+logk$                

Comparing with y = mx + C

Slope = $\frac{1}{n}$  = 1

Intercept, log k = 0.602

log k = log 4

k = 4

from equation (i)

  $\frac{x}{m}=4\times {\left(0.03\right)}^1$              

= 0.12

= 12 × 10^-2

So, 12 × 10^-2 g of gas is adsorbed per gram of adsorbent,

Lindlar's catalyst reduces alkynes to cis-alkene.

Lindlar's catalyst reduces alkynes to cis-alkene.

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x\times \left(+2\right)+\left(1-x\right)\times \left(+3\right)=$ O.S. of Fe atom

$or2x+3-3x=\frac{+200}{93}$

$\therefore \mathrm{\%}ofF{e}^{2+}=85\mathrm{\%}and\mathrm{\%}ofF{e}^{3+}=15\mathrm{\%}$

$\Delta V=2.4\times 10^{-26}m/sec$

And $\Delta x=10^{-7}m$  

According to Heisenberg’s uncertainty

$\Delta v\times \Delta x=\frac{h}{4\pi \times m}$  

$\therefore m=\frac{h}{4\pi \times \Delta v\times \Delta x}=\frac{6.626\times 10^{-34}}{4\times 3.14\times 2.4\times 10^{-26}\times 10^{-7}}$

$\therefore$  m = 0.02198 Kg

Or 21.98 gm

Or 22 gm in the nearest integer.

  $\Delta T_b=i\times K_b\times m$

 or $\Delta T_b=K_b\times m$   (i = 1 for non-electrolyte)

  $or\frac{{\left(\Delta T_b\right)}_A}{{\left(\Delta T_b\right)}_B}=\frac{{\left(K_b\right)}_A}{{\left(K_b\right)}_B}$

  $or\frac{x}{y}=\frac{1}{8}\left|\therefore y=8\right|$              

$2NOC{l}_{\left(g\right)}?2NO\left(g\right)+C{l}_2\left(g\right)$

t = 0       2 mol/Lit             0            0

t_eq          2 – x                    x            x/2

$K_C=\frac{{\left[NO\right]}^2\times \left[C{l}_2\right]}{{\left[NOCl\right]}^2}=\frac{x^2\times x}{{\left(2-x\right)}^2\times 2}=\frac{x^3}{{\left(2-x\right)}^2\times 2}$      

$=\frac{0.4\times 0.4\times 0.4}{1.6\times 1.6\times 2}$ = 0.0125 = 125 × 10^-4

${\lambda }_m^{\infty }$  of Nal = 12.7 mSm² mol^-1

${\lambda }_m^{\infty }$ of NaNO₃ = 12.0 m Sm² mol^-1