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

19. Option (i) Same in all directions is correct since quartz glass is an amorphous solid showing isotropic properties and hence exhibits the same values of refractive index when measured along different directions.

18. Option (iv) is correct since in antiferromagnetic substances the domains  are oppositely oriented and hence they cancel out each other's magnetic moments. 

21. Option (ii) a regular arrangement of constituent particles observed over a long distance in the crystal lattice is correct since the regularity of the crystalline lattice creates local environments that are the same and hence crystals exhibit sharp melting point.

23. Option (iii) Diamond  is correct since in diamond, the carbon atoms are held together by strong covalent bonds. It is a giant molecule. Thus, it is a solid network.

29. Option (iii) electron is correct since lattice sites can only be occupied by electrons when there is imperfection in a solid and do not occur in pure crystals. Hence, the lattice site in a pure crystal cannot be occupied by electrons.

49. Option (iv) 8 tetrahedral voids within the unit cells is correct since in ccp structure there are 4 atoms per unit cell and the number of tetrahedral voids is twice the number of atoms (i.e. eight tetrahedral voids per unit cell of cubic close packing.

50. Option (i) $\sqrt[2]{2r}$ , $\frac{4r}{\surd 3}$  ,2r

In FCC crystal - the atoms touch each other at the face diagonals. If a is the edge length of the cube and r is the radius of the atom 

Therefore, $\sqrt[]{2a}=4r$

a = $\frac{4r}{\surd 3}$

In a simple cubic crystal - The atoms at the corners touch each other.

Therefore, a = 2r

66. Options (i) and (ii) are correct since in this defect an atom is displaced from its lattice point to an interstitial site and thus creating a vacancy at the lattice point. This usually occurs when the size of anion is quite large as compared to that of the cation due to which it leads to the dislocation of the cation to the interstitial site and hence it is known as dislocation defect and during this defect there is no change in the stoichiometry of the crystal so, it is also known as stoichiometric defect.

67. Options (ii) and (iv) are correct since Vacancy and Schottky, both the defects lead to the decrease in the density while in the Frenkel and Interstitial defect there is no change in the density of the substance.

76. Option (ii) Assertion and reason both are correct statements, but reason is not correct explanation for assertion is the answer since in fcc structure the number of atoms present = 4 per unit cell which provides a maximum packing efficiency of 74%. 

77. Option (iii) Assertion is correct statement, but reason is wrong statement is the answer since intermediate conductivity in semiconductor is due to the small energy gap between valence band and conduction band and hence the assertion is correct, but the reason is wrong.

1. When any atom is surrounded by six atoms it creates an octahedral void. In fcc, body centre is surrounded by six atoms present at the face centre. And hence one octahedral void is present at the body centre of each unit cell.

No. of octahedral void at the centre of 12 edge = 12 * $\frac{1}{4}$  = 3

No. of octahedral void at body centre = 1 

Total no. of octahedral void in ccp = 3+1+4

Location of octahedral voids per unit cell of ccp or fcc lattice (a) at body centre of the cube and (b) at the centre of each edge (only one such void is shown)

2. A ccp structure unit cell is divided into 8 small cubes. Each small cube has atoms at alternate corners as shown in the figure. In all, each small cube has 4 atoms. When these are joined to one another, they make a regular tetrahedron. Thus, there is one tetrahedral void in each small cube and 8 tetrahedral voids in total. Each of the eight small cubes has one void in one unit cell of ccp structure. We know that ccp structure has four atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.

3. The conductivity of a semiconductor is very low and it can be increased by adding small impurities and this process is known as doping. 

Doping can be done with an impurity which is electron rich or electron deficient-

n-type semiconductors- Si or Ge (group-14 elements)  are doped with electron rich impurity (group-15 elements like P or As)  and are known as n-type semiconductors. In them the conductivity is due to the extra electron or delocalized electron. 

When intrinsic semiconductors like Si or Ge are doped with pentavalent elements like P or As, they occupy some of the lattice sites in silicon or germanium crystal . Four out of five electrons are used in formation of four covalent bonds with four neighbouring silicon atoms. The fifth electron is extra and gets delocalised. These delocalised electrons increase the conductivity of doped silicon ( or germanium).

p-type semiconductors- Si or Ge (group-14 elements) doped with electron deficient impurity (group 13 elements like B or Al or Ga) and are termed as  p-type semiconductors. Here the conductivity is due to the positively charged electron holes. 

In this case the doping of intrinsic semiconductors like silicon or germanium with trivalent elements like B / In / Ga, three out of four electrons in silicon or germanium form bonds with doping impurity ( i.e. B/ In / Ga ). The fourth electron remains unbonded. The place where the fourth valence electron is missing is known as electron hole or electron vacancy.

An electron from a neighbouring atom can come and fill up the electron hole, but in doing so it would leave an electron hole at its original position. If this happens it would appear that the electron hole has moved in a direction opposite to that of the electron that filled it. Under the influence of electric field electrons would move towards the positively charged plate through electron holes, but it would appear as if electrons were positively charged and are moving towards the negatively charged plate.

4. Let the formula of sample be (Fe²⁺)x (Fe³⁺) yO

From the formula of the compound the equation can be formed as-

x+ y= 0.93 . (i) 

Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen. Hence,  

2x+ 3y= 2 . (ii) 

⇒ x+ $\frac{3}{2}$  y=1. (iii)

On subtracting equation (i) from equation (iii) we have

$\frac{3}{2}$ y-y=1 - 0.93

12y=0.07

y= 0.14 

On putting the value of y in equation (i)

 we get, x+ 0.14 = 0.93 

⇒x = 0.93 – 0.14 

⇒x = 0.79 

Fraction of Fe²⁺ ions present in the sample is  $\frac{0.79}{0.93}$  = 0.81

Metal deficiency defect is found to be present in the sample as iron is less in amount than that required for stoichiometric composition.

5. The liquids and gases both are categorised as fluids as they show the ability to flow and also their molecules can move past one another freely. 

6. In solids the constituent particles are found to possess fixed positions and can only oscillate about their mean positions and hence they are said to be incompressible and rigid. 

7. Crystals have long range ordered arrangement of  their constituent particles but usually these crystals are not perfect as during the process of crystallisation some deviations as compared to such ideal arrangement set in depending upon the rate of cooling or presence of impurities in solution  also this process occurs at such a rate that the constituent particles may not get the sufficient time to arrange themselves in a perfect order  and hence these deviations or irregularities in arrangement is being termed as defects or imperfections. Therefore, crystals are usually not perfect. 

8. Yellow colour in NaCl is due to the defect called as metal excess defect. In this defect anionic vacancies get created due to the diffusion of Cl-ions to the surface of the crystal and there after unpaired electrons occupy anionic sites. These sites are known as F-centres. The electrons at F-centres then absorb energy from the visible region and undergo excitation which makes the crystal appear yellow.

9. In FeO crystal, some of the Fe²⁺ ions are replaced by Fe³⁺ ions i.e., 3Fe²⁺ ions are replaced by 2Fe³⁺ ions to make up for the loss of positive charge. As a result of which it leads to lesser amount of metal as compared to the stoichiometric proportion.

10. The ZnO crystal becomes yellow oh heating because of the metal excess defect  which is caused due to the presence of extra cations at the interstitial sites and on heating this white crystal it loses oxygen and turns yellow. The reaction involved is given as-

ZnO  Zn²⁺ + $\frac{1}{2}$ O₂ + 2^e-

Here the excess of Zn²⁺ ions move to the interstitial sites and electrons to neighbouring interstitial sites.

11. In semiconductors the gap between conduction band and valence band is small and hence some of the electrons from the valence band can easily jump to the conduction band and shows some conductivity but with rise in the temperature more of the electrons gets jump to the conduction band and thus, their electrical conductivity increases with rise in the temperature.

12. When tetravalent germanium is doped with trivalent gallium  then some of the positions of the lattice of germanium gets occupied by the gallium. Since the gallium atom has only three valence electrons, the fourth valency of the nearby germanium atom does not get satisfied and hence this place remains vacant. This place is deficient of electrons and is therefore called an electron hole or electron vacancy.

Now, the electron from the neighbouring atom comes and fills the gap and leads to the formation of a hole in its original position. Under the influence of electric fields, the electrons move towards the positively charged plates through these holes and lead to the conduction of electricity while the holes appear to move towards the negatively charged plates. 

13. In ccp lattice the number atoms per unit cell = 4 

The number of tetrahedral voids is given as = 2 (n) = 2 x 4 = 8

Only one-third of tetrahedral voids are occupied by metal M so, the ratio of atoms of element M to that of element N = 1/3 (8) : (4) = 2 : 3 

or M : N = 2 : 3 

Hence, the formula of the compound is M₂N₃.

14. On heating the amorphous substance it gets changed to the crystalline form at some temperature. This is due to the process called crystallization. As on heating at some temperature it may become crystalline since slow heating and cooling over a longer period of time makes these changes.

15. Option (ii) Low temperature is correct since at sufficiently low temperature, the thermal energy is low, so the intermolecular forces bring the molecules of a substance closer so that they cling to one another and occupy fixed positions. They keep on vibrating about their fixed positions. Such conditions favours the existence of the substance in solid state. 

16. Option (ii) Isotropic nature is correct since crystalline solids exhibit anisotropic properties like refractive index, electrical resistance etc. Since these are found to have different values when measured along different directions in the same crystal and hence they are not isotropic in nature. 

17. Option (ii) Quartz glass (SiO₂) is correct since quartz glass (SiO₂)is amorphous in nature as there is no long range ordered arrangement of the constituent particles being present in it and hence it is an amorphous solid.

20.  (iv) They are anisotropic in nature since amorphous solid shows isotropic properties as they exhibit same values of the properties like refractive index, electrical resistance when measured along different directions.

22. Option (i) London forces is correct since iodine molecules are nonpolar  and covalent in nature. These molecules are found to be electrically symmetrical and have no dipole moment. The molecules in a crystal lattice of iodine are thus attracted together by weak London forces.

24. Option (iii) (C) and (D) is correct since Iodine is a non-polar molecular solid which acts as an insulator while H₂O is a hydrogen bonded molecular solid which also acts as an insulator. Whereas Mg is a metal and acts as a conductor in solid and in its molten state and TiO is also an ionic solid which acts as an insulator in solid state but acts as a conductor in molten and in aqueous solutions. 

25. Option (i) Very low value of electrical conductivity in the molten state is correct since in molten state, the ionic solids possess free ions which help in the conduction of electricity. Due to the availability of these free ions the electrical conductivity of ionic compounds in molten state is found to be very high. 

26.  (ii) Free valence electrons 

27. Option (iii) TiO₃  is correct since certain metal oxides such as VO, VO₂, VO₃and TiO₃  show either metallic or insulating properties depending upon the temperature. 

28. Option (iv) CrO₂ is correct since it shows electrical conductivity similar to the metals. 

30. Option (iv) ionic solid is correct since in graphite each atom is linked to the other through a covalent bond and hence it is a covalent solid and cannot fall under the category of ionic solid. 

31. Option (i) Frenkel defect is correct since it is a type of defect in which an atom is dislocated from its lattice position to an interstitial site. 

32. Option (ii) equal number of cations and anions are missing from the lattice is correct since Schottky defect in crystals occurs when equal numbers of cations and anions are missing from the lattice.

33. Option (ii) neutral is correct since p-type semiconductor mainly contains positively charged carriers ( holes ) which are able to move freely. It is still considered to be neutral as the fixed acceptor atoms having accepted electrons are negative. 

34. Option (iv) 5 is correct since silicon is doped with a pentavalent substance like Phosphorus or Arsenic so, four out of five electrons are used in the formation of four covalent bonds with four of its neighbouring silicon atoms. The fifth electron is extra and becomes delocalised. These delocalised electrons increase the conductivity of doped silicon. 

35. Option (ii) 8 is correct since in face centred cubic unit cell the number of atoms per unit cell is 4 atoms and hence the number of tetrahedral voids in fcc unit cell is found to be 8

36. Option (i) (A) and (B) is correct since AgBr shows both Frenkel as well as Schottky defects. 

37. Option (ii) hcp and ccp is correct since in hcp and ccp the packing efficiency is 74% ( maximum ), hence this pair of packing is considered as the most efficient packing. 

38. Option (iii) 32 is correct  since the packing efficiency in bcc arrangement is 68%, therefore the percentage empty space  can be calculated as – 100 - 68 = 32 

39. Option (iv) since in this arrangement the spheres of fourth layer are exactly aligned with those of the first layer is correct since in hexagonal close packed structure there is ABAB.type arrangement which means that in this case spheres of third layer are exactly aligned with that of the first layer. Hence, the pattern of spheres is repeated in alternate layers. Thus, the statement at (iv) is not true about hexagonal close packing. 

40. Option (i) Cl-ion form fcc lattice and Na⁺ ions occupy all octahedral voids of the unit cell is correct since In NaCl,  chloride ion form fcc lattice and sodium ions occupy all the octahedral voids of the unit cell. The coordination number of both the cation and anion is found to be 6.

41. Option (iii) 4 is correct since in  2-D scope structure each sphere is in contact with four of its neighbours. Thus, its coordination number is found to be 4. 

42. Option (iv) Electronic defects is correct since doping can be done with an impurity which is electron rich or electron deficient as compared to the intrinsic semiconductor like Si or Ge . Such impurities lead to the electronic defects in them. 

43. Option (ii) n-type semiconductor is correct since silicon is doped with electron rich impurities like Phosphorus or Arsenic and this leads to the increase in the conductivity due to the negatively charged electron . Hence, silicon doped with electron rich impurity is termed as n -type semiconductor. 

44. Option (ii) Ferromagnetic substances cannot be magnetised permanently is correct since ferromagnetic substances are those which are very strongly attracted by the magnetic field and can be magnetized permanently.

45. Option (iv) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids is not true since the occupation of the voids depends upon the stoichiometry of the compound instead of the radii of the ions. 

46. Option (i) all the domains get oriented in the direction of magnetic field is correct since whenever a ferromagnetic substance is placed in a magnetic field it becomes a permanent magnet as all the domains get oriented in the directions of the magnetic field even after removal of  the applied magnetic field.

47. Option (ii) The correct order of the packing efficiency is fcc > bcc > simple cubic

Packing Efficiency (fcc) - 74% 

Packing Efficiency (bcc) - 68% 

Packing Efficiency (simple cubic) - 52% 

48. Option (i) Frenkel defect is correct since in this defect an atom is displaced from its lattice point to an interstitial site thus creating a vacancy at the lattice point. This usually occurs in those ionic crystals where the size of anion is quite large as compared to that of the cation. Here dislocation of atoms occurs and hence it is known as dislocation defect.

51. Option (i)? metals >>? insulators < ? semiconductors is correct since metals are the ones which shows the highest conductivity due to the presence of free electrons in them which helps in the conduction of electricity while semiconductors are the ones whose conductivity depends upon the temperature.

Since at normal temperatures, they act as insulators whereas if the temperature rises, the conductivity of the semiconductor also increases. Insulators are the ones which do not have free electrons in them. Due to which they act as poor conductors and hence their conductivity is the lowest.

52. Option (iii) and (iv) are correct since tetrahedral voids are formed when the triangular void in the second layer lie exactly above the triangular voids in the first layer and the triangular shape of these voids oppositely overlap while the octahedral voids are formed when triangular void of second layer is not exactly overlap with similar void in first layer. This can be shown as-

53. Options (iii) and (iv) are correct since in anti-ferromagnetic substances the magnetic moments and domains are oppositely oriented, and they cancel out each other's magnetic moment. For example,   the anti-ferromagnetic substance-MnO.

54. Options (iii) and (iv) are the statements which are not true since in impurity defect there is a replacement of ions by the ions of another compound. By doing so, even though the electrical neutrality of the solid is maintained, a vacancy defect is caused due to creation of vacant sites in the ionic solid. This results into change in density of the solid. For example, if molten NaCl containing a little amount of SrCl₂ is crystallised some of the sites of Na⁺ are occupied by Sr²⁺ . 

Each Sr²⁺ replaces two Na⁺ ions and hence this affects density. Frenkel defect causes vacancy defect at its original site and an interstitial defect at its new location. Therefore, it does not change the density of the solid. Thus, the statement at option (iv) is also not true. 

55.  (i), (ii) and (iv) is correct since gap between valence band and conduction band determines the conductivity of the metals. The electrical conduction of metals depend upon the type of valence band & also the gap between the valence band and conduction band while the valence band may remain partially filled.

56. Options (i) and (ii) are correct since Si or Ge of group-14 are doped with electron deficient impurity i.e.,   group-13 elements like B, Al, or Ga containing only three valence electrons and hence is called as p-type semiconductors while the electron holes are created whenever an atom of such trivalent metal gets bonded with three of its neighbouring tetravalent silicon or germanium atoms.

The fourth valence electron of Ge or Si moves through the holes and under the influence of the electric field electrons move towards the positively charged plate through these holes. But it would appear as if the electron holes are positively charged and are moving towards the negatively charged plate. 

57. Options (ii) and (iii) are correct since silicon has 4 valence electrons and is doped with electron rich impurity is an n-type semiconductor due to extra electrons and hence the availability of the delocalised electrons increase the conductivity of this.

58. Options (i) and (iv) are correct since an excess of K+ ions make KCl crystals appear violet due to the defect called metal excess defect . The anionic vacancies are found to be created due to the movement of potassium atoms to the surface of the crystal when KCl is heated in an atmosphere of potassium vapours and the Cl-ions then diffuse to the surface.

This happens due to the loss of the electron by the potassium atom to form K+ ions. In this case the electrons released diffuse to the crystal and occupy anionic sites being created. The anionic sites occupied by unpaired electrons are known as F-centres. These F-centres impart colour to the crystals for instance - yellow to NaCl and violet to KCl due to the excitation of these electrons by absorbing energy from the light falling on these crystals. 

59. Options (ii) and (iii) are correct since the number of tetrahedral voids per unit cell in NaCl is calculated as-

Since NaCl crystal forms fcc lattice so,  

The number of atoms (n) per unit cell = 4 

So, the number of octahedral voids = n = 4 

While the number of tetrahedral voids = 2n = 2*4=8  

60. Options (i) and (iii) are correct since the structure of the amorphous solids is similar to that of the liquids and hence they tend to flow like liquids, although very slowly. So, sometimes they are known as super cooled liquids or pseudo solids. 

61. Options (i) and (iii) are correct since n-type semiconductors are prepared by doping silicon with pentavalent elements of group 15 (like P or As). Some of the lattice sites in silicon crystals are occupied by the fifth electron which is left over as extra electron and couldn't form a bond with the tetravalent Silicon and hence the images (i) and (iii) are showing this situation and hence representing n-type semiconductor. 

62. Options (i) and (iv) are correct since these statements agrees with the ferrimagnetic and ferromagnetic nature of the substances respectively. 

63. Options (i) and (iii) are correct since quartz glass is an amorphous solid so, it does not exhibit a definite heat of fusion. 

64. Options (i), (ii) and (iii) are correct since SiC, AlN, and diamond are not regarded as molecular solids, instead they come under network or covalent solids. 

65. Options (i) and (iv) are correct since when the tetrahedral void of the first layer and the tetrahedral void of the second layer align together, they form an octahedral void, and this is possible in case of hep and fcc arrangements.

68. (i) → (c); (ii) → (a); (iii) → (d); (iv) → (b)

69. (i) → (b); (ii) → (c); (iii) → (c); (iv) → (a)

70. (i) → (c); (ii) → (a); (iii) → (b)

71.  (i) → (d); (ii) → (c); (iii) → (b); (iv) → (a)

72. (i) → (c); (ii) → (a); (iii) → (d); (iv) → (b)

73. Option (i) Assertion and reason both are correct statements and reason is correct explanation for assertion is the answer since in simple cubic crystal unit cell, the total number of atoms are given as - 8 * 18=1

74. Option (ii) Assertion and reason both are correct statements, but reason is not correct explanation for assertion is the answer since in graphite carbon atoms are arranged in different layers and each atom is covalently bonded to three of its neighbouring atoms in the same layer due to which the fourth valence electron of each atom, present between the different layers is free to move and this makes graphite a good conductor of electricity.

As different layers can slide over each other which makes it soft. while in case of diamond C atom is bonded covalently with their adjacent atoms throughout the crystal. Thus, all the four valencies are satisfied. Covalent bonds are strong and directional in nature ; therefore, atoms are held very strongly at their positions. This is why diamond is hard and has a very high melting point but decomposes before melting, also acting as an insulator.

75. Option (iii) Assertion is the correct statement, but the reason is wrong is the answer since besides the body centre there is one octahedral void at the centre of each 12 edges which is shared between four adjacent unit cells. Thus, Octahedral voids present at the body centre of the cube = 1 

12 octahedral voids are located at each edge and shared between four-unit cells = 12 * $\frac{1}{2}$  = 3

Total number of octahedral voids = 4. 

BrF₅       -             Square pyramid

[CrF₆]^3- -             Octahedral

O₃          -             Bent

PCl₅       -             Trigonal Bipyramid

 $2S{O}_2\left(g\right)+O_2\left(g\right)\underset{}{\overset{V_2{O}_5}{\to }}2S{O}_3\left(g\right)contactprocess$

$4N{H}_3\left(g\right)+5O_2\left(g\right)\underset{}{\overset{A\left(s\right)-Rh\left(s\right)}{\to }}4NO\left(g\right)+6H_{2}O\left(g\right)$ ostwald’s process

$N_2\left(g\right)+3H_2\left(g\right)\underset{}{\overset{Fe\left(s\right)}{\to }}2N{H}_3\left(g\right)Haber\text{'}sprocess$

Vegetable oil (l) + H₂ $\underset{}{\overset{Ni\left(s\right)}{\to }}$ vegetable ghee; Hydrogenation

Ionic radius of Cl^- = 167 pm

Covalne radius of Cl = 99 pm

Liquation process is used for purification of that metal, whose melting point is lesser then that of impurities.

Urea acts as negative catalyst for decomposition of H₂O₂.

$2BeC{l}_2+LiAl{H}_4\to 2Be{H}_2+LiCl+AlC{l}_3$

$B_2{H}_6+N{H}_3\underset{}{\overset{HighTemperature}{\to }}{B}_3{N}_3{H}_6\left(Borazine\right)$

It is reduction process, while rest are examples of disproportionation reaction.

 $\stackrel{+7}{M}n{O}_4^{-}+8H^{+}+5e\to \stackrel{+2}{M}{n}^{2+}+4H_{2}O$ change in oxidation number of Mn = 7 – 2 = 5

$\stackrel{•}{C}l+C{H}_3-H\to \stackrel{•}{C}{H}_3+HCl$

Solvent polarity is related to R_f value of nitro compounds

100 mg P-nitrophenol and picric acid have different R_f value on silica gel plate.

Preparative TLC is best to separate 100 mg of P-nitro phenol and picric acid.

Both indicated hydrogen atoms repel each other so due steric hinderance the given compound becomes non planar. It is non-aromatic.

Kindly consider the following figure

Kindly consider the following figure

Birch reduction takes place of carbon connected with electron withdrawing group (-NO₂)

Electron donating group increases stability of diazonium salt. Whereas electron with drawing group decrease it.

-OCH₃ (+M), -NO₂ & -CN (-M)

Kindly consider the following figure

Reducing sugar must have hemi-acetal group and option (A) has that group.

Aniline makes adduct with AlCl₃ and thus gets deactivated for electrophilic aromatic substitution and do not react with CH₃-Cl

Mass of solute = 48 * 2 * 10^-3 gram

Atoms of Mg = 24.08 * 10²⁰

 $x_{H_2}=\frac{n_{H_2}}{n_{H_2}+n_{o_2}}=\frac{\left(40/2\right)}{\left(40/2\right)+\left(60/32\right)}$ = 0.914

$P_{H_2}=x_{H_2}\times P_T=0.914\times 2.2\sim 2bar$

Xe = Xn

MeVe = MnVn

$\frac{Ve}{Vn}=\frac{Mn}{Me}\sim 1758.24$

 $C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)\Delta ng=0$

$q_C=20\times 2=40kJ$

Number of moles of C = $\frac{2.4}{12}=0.2moles$

$\Delta H_C=\frac{40kJ}{0.2mole}=200kJ/mole$

Remaining volume of solution = 400 ml

Mass of HNO3 = 25.2 – 11.5 = 13.7

Molarity = $\frac{\left(13.7/63\right)}{\left(400/1000\right)}$ = 0.54 M = 54 × 10^-2 M

$K_{eq}={\left(\frac{1}{2*10^{15}}\right)}^{1/2}=2.24*10^{?8}$

Fe₃O₄ + 8H⁺ + 8e -> 3Fe + 4H₂O

 $t_{1/2}\alpha {\left[Reactant\right]}_{t=0}^{\left(1-n\right)},$ where n = order of reaction.

 $\left[Co{\left(H_{2}O\right)}_6\right]C{l}_2\Rightarrow C{o}^{2+}\Rightarrow 3d^7\Rightarrow {t_{2g}}^{5}e{g}^2$

Number of unpaired electrons = 3

$\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3\Rightarrow C{r}^{3+}\Rightarrow 3d^3\Rightarrow {t_{2g}}^{3}e{g}^0$

Number of unpaired electrons = 3

Kindly consider the following figure