IGNOU BCHCT-137 Solved Assignment 2024 B.Sc. CBCS Chemistry cover page

IGNOU BCHCT-137 Solved Assignment 2024 | B.Sc. CBCS Chemistry

Solved By – Anjali Patel – Bachelor of Science (B.Sc) from Mumbai University

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IGNOU BCHCT-137 Assignment Question Paper 2024

bchct-137-solved-assignment-2024-qp-0b03831b-9f6e-4f3d-bad2-2f85f362edb1

bchct-137-solved-assignment-2024-qp-0b03831b-9f6e-4f3d-bad2-2f85f362edb1

PART A: COORDINATION CHEMISTRY
  1. a) What is the ground state configuration of S c + S c + Sc^(+)\mathrm{Sc}^{+}Sc+ion? Justify.
    b) Why zine and cadmium are soft metals?
  2. What is the composition of brass? Is it harder than pure copper? Give its uses.
  3. Why do lanthanoids mainly show ionic bonding? What is the nature of their melting and boiling points?
  4. What are the possible number of isomers for the octahedral complex ion [ C o ( N H 3 ) C l 2 ] + C o N H 3 C l 2 + [Co(NH_(3))Cl_(2)]^(+)\left[\mathrm{Co}\left(\mathrm{NH}_3\right) \mathrm{Cl}_2\right]^{+}[Co(NH3)Cl2]+?
  5. What type of isomerism is exhibited in the complexes [ C o ( N H 3 ) 5 ( S O 4 ) ] B r C o N H 3 5 S O 4 B r [Co(NH_(3))_(5)(SO_(4))]Br\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{Br}[Co(NH3)5(SO4)]Br and [ C o ( N H 3 ) 5 B r ] S O 4 C o N H 3 5 B r S O 4 [Co(NH_(3))_(5)Br]SO_(4)\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Br}\right] \mathrm{SO}_4[Co(NH3)5Br]SO4 ?
  6. Give the hybridized orbitals and the corresponding geometries of [ T i ( H 2 O ) 6 ] 3 + T i H 2 O 6 3 + [Ti(H_(2)O)_(6)]^(3+)\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}[Ti(H2O)6]3+ and [ C o C l 4 ] 2 = C o C l 4 2 = [CoCl_(4)]^(2=)\left[\mathrm{CoCl}_4\right]^{2=}[CoCl4]2=.
  7. Explain the directional properties of the five d d ddd orbitals in a free transition metal ion with the help of suitable diagrams.
  8. Give the CFSE of an octahedral complex with seven electrons in the d d ddd orbitals.
    9 Give the possible electronic configurations for d 3 d 3 d^(3)d^3d3 and d 6 d 6 d^(6)d^6d6 systems in a tetrahedral crystal field.
  9. Which are the cases when tetrahedral geometry is favoured over octahedral geometry for metal complexes? Why are they so?
PART B: STATES OF MATTER & CHEMICAL KINETICS
  1. Calculate the i) average speed ii) root mean square speed and iii) most probable speed of oxygen molecules at 515 K 515 K 515K515 \mathrm{~K}515 K. (Given M m ( O 2 ) = 0.016 k g m o l 1 ) M m O 2 = 0.016 k g m o l 1 {:M_(m)(O_(2))=0.016(kg)mol^(-1))\left.M_m\left(\mathrm{O}_2\right)=0.016 \mathrm{~kg} \mathrm{~mol}^{-1}\right)Mm(O2)=0.016 kg mol1)
  2. State the Dalton’s law of partial pressure and give its significance.
  3. Explain the pressure and volume correction terms to the ideal gas equation, and deduce van der Waals equation.
  4. What is meant by rate of a reaction? List and explain different types of rates used in chemical kinetics.
  5. With suitable example and figure explain the integrated rate law for first order reaction.
  6. For the reaction, C l 2 ( g ) + 2 N O ( g ) 2 N O C l ( g ) C l 2 ( g ) + 2 N O ( g ) 2 N O C l ( g ) Cl_(2)(g)+2NO(g)rarr2NOCl(g)\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow 2 \mathrm{NOCl}(\mathrm{g})Cl2( g)+2NO(g)2NOCl(g), the initial concentrations, [ C l 2 ] 0 C l 2 0 [Cl_(2)]_(0)\left[\mathrm{Cl}_2\right]_0[Cl2]0 and [ N O ] 0 [ N O ] 0 [NO]_(0)[\mathrm{NO}]_0[NO]0 are given below along the corresponding with initial rates.
[ C l 2 ] 0 / M C l 2 0 / M [Cl_(2)]_(0)//M\left[\mathrm{Cl}_2\right]_0 / \mathrm{M}[Cl2]0/M [ N O ] 0 / M [ N O ] 0 / M [NO]_(0)//M[\mathrm{NO}]_0 / \mathrm{M}[NO]0/M Initial rate / M s 1 / M s 1 //Ms^(-1)/ \mathrm{Ms}^{-1}/Ms1
0.10 0.10 3.0 × 10 3 3.0 × 10 3 3.0 xx10^(-3)3.0 \times 10^{-3}3.0×103
0.20 0.10 6.0 × 10 3 6.0 × 10 3 6.0 xx10^(-3)6.0 \times 10^{-3}6.0×103
0.20 0.20 2.4 × 10 2 2.4 × 10 2 2.4 xx10^(-2)2.4 \times 10^{-2}2.4×102
[Cl_(2)]_(0)//M [NO]_(0)//M Initial rate //Ms^(-1) 0.10 0.10 3.0 xx10^(-3) 0.20 0.10 6.0 xx10^(-3) 0.20 0.20 2.4 xx10^(-2)| $\left[\mathrm{Cl}_2\right]_0 / \mathrm{M}$ | $[\mathrm{NO}]_0 / \mathrm{M}$ | Initial rate $/ \mathrm{Ms}^{-1}$ | | :— | :— | :— | | 0.10 | 0.10 | $3.0 \times 10^{-3}$ | | 0.20 | 0.10 | $6.0 \times 10^{-3}$ | | 0.20 | 0.20 | $2.4 \times 10^{-2}$ |
Determine (i) the order of the reaction with respect to N O N O NO\mathrm{NO}NO and C l 2 C l 2 Cl_(2)\mathrm{Cl}_2Cl2; (ii) the rate law; and (iii) the rate constant
17. Give the reasons for the striking contrast in the boiling points of ethanol ( 351 K ) ( 351 K ) (351K)(351 \mathrm{~K})(351 K) and dimethyl ether (249 K). Give suitable diagrams to illustrate this.
18. What are the elements of symmetry? Give the suitable diagrams for any one of them.
19. Explain close packing in two dimensions for a crystalline solid with the help of a suitable diagram.
20. Give the consequences of Schottky and Frenkel defect.
\(cos\:2\theta =1-2\:sin^2\theta \)

BCHCT-137 Sample Solution 2024

bchct-137-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

bchct-137-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

PART A: COORDINATION CHEMISTRY
  1. a) What is the ground state configuration of S c + S c + Sc^(+)\mathrm{Sc}^{+}Sc+ion? Justify.
Answer:
The ground state electron configuration of a neutral scandium ( S c S c Sc\mathrm{Sc}Sc) atom is [ A r ] 3 d 1 4 s 2 [ A r ] 3 d 1 4 s 2 [Ar]3d^(1)4s^(2)[\mathrm{Ar}]\,3d^1\,4s^2[Ar]3d14s2, where [ A r ] [ A r ] [Ar][\mathrm{Ar}][Ar] represents the electron configuration of the noble gas argon ( 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)1s^2\,2s^2\,2p^6\,3s^2\,3p^61s22s22p63s23p6). Scandium is the first element in the transition metals series, with an atomic number of 21, indicating it has 21 electrons as a neutral atom.
When scandium forms a + 1 + 1 +1+1+1 ion ( S c + S c + Sc^(+)\mathrm{Sc}^+Sc+), it loses one electron. The removal of an electron from a transition metal typically starts with the outermost s s sss electrons before the d d ddd electrons. This is because the s s sss electrons are generally at a higher energy level than the d d ddd electrons of the same period in the transition metals, making them easier to remove.
Therefore, the electron configuration of the S c + S c + Sc^(+)\mathrm{Sc}^+Sc+ ion, after losing one electron, would be:
[ A r ] 3 d 1 4 s 1 [ A r ] 3 d 1 4 s 1 [Ar]3d^(1)4s^(1)[\mathrm{Ar}]\,3d^1\,4s^1[Ar]3d14s1
This configuration shows that the S c + S c + Sc^(+)\mathrm{Sc}^+Sc+ ion has one electron in the 3 d 3 d 3d3d3d orbital and one electron in the 4 s 4 s 4s4s4s orbital after the loss of one electron from the 4 s 4 s 4s4s4s orbital of the neutral scandium atom. This configuration is justified based on the general principle of electron removal from the outermost shell first, which, for transition metals, means the 4 s 4 s 4s4s4s electrons are removed before the 3 d 3 d 3d3d3d electrons.
b) Why zine and cadmium are soft metals?
Answer:
Zinc and cadmium are considered relatively soft metals compared to other transition metals. This softness can be attributed to several factors related to their electronic structure and bonding:

1. Electronic Configuration

  • Zinc ( Z n Z n ZnZnZn) has the electronic configuration [ A r ] 3 d 10 4 s 2 [ A r ] 3 d 10 4 s 2 [Ar]3d^(10)4s^(2)[Ar] 3d^{10} 4s^2[Ar]3d104s2, and cadmium ( C d C d CdCdCd) has the electronic configuration [ K r ] 4 d 10 5 s 2 [ K r ] 4 d 10 5 s 2 [Kr]4d^(10)5s^(2)[Kr] 4d^{10} 5s^2[Kr]4d105s2. In both cases, the d d ddd orbitals are completely filled. The filled d d ddd orbitals contribute to the stability of the electron configuration but do not significantly contribute to metallic bonding compared to elements with partially filled d d ddd orbitals. The effectiveness of metallic bonding is a key factor in determining a metal’s hardness; stronger metallic bonds usually result in harder metals.

2. Metallic Bonding

  • The strength of metallic bonds in zinc and cadmium is relatively weaker than in other transition metals with partially filled d d ddd orbitals. This is because the delocalized electrons in zinc and cadmium are primarily from the s s sss orbital, as the d d ddd orbitals are fully occupied and do not contribute much to the bonding. The weaker metallic bonds make these metals softer and less rigid.

3. Crystal Structure

  • Both zinc and cadmium crystallize in the hexagonal close-packed (hcp) structure. While this structure is quite efficient in packing atoms, the particular arrangement in zinc and cadmium, combined with their electronic configurations, does not favor the formation of very strong metallic bonds. The directional nature of some of the metallic bonds in the hcp structure can also contribute to anisotropy in mechanical properties, which might influence the perceived softness.

4. Cohesive Energy

  • The cohesive energy, which is a measure of the strength of the bonds holding a metal together, is relatively lower for zinc and cadmium compared to other transition metals. Lower cohesive energy correlates with lower melting points and softer materials.

Conclusion

The relative softness of zinc and cadmium compared to other metals is primarily due to their filled d d ddd orbital configurations, which lead to weaker metallic bonding and lower cohesive energy. Their crystal structures also do not favor the formation of very strong metallic bonds, contributing to their softness. These properties make zinc and cadmium suitable for applications where malleability and ductility are advantageous, but they are less suited for applications requiring high strength and hardness.
  1. What is the composition of brass? Is it harder than pure copper? Give its uses.
Answer:
Brass is an alloy primarily composed of copper (Cu) and zinc (Zn). The proportions of copper and zinc can vary depending on the specific type of brass and its intended use, with copper content typically ranging from about 55% to 90%. The addition of zinc to copper results in an alloy with enhanced properties compared to pure copper, such as increased strength, hardness, and corrosion resistance.

Composition

  • Standard Brass: The most common form of brass, known as "cartridge brass," contains about 70% copper and 30% zinc. This composition is well-balanced, offering good strength and ductility, making it suitable for a wide range of applications.
  • Special Brass Alloys: Other elements may be added to brass in small amounts to achieve specific properties. For example, lead (Pb) is often added (up to 3%) to improve machinability, while aluminum (Al), tin (Sn), and nickel (Ni) can be added to enhance corrosion resistance, strength, and color.

Hardness Compared to Pure Copper

Brass is generally harder than pure copper. The addition of zinc to copper results in a material that is stronger and more resistant to wear and deformation under stress. The exact hardness of brass can vary depending on its zinc content and any additional elements that have been added. The hardening effect is due to the solid solution strengthening and, in some cases, the formation of a more complex microstructure than that of pure copper.

Uses of Brass

Brass is used in a wide variety of applications due to its excellent machinability, corrosion resistance, and attractive appearance. Some common uses include:
  • Musical Instruments: Brass is used to make horns, trumpets, trombones, and other brass instruments due to its acoustic properties.
  • Decorative Items: Its gold-like appearance makes brass a popular choice for decorative items such as door handles, plaques, and jewelry.
  • Plumbing and Water Fittings: Brass’s corrosion resistance makes it suitable for water pipes, faucets, and fittings.
  • Ammunition Casings: The alloy known as cartridge brass (70% copper, 30% zinc) is commonly used for ammunition casings due to its workability and durability.
  • Gears and Bearings: Brass is used in gears, bearings, and valves where low friction is required.
  • Marine Hardware: The addition of elements like tin can make brass resistant to saltwater corrosion, making it suitable for marine hardware.
The versatility of brass, stemming from its various compositions, allows it to meet the requirements of many industrial, decorative, and musical applications, making it a valuable material in numerous fields.

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