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Redox Reactions

🎓 Class 11📖 Chemistry Part II📖 10 notes🧠 15 Q&A⏱️ ~15 min
Chapter 1 of 3chemistry

Redox ReactionsStudy Notes

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Introduction

Explanation

Introduction

Redox reactions, short for reduction-oxidation reactions, constitute a fundamental class of chemical reactions where oxidation and reduction processes occur simultaneously. Chemistry fundamentally studies matter and its transformations, and redox reactions represent a significant category of such transformations. These reactions involve electron transfer between species and are pivotal in various physical, biological, and industrial processes. Examples include combustion of fuels for energy, electrochemical extraction of reactive metals and non-metals, manufacturing of chemicals like caustic soda, battery operation, and corrosion of metals. Environmental phenomena such as the hydrogen economy and ozone layer depletion also involve redox processes. Understanding redox reactions is essential for grasping many chemical and biological phenomena and their applications.

  • Redox reactions involve simultaneous oxidation and reduction.
  • They are central to chemical transformations in nature and industry.
  • Electron transfer is the core mechanism in redox reactions.
  • Applications include energy production, metallurgy, and environmental chemistry.
  • Redox reactions underpin processes like combustion, corrosion, and battery function.
  • 📌 Redox reactions: Chemical reactions involving simultaneous oxidation and reduction.
  • 📌 Oxidation: Process involving loss of electrons or increase in oxidation state.
  • 📌 Reduction: Process involving gain of electrons or decrease in oxidation state.

7.1 CLASSICAL IDEA OF REDOX REACTIONS - OXIDATION AND REDUCTION REACTIONS

Explanation

7.1 CLASSICAL IDEA OF REDOX REACTIONS - OXIDATION AND REDUCTION REACTIONS

Initially, oxidation was defined as the addition of oxygen to a substance, as many elements combine with atmospheric oxygen to form oxides. For example, magnesium reacts with oxygen to form magnesium oxide (2Mg(s) + O2(g) → 2MgO(s)) [formula_1], and sulphur forms sulphur dioxide (S(s) + O2(g) → SO2(g)) [formula_2]. Methane combustion also illustrates oxidation by oxygen addition (CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)) [formula_3]. Subsequently, oxidation was broadened to include removal of hydrogen from a substance, as seen in the reaction 2H2S(g) + O2(g) → 2S(s) + 2H2O(l) [formula_4]. Further extension included reactions involving electronegative elements other than oxygen, such as fluorine, chlorine, and sulphur reacting with magnesium to form MgF2 [formula_5], MgCl2 [formula_6], and MgS [formula_7]. Oxidation thus encompasses addition of oxygen or electronegative elements, or removal of hydrogen or electropositive elements. Reduction, originally defined as removal of oxygen, now includes removal of electronegative elements or addition of hydrogen or electropositive elements. Examples include decomposition of mercuric oxide (2HgO(s) → 2Hg(l) + O2(g)) [formula_9], reduction of ferric chloride by hydrogen (2FeCl3(aq) + H2(g) → 2FeCl2(aq) + 2HCl(aq)) [formula_10], hydrogenation of ethene (CH2=CH2(g) + H2(g) → C2H6(g)) and reduction of mercuric chloride by stannous chloride (2HgCl2(aq) + SnCl2(aq) → Hg2Cl2(s) + SnCl4(aq)). It is important to note that oxidation and reduction always occur simultaneously, leading to the term 'redox' reactions. For example, in the reaction between hydrogen sulphide and chlorine gas (H2S + Cl2 → 2HCl + S), H2S is oxidised and Cl2 is reduced. Similarly, in the reaction of Fe3O4 with aluminium (3Fe3O4 + 8Al → 9Fe + 4Al2O3), aluminium is oxidised and Fe3O4 is reduced. These reactions illustrate the classical concept of oxidation and reduction based on addition or removal of oxygen, hydrogen, or electronegative/electropositive elements.

  • Oxidation originally meant addition of oxygen to a substance.
  • Definition expanded to include removal of hydrogen or electropositive elements.
  • Reduction originally meant removal of oxygen, now includes addition of hydrogen or electropositive elements.
  • Oxidation and reduction always occur simultaneously in redox reactions.
  • Examples include combustion, hydrogenation, and metal displacement reactions.
  • 📌 Oxidation: Addition of oxygen or electronegative element, or removal of hydrogen or electropositive element.
  • 📌 Reduction: Removal of oxygen or electronegative element, or addition of hydrogen or electropositive element.
  • 📌 Redox reaction: Chemical reaction involving simultaneous oxidation and reduction.

7.2 REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS

Explanation

7.2 REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS

Redox reactions can be fundamentally understood as electron transfer processes. For example, sodium reacts with chlorine, oxygen, or sulphur to form ionic compounds: sodium chloride, sodium oxide, and sodium sulphide respectively [formulas 7.12 to 7.

Practice QuestionsRedox Reactions

Includes NCERT exercise questions with answers

Q1.When hydrogen fluoride is dissolved in water, it dissociates to give______.
A.difluoride ion
B.fluoride ion
C.hydrogen and fluorine atom
D.none of the above

Answer:

difluoride ion

Explanation:

[{"id": "a26c0e18-5df3-7d50-3471-e4b4379adb73", "type": "html", "value": " This is the abnormal behaviour of HF which is due to hydrogen bond. This behaviour is responsible for the existence of KHF 2 . 2HF ⇋ + "}]

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Q2.The conditions for the formation of hydrogen bond are____
A.The molecule must have a highly electronegative atom linked to H-atoms and electronegative atoms should have a small size.
B.The molecule must have a highly electronegative atom linked to H-atom and the electronegative atom should have a larger size.
C.The molecule must have a less electronegative atom linked to H-atoms and electronegative atoms should have a small size.
D.The molecule must have a less electronegative atom linked to H-atom and the electronegative atom should have a larger size.

Answer:

The molecule must have a highly electronegative atom linked to H-atoms and electronegative atoms should have a small size.

Explanation:

[{"id": "280e181d-81be-33e4-efaa-480c2ccea64a", "type": "html", "value": " Solution: (Optional): The higher the electronegativity, the more is the polarization of the molecule. The smaller is the size, the greater is the electrostatic force of attraction. High electronegativity and small size of atom bonded to H atom leads to formation of hydrogen bond. "}]

MediumNCERT
Q3.The molecular orbital electronic configuration of an oxygen molecule is___.
A.(σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (σ2p z ) 2 (π2p x 2 = π2p y 2 )(π*2p x 1 =π*2p y 1 )
B.(σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (π2p x 2 = π2p y 2 )(π*2p x 1 =π*2p y 1 )(σ2p z ) 2
C.(σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (π2p x 2 = π2p y 2 )(π*2p x 1 =π*2p y 1 )(σ2p z ) 1
D.(σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (π2p x 2 = π2p y 2 )(σ2p z ) 2 (π*2p x 1 =π*2p y 1 )

Answer:

(σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (σ2p z ) 2 (π2p x 2 = π2p y 2 )(π*2p x 1 =π*2p y 1 )

Explanation:

[{"id": "e5293c6c-28ea-d478-67ce-5fd9fb8321cc", "type": "html", "value": " Aufbau Rule, Pauli’s Exclusion principle and Hund’s rule of maximum multiplicity are followed while filling molecular orbitals. Bonding molecular orbitals are lower energy orbitals. So, they are filled first. "}]

MediumNCERT
Q4.The number of bonding and antibonding electrons in a fluorine molecule are______and _____ respectively.
A.6, 8
B.8, 6
C.10, 6
D.8, 10

Answer:

8, 6

Explanation:

[{"id": "5620e90a-da61-fb58-e490-bd0c56f4031e", "type": "html", "value": " The molecular orbital electronic configuration of a fluorine molecule is. (σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (σ2p z ) 2 (π2p x 2 = π2p y 2 )(π*2p x 2 =π*2p y 2 ) 4 electrons in σ1s and σ*1s orbitals i.e.; K-K shells are nonbonding and 14 electrons are allotted to molecular orbitals. Bonding electrons are 8 and antibonding electrons are 6. "}]

MediumNCERT
Q5.The formation of molecular orbitals by linear combinations of atomic orbitals take place if_____
A.The combining orbital must have the same or nearly the same energy, i.e.; comparable energy.
B.The atomic orbitals must have proper orientation, i.e.; same symmetry about the molecular axis.
C.The atomic orbitals must overlap to the maximum extent.
D.All the above.

Answer:

All the above.

Explanation:

[{"id": "3249331c-57d1-4814-fdf4-fa7e0eb77600", "type": "html", "value": " Molecular orbitals are formed when combining atomic orbitals do have comparable energy, same symmetry about the molecular axis, and maximum overlap. "}]

MediumNCERT
Q6.If N b > N a i.e.; the number of bonding electrons > the number of antibonding electrons. Then the molecule is____
A.unstable as the influence of antibonding electrons is slightly more than that of bonding electrons resulting in a net force of repulsion.
B.unstable as the antibody electron influence is stronger resulting in a net force of repulsion
C.Stable due to the net force of attraction.
D.Stable due to the net force of repulsion.

Answer:

Stable due to the net force of attraction.

Explanation:

[{"id": "6fe9b2ef-d36b-e7fb-b381-b8f0223f5a45", "type": "html", "value": " A molecule is formed only when the total energy of the system decreases which is only possible when the number of bonding electrons is more than the number of antibonding electrons. Since bonding molecular orbitals are formed by the linear combination of atomic orbitals when their wave functions are added (ɸ = ψ A + ψ B ). The electron density increases in the region between their nuclei of bonded atoms. It stabilizes the molecule. "}]

MediumNCERT
Q7.Which of the following geometry and angle corresponds to dsp 2 hybridization?
A.square planar, 90 0
B.tetrahedral, 109 0 28’
C.octahedral , 90 0
D.Trigonal planar, 120 0

Answer:

square planar, 90 0

Explanation:

[{"id": "786c295c-86d2-b8e3-88c8-5376efb691e5", "type": "html", "value": " 4 atomic orbitals i.e.; one d, one s and two p orbitals undergo dsp 2 to form four dsp 2 orbitals which orient themselves at 90 0 to give square planar geometry to the molecule. "}]

MediumNCERT
Q8.The type of hybrid orbitals of S in SF 6 and I in IF 7 are______ and _______ respectively
A.SP 3 d 3 , SP 3 d 2
B.SP 3 d 2 , SP 3 d 3
C.SP 3 d , SP 3 d 3
D.SP 3 d 3 , SP 3 d

Answer:

SP 3 d 2 , SP 3 d 3

Explanation:

[{"id": "05e8be7d-f0cf-47e9-004f-1262c97a68b9", "type": "html", "value": " Sulphurin SF 6 forms 6 SP 3 d 2 hybrid orbitals. Each of them overlaps the p orbital of a fluorine atom to form a sigma bond. This leads to the octahedral geometry of the molecule whereas Iodine in IF 7 forms 7 SP 3 d 3 hybrid orbitals. Each of them overlaps the p orbital of a fluorine atom to form a sigma bond. This leads to the pentagonal bi-pyramidal geometry of the molecule. SF 6 IF 7 "}]

MediumNCERT