Inorganic Chemistry

91. In case of a standard hydrogen electrode

In case of a standard hydrogen electrode

[amp_mcq option1=”absolute electrode potential is not zero” option2=”absolute electrode potential is zero” option3=”both absolute and standard electrode potential values are zero” option4=”electrode potential is zero only at 25°C” correct=”option1″]

This question was previously asked in
UPSC CDS-2 – 2016
Download PDFAttempt Online
The absolute electrode potential of a standard hydrogen electrode (SHE) is not zero.
By international convention, the *standard electrode potential* of the SHE is *defined* as zero volts (0 V) at standard conditions (typically 25°C, 1 bar/atm H₂ pressure, and 1 M H⁺ concentration). This defines a relative scale for measuring the standard potentials of other half-cells. However, the absolute potential of the SHE, which represents the potential difference between the electrode and a point infinitely far away in vacuum, is not zero and is estimated to be around +4.44 V at 25°C.
The ability to measure absolute electrode potentials is limited; therefore, a reference electrode with a defined potential, like the SHE, is used to create a relative potential series. Option D is misleading as the zero potential is defined at standard conditions including 25°C, but the phrasing “only at 25°C” is inaccurate and incomplete regarding the conditions.

92. Match List-I with List-II and select the correct answer using the code

Match List-I with List-II and select the correct answer using the code given below the Lists :

List-I
(Element)		List-II
(Property/Use)
A. Mg1. Gives red colour to flame
B. Ca2. Sulphate compound used in medicine to examine the alimentary canal of a patient
C. Sr3. Traps the energy of sunlight in photosynthesis
D. Ba4. Control of muscle contraction

Code :

[amp_mcq option1=”A-2, B-4, C-1, D-3″ option2=”A-2, B-1, C-4, D-3″ option3=”A-3, B-4, C-1, D-2″ option4=”A-3, B-1, C-4, D-2″ correct=”option3″]

This question was previously asked in
UPSC CDS-2 – 2016
Download PDFAttempt Online
Matching the elements with their properties/uses yields A-3, B-4, C-1, D-2.
Magnesium (Mg) is a key component of chlorophyll, which traps sunlight energy in photosynthesis (A-3). Calcium (Ca) ions play a vital role in muscle contraction (B-4). Strontium (Sr) salts are used to impart a red colour to flames, such as in fireworks (C-1). Barium sulfate (BaSO4) is an opaque compound used in medicine as a contrast agent (barium meal) to examine the alimentary canal of a patient (D-2).
These are common properties and applications of elements from Group 2 (alkaline earth metals). While other calcium compounds are used medicinally, calcium ions’ role in muscle function is a fundamental biological property. Barium sulfate is non-toxic due to its extreme insolubility, unlike most other barium compounds.

93. Which one of the following elements does not form solid hydrogen

Which one of the following elements does not form solid hydrogen carbonate?

[amp_mcq option1=”Sodium” option2=”Potassium” option3=”Caesium” option4=”Lithium” correct=”option4″]

This question was previously asked in
UPSC CDS-2 – 2016
Download PDFAttempt Online
Lithium (Li) is the first element in the alkali metal group (Group 1). While other alkali metals (Sodium, Potassium, Rubidium, Caesium) form stable solid hydrogen carbonates (MΗCO₃), Lithium hydrogen carbonate (LiHCO₃) is significantly less stable as a solid.
LiHCO₃ exists in aqueous solution, but when attempts are made to isolate it as a solid, it readily decomposes upon heating or dehydration into lithium carbonate (Li₂CO₃), water (H₂O), and carbon dioxide (CO₂). This instability prevents the formation of a stable solid crystalline LiHCO₃ under normal conditions, unlike NaHCO₃, KHCO₃, CsHCO₃, etc., which are stable solids. This difference is often attributed to Lithium’s small ionic size and relatively high polarizing power (diagonal relationship with Magnesium).
The stability of alkali metal bicarbonates increases down the group. Sodium bicarbonate (baking soda) and Potassium bicarbonate are common solid compounds.

94. In paper manufacturing, degumming of the raw material is done using

In paper manufacturing, degumming of the raw material is done using

[amp_mcq option1=”sulphuric acid” option2=”bleaching powder” option3=”caustic soda” option4=”nitric acid” correct=”option3″]

This question was previously asked in
UPSC CDS-2 – 2016
Download PDFAttempt Online
In paper manufacturing, the raw material (typically wood or plant fibres) contains not just cellulose but also lignin, hemicellulose, waxes, and other substances. “Degumming” is a process aimed at removing or breaking down unwanted non-cellulose components to obtain clean fibres suitable for papermaking. Alkaline solutions, such as caustic soda (sodium hydroxide, NaOH), are widely used in various pulping processes (like the Kraft process) to break down lignin and extract other materials, including some gummy substances or waxes, from the cellulose fibres.
Caustic soda is a strong alkali used in the pulping process to dissolve lignin and other impurities, separating the cellulose fibres. This process effectively “degums” or cleans the fibres before they are processed into paper. Bleaching powder is used for whitening the pulp after pulping. Sulphuric acid and nitric acid are strong acids with different applications (e.g., in chemical processing or nitration) and are not typically the primary agents for degumming in standard paper pulping, although acidic processes (like the sulfite process) exist, they target lignin and hemicellulose differently than the general removal of gummy substances often associated with alkaline treatments or specific fibre degumming steps.
Pulping processes, whether alkaline (like Kraft) or acidic (like Sulfite), aim to liberate cellulose fibers by removing lignin and other matrix components. While the term “degumming” is sometimes specifically used for fibres like cotton or hemp, the underlying chemical principle of using alkalis to remove impurities containing waxes, pectins, etc., is relevant to how caustic soda functions in cleaning paper pulp fibres.

95. Which one of the following has the highest ionic character?

Which one of the following has the highest ionic character?

[amp_mcq option1=”BeF₃” option2=”SiO₂” option3=”NCl₃” option4=”K₂S” correct=”option4″]

This question was previously asked in
UPSC CDS-1 – 2024
Download PDFAttempt Online
Among the given options, K₂S has the highest ionic character.
Ionic character is primarily determined by the electronegativity difference between the bonded atoms. A larger electronegativity difference leads to a more ionic bond. Bonds between highly electropositive metals (Group 1 and 2) and highly electronegative non-metals (Group 16 and 17) are generally ionic.
Approximate electronegativity values: K (0.8), S (2.5), Be (1.5), F (4.0), Si (1.9), O (3.4), N (3.0), Cl (3.2).
Electronegativity differences:
– BeF₃ (assuming Be-F bond in a compound): |4.0 – 1.5| = 2.5. Be compounds often show significant covalent character due to Be²⁺’s high charge density.
– SiO₂ (Si-O bond): |3.4 – 1.9| = 1.5. SiO₂ is a covalent network solid.
– NCl₃ (N-Cl bond): |3.2 – 3.0| = 0.2. This is a covalent molecule.
– K₂S (K-S bond): |2.5 – 0.8| = 1.7. K is a Group 1 metal, S is a non-metal. The bond is predominantly ionic.
While Be-F bond has a larger electronegativity difference, K₂S is formed between a highly electropositive alkali metal (K) and a chalcogen (S), resulting in a typical ionic compound with K⁺ and S²⁻ ions in an ionic lattice. K₂S exhibits more characteristics of an ionic compound compared to BeF₂ (often described as having significant covalent character or being polymeric/covalent in gaseous state) or the definitively covalent SiO₂ and NCl₃.

96. Which one of the following is an amphoteric oxide?

Which one of the following is an amphoteric oxide?

[amp_mcq option1=”MgO” option2=”P₄O₁₀” option3=”Na₂O” option4=”Al₂O₃” correct=”option4″]

This question was previously asked in
UPSC CDS-1 – 2024
Download PDFAttempt Online
An amphoteric oxide is one that exhibits both acidic and basic properties, meaning it can react with both acids and bases to form salts. Aluminium oxide (Al₂O₃) is a common example of an amphoteric oxide. It reacts with acids like HCl to form aluminium chloride and water (basic behaviour) and with bases like NaOH to form sodium aluminate (acidic behaviour).
Amphoteric oxides react with both acids and bases. Aluminium oxide (Al₂O₃) is an amphoteric oxide.
MgO (Magnesium oxide) and Na₂O (Sodium oxide) are basic oxides, typical of alkali and alkaline earth metal oxides. P₄O₁₀ (Phosphorus pentoxide) is an acidic oxide, typical of non-metal oxides, reacting with water to form phosphoric acid and with bases to form phosphates.

97. Which of the following statements with regard to the reaction given be

Which of the following statements with regard to the reaction given below are correct? CaO(s) + H₂O(l) → Ca(OH)₂(aq) + Heat

  • 1. Quicklime is used for white-washing of walls.
  • 2. The solution of slaked lime is used for whitewashing of walls.
  • 3. CaO reacts slowly with CO₂ in air to form a thin layer of CaCO₃ on walls.
  • 4. Calcium hydroxide called ‘slaked lime’ is an inorganic compound.

Select the correct answer using the code given below.

[amp_mcq option1=”1 and 4 only” option2=”2 and 4″ option3=”2 and 3″ option4=”1, 3 and 4″ correct=”option2″]

This question was previously asked in
UPSC CDS-1 – 2023
Download PDFAttempt Online
The reaction given is CaO(s) + H₂O(l) → Ca(OH)₂(aq) + Heat. This is the slaking of quicklime (CaO) to produce slaked lime (Ca(OH)₂).
Statement 1: Quicklime (CaO) is used for white-washing of walls. While quicklime is the starting material, it is typically slaked with water to form calcium hydroxide (slaked lime) before being applied for whitewashing. So, this statement is not strictly accurate as CaO itself is not directly applied.
Statement 2: The solution of slaked lime (Ca(OH)₂) is used for whitewashing of walls. Correct. A suspension or solution of calcium hydroxide (slaked lime) is the material commonly applied as whitewash.
Statement 3: CaO reacts slowly with CO₂ in air to form a thin layer of CaCO₃ on walls. Incorrect. It is calcium hydroxide (slaked lime, Ca(OH)₂) that reacts with carbon dioxide in the air over time to form calcium carbonate (CaCO₃), which hardens and gives a shiny finish to the whitewash. CaO itself reacts readily with CO₂, but the reaction relevant to the setting of whitewash is between Ca(OH)₂ and CO₂.
Statement 4: Calcium hydroxide called ‘slaked lime’ is an inorganic compound. Correct. Calcium hydroxide is a compound of calcium, oxygen, and hydrogen, without carbon-hydrogen bonds, classifying it as an inorganic compound.
Based on the analysis, statements 2 and 4 are correct, and statements 1 and 3 are incorrect or less accurate.
Slaked lime (Ca(OH)₂) solution is used for whitewashing. Slaked lime reacts with CO₂ in air to form CaCO₃.
The process of whitewashing involves applying slaked lime (calcium hydroxide) to walls. The applied calcium hydroxide then reacts slowly with atmospheric carbon dioxide to form a hard layer of calcium carbonate, which is durable and provides a good white finish. CaO is also known as quicklime or burnt lime, and Ca(OH)₂ is also known as hydrated lime or slaked lime.

98. When we heat lead nitrate [Pb(NO₃)₂] in a boiling tube, we observe the

When we heat lead nitrate [Pb(NO₃)₂] in a boiling tube, we observe the emission of brown fumes. Which one of the following is the brown gas ?

[amp_mcq option1=”NO” option2=”N₂O₂” option3=”NO₃” option4=”NO₂” correct=”option4″]

This question was previously asked in
UPSC CDS-1 – 2021
Download PDFAttempt Online
When lead nitrate [Pb(NO₃)₂] is heated, it undergoes thermal decomposition, producing lead(II) oxide (PbO), oxygen gas (O₂), and nitrogen dioxide gas (NO₂). The brown fumes observed are due to nitrogen dioxide (NO₂).
– The balanced chemical equation for the decomposition of lead nitrate is: 2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g).
– Nitrogen dioxide (NO₂) is a toxic, reddish-brown gas.
– Nitrogen monoxide (NO) is a colourless gas.
– N₂O₂ is a dimer of NO, and while related to nitrogen oxides, it is not the species responsible for the prominent brown fumes in this decomposition.
– NO₃ is a highly reactive radical and not the stable product gas observed as brown fumes.

99. The PCl$_{5}$ molecule has trigonal bipyramidal structure. Therefore,

The PCl$_{5}$ molecule has trigonal bipyramidal structure. Therefore, the hybridization of p orbitals should be

[amp_mcq option1=”sp$^{2}$” option2=”sp$^{3}$” option3=”dsp$^{2}$” option4=”dsp$^{3}$” correct=”option4″]

This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The PCl$_{5}$ molecule has a trigonal bipyramidal structure with the Phosphorus atom as the central atom. In PCl$_{5}$, the central Phosphorus atom forms five single bonds with the five Chlorine atoms and has no lone pairs. According to VSEPR theory, a central atom with five electron domains (five bonding pairs) has a trigonal bipyramidal electron geometry and molecular geometry. This geometry is associated with sp$^3$d hybridization of the central atom. The sp$^3$d hybridization involves mixing one s atomic orbital, three p atomic orbitals, and one d atomic orbital to form five hybrid orbitals. The option dsp$^3$ refers to the same type of hybridization, where one d, one s, and three p orbitals are mixed, resulting in five hybrid orbitals oriented in a trigonal bipyramidal arrangement. Among the given options, dsp$^3$ is the only hybridization scheme that corresponds to the trigonal bipyramidal structure of PCl$_{5}$.
– PCl$_{5}$ has a trigonal bipyramidal structure.
– This structure arises from the hybridization of the central atom’s valence orbitals.
– A trigonal bipyramidal geometry corresponds to sp$^3$d (or dsp$^3$) hybridization.
– The central P atom has 5 bonding pairs and 0 lone pairs.
The common notation for main group elements with 5 coordination is sp$^3$d hybridization, involving one s, three p, and one d orbital. The dsp$^3$ notation, while also representing a mix of one d, one s, and three p orbitals, is sometimes used depending on whether the d-orbital is an inner or outer orbital, or simply as an alternative way to list the contributing orbitals. In the context of the given options, dsp$^3$ is the correct choice corresponding to the PCl$_{5}$ structure.

100. Which one of the following is not a characteristic of a compound?

Which one of the following is not a characteristic of a compound?

[amp_mcq option1=”Composition is variable.” option2=”All particles of compound are of only one type.” option3=”Particles of compound have two or more elements.” option4=”Its constituents cannot be separated by simple physical methods.” correct=”option1″]

This question was previously asked in
UPSC CDS-1 – 2018
Download PDFAttempt Online
A compound is a pure substance formed when two or more different elements are chemically bonded together in a fixed ratio. Therefore, the composition of a compound is constant and definite, not variable. Mixtures, on the other hand, have variable compositions.
A defining characteristic of a compound is its fixed and definite composition by mass.
Other characteristics of compounds include that all particles (molecules or formula units) are identical, they contain atoms of two or more elements chemically joined, and their constituent elements lose their individual properties when forming the compound. The elements in a compound can only be separated by chemical means, not simple physical methods.
Page 1Page 2Page 3Page 4Page 5