Inorganic Chemistry

61. Which one of the following is not used as a raw material in the manufa

Which one of the following is not used as a raw material in the manufacture of glass ?

[amp_mcq option1=”Soda” option2=”Alumina” option3=”Borax” option4=”Gypsum” correct=”option4″]

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Gypsum is not typically used as a primary raw material in the manufacture of standard glass.
The main raw materials for manufacturing common soda-lime glass are silica (sand), soda ash (sodium carbonate), and limestone (calcium carbonate). Alumina and borax are used in the production of specialized glasses like aluminosilicate glass and borosilicate glass, respectively. Gypsum (calcium sulfate) is primarily used in plaster, drywall, and cement production.
Minor components might be added to glass batches as fining agents (to remove bubbles), colorants, or flux enhancers, but gypsum is not a standard ingredient for bulk glass manufacturing.

62. Which one of the following oxides dissolves in water ?

Which one of the following oxides dissolves in water ?

[amp_mcq option1=”CuO” option2=”Al2O3” option3=”Fe2O3” option4=”Na2O” correct=”option4″]

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The correct answer is D) Na₂O.
Metallic oxides generally react with water to form bases. The solubility of metallic oxides in water varies. Oxides of alkali metals (Group 1) and some alkaline earth metals (Group 2, heavier ones) are soluble in water, reacting to form strong bases. Oxides of transition metals and other metals are typically insoluble or only slightly soluble in water.
– CuO is a copper(II) oxide, a transition metal oxide, which is insoluble in water. It is a basic oxide but dissolves in acids.
– Al₂O₃ is aluminium oxide, an amphoteric oxide (reacts with both acids and bases), and is generally insoluble in water.
– Fe₂O₃ is iron(III) oxide, an iron oxide, which is insoluble in water. It is a basic oxide.
– Na₂O is sodium oxide, an alkali metal oxide. It reacts vigorously with water to form sodium hydroxide (NaOH), a soluble base: Na₂O(s) + H₂O(l) → 2NaOH(aq). Thus, Na₂O effectively dissolves by reacting with water.

63. Statement-I: Oxygen gas is easily produced at a faster rate by heating

Statement-I: Oxygen gas is easily produced at a faster rate by heating a mixture of potassium chlorate and manganese dioxide than heating potassium chlorate alone.
Statement-II: Manganese dioxide acts as a negative catalyst.

[amp_mcq option1=”Both the statements are individually true and Statement II is the correct explanation of Statement I” option2=”Both the statements are individually true but Statement II is not the correct explanation of Statement I” option3=”Statement I is true but Statement II is false” option4=”Statement I is false but Statement II is true” correct=”option3″]

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The correct option is C. Statement I is true, but Statement II is false.
Statement I correctly describes the effect of adding manganese dioxide (MnO₂) to potassium chlorate (KClO₃) when heated. MnO₂ acts as a catalyst, speeding up the decomposition of KClO₃ into potassium chloride (KCl) and oxygen gas (O₂). The reaction is 2KClO₃ (s) → 2KCl (s) + 3O₂ (g).
Statement II claims that manganese dioxide acts as a negative catalyst. A negative catalyst (or inhibitor) slows down a reaction. Since MnO₂ increases the rate of decomposition of KClO₃, it is a positive catalyst.
Catalysts increase the rate of a chemical reaction without being consumed in the process. Positive catalysts increase the rate, while negative catalysts decrease it. Manganese dioxide is a common positive catalyst used in the laboratory preparation of oxygen from potassium chlorate.

64. Which one among the following trace elements is not highly toxic ?

Which one among the following trace elements is not highly toxic ?

[amp_mcq option1=”Arsenic” option2=”Chromium” option3=”Copper” option4=”Mercury” correct=”option3″]

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The question asks which trace element among the options is *not* highly toxic. While toxicity is dose-dependent for all substances, some trace elements are generally considered more acutely or cumulatively toxic than others, even at low concentrations.
– Arsenic (As) is a metalloid widely recognized as highly toxic and carcinogenic.
– Chromium (Cr) has both essential and toxic forms; Hexavalent chromium (Cr(VI)) is highly toxic and carcinogenic.
– Mercury (Hg) is a heavy metal that is highly toxic, particularly in its organic form (methylmercury), affecting the nervous system.
– Copper (Cu) is an essential trace element necessary for many biological processes. While excessive intake can cause toxicity, it is less likely to be classified as *highly toxic* compared to arsenic, mercury, and certain forms of chromium in common environmental exposures, as the body has mechanisms to regulate copper levels and it is required for life.
Essential trace elements like copper are required in small amounts, and deficiency or excess can lead to health problems. Non-essential trace elements like lead, cadmium, arsenic, and mercury are generally toxic even at low concentrations. Among the given options, copper stands out as being an essential nutrient, making it relatively less “highly toxic” in general contexts compared to the others which are primarily known for their toxicity.

65. Which one of the following Ni(II) complexes is diamagnetic and has tet

Which one of the following Ni(II) complexes is diamagnetic and has tetrahedral geometry ?

[amp_mcq option1=”[NiCl₄]²⁻” option2=”[Ni(CO)₄]” option3=”[Ni(NH₃)₄]²⁺” option4=”[Ni(CN)₄]²⁻” correct=”option2″]

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Among the given options, [Ni(CO)₄] is both diamagnetic and has tetrahedral geometry.
[Ni(CO)₄]: In this complex, Nickel is in the 0 oxidation state (Ni(0)). Its electron configuration is d¹⁰. All 10 d electrons are paired. The hybridization is sp³, resulting in a tetrahedral geometry. Since all electrons are paired, the complex is diamagnetic.
For Ni(II) complexes (d⁸):
– [NiCl₄]²⁻: Cl⁻ is a weak field ligand. Forms a tetrahedral complex (sp³ hybridization) which is paramagnetic (2 unpaired electrons).
– [Ni(NH₃)₄]²⁺: NH₃ is a moderate field ligand. Can form tetrahedral (sp³, paramagnetic) or square planar (dsp², diamagnetic) complexes depending on conditions, but the tetrahedral form is paramagnetic.
– [Ni(CN)₄]²⁻: CN⁻ is a strong field ligand. Forms a square planar complex (dsp² hybridization) which is diamagnetic (all electrons paired).
The question specifically asks for a Ni(II) complex that is diamagnetic and tetrahedral. Based on standard crystal field and valence bond theories, Ni(II) (d⁸) tetrahedral complexes are typically paramagnetic, while diamagnetic Ni(II) complexes are typically square planar. Therefore, none of the Ni(II) options (A, C, D) are both diamagnetic and tetrahedral. However, option B, [Ni(CO)₄], is tetrahedral and diamagnetic, fitting these two criteria perfectly, although it is a Ni(0) complex. This suggests a likely error in the question’s specification of “Ni(II) complexes”, and that option B was the intended answer despite this discrepancy. As per the provided options, B is the only complex listed that possesses both tetrahedral geometry and diamagnetic character.

66. Which one of the following types of isomerisms is shown by [CoCl₂(en)₂

Which one of the following types of isomerisms is shown by [CoCl₂(en)₂]Cl ?

[amp_mcq option1=”Optical isomerism only” option2=”Geometrical isomerism only” option3=”Both ionisation and geometrical isomerisms” option4=”Both geometrical and optical isomerisms” correct=”option4″]

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The complex ion is [CoCl₂(en)₂]⁺, which is an octahedral complex with coordination number 6. It exhibits both geometrical and optical isomerism.
The complex [CoCl₂(en)₂]⁺ is of the type [Ma₂b₂] (where ‘a’ is Cl and ‘b’ is the bidentate ligand ‘en’ represented by two coordination points). Such octahedral complexes can exist as cis and trans geometrical isomers. The trans isomer (where the two Cl ligands are opposite to each other) has a plane of symmetry and is optically inactive. The cis isomer (where the two Cl ligands are adjacent) lacks a plane of symmetry and is chiral, existing as a pair of enantiomers (non-superimposable mirror images), which are optically active.
Geometrical isomerism is shown by the existence of cis and trans forms. Optical isomerism is shown by the cis isomer due to its chirality. Ionisation isomerism involves the exchange of an ion inside the coordination sphere with a counter ion outside, which is not applicable here as the counter ion (Cl⁻) does not correspond to a ligand inside the sphere that can be exchanged. Linkage isomerism requires ambidentate ligands, which are not present here (en and Cl are not ambidentate).

67. Which one of the following is the IUPAC name of [Ag(NH₃)₂] [Ag(CN)₂] ?

Which one of the following is the IUPAC name of [Ag(NH₃)₂] [Ag(CN)₂] ?

[amp_mcq option1=”Diammine dicyanidodisilver(I)” option2=”Diammine dicyanidoargentate(I)” option3=”Diamminesilver(I) dicyanidoargentate(I)” option4=”Dicyanidoargentate(I) diamminesilver(I)” correct=”option3″]

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The compound is a salt consisting of a complex cation and a complex anion: [Ag(NH₃)₂]⁺ [Ag(CN)₂]⁻. The IUPAC name is derived by naming the cation complex followed by the anion complex.
For the cation [Ag(NH₃)₂]⁺: The metal is Silver (Ag). Ammonia (NH₃) is an neutral ligand named ‘ammine’. There are two ammine ligands (diammine). The charge of the complex is +1. Since NH₃ is neutral, the oxidation state of Ag is +1. As it is a cation, the metal name is used as is, followed by the oxidation state in Roman numerals. Name: Diamminesilver(I).
For the anion [Ag(CN)₂]⁻: The metal is Silver (Ag). Cyanide (CN⁻) is an anionic ligand named ‘cyanido’ or ‘cyano’. There are two cyanido ligands (dicyanido). The charge of the complex is -1. Since CN⁻ has a charge of -1, Ag + 2(-1) = -1, so the oxidation state of Ag is +1. As it is an anion, the suffix ‘-ate’ is added to the Latin name of the metal (Argentum for Silver), followed by the oxidation state. Name: Dicyanidoargentate(I).
Combining the cation and anion names gives the full IUPAC name: Diamminesilver(I) dicyanidoargentate(I). This matches option C. Option A incorrectly uses ‘disilver’ and doesn’t separate the cation and anion naming properly. Option B combines parts of both names but doesn’t represent the ionic structure accurately. Option D reverses the order of cation and anion names.

68. What is the colour of anhydrous CuSO₄ ?

What is the colour of anhydrous CuSO₄ ?

[amp_mcq option1=”White” option2=”Blue” option3=”Green” option4=”Yellow” correct=”option1″]

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The colour of anhydrous CuSO₄ is white.
Copper(II) sulfate exists in several hydration states. The most common form is copper(II) sulfate pentahydrate (CuSO₄·5H₂O), which is a bright blue crystalline solid. The blue colour is due to the presence of hydrated copper(II) ions, [Cu(H₂O)₄]²⁺, where water molecules act as ligands coordinated to the Cu²⁺ ion. The d-d electronic transitions within the Cu²⁺ ion, influenced by the surrounding water ligands, cause the absorption of certain wavelengths of visible light (specifically red-orange light), resulting in the complementary blue colour being observed.
When copper(II) sulfate pentahydrate is heated, it loses its water of crystallization. The blue pentahydrate first turns into pale blue CuSO₄·3H₂O, then blue-green CuSO₄·H₂O, and finally becomes anhydrous copper(II) sulfate (CuSO₄), which is a white powder. In the anhydrous state, the crystal structure changes, and the Cu²⁺ ion is no longer surrounded by water ligands in the same way. This alters the energy levels of the d orbitals, and the absorption characteristics change such that no visible light is strongly absorbed, causing the compound to appear white.
The reaction of anhydrous copper sulfate with water is exothermic and results in the formation of the blue hydrated salt. This reaction is often used as a test for the presence of water.

69. Which one among the following elements has the most negative electron

Which one among the following elements has the most negative electron gain enthalpy ?

[amp_mcq option1=”F” option2=”Cl” option3=”P” option4=”S” correct=”option2″]

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Among the given elements, Chlorine (Cl) has the most negative electron gain enthalpy.
Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form a uninegative anion. A more negative value indicates a greater tendency to accept an electron.
Fluorine (F) and Chlorine (Cl) are halogens (Group 17), which generally have high (very negative) electron gain enthalpies because adding an electron gives them a stable noble gas configuration. Phosphorus (P) and Sulfur (S) are in periods 3. P is in Group 15, S is in Group 16.
Electron gain enthalpy generally becomes more negative across a period and less negative down a group.
Halogens (Group 17) have the most negative electron gain enthalpies. Comparing F and Cl, due to the small size of F, the added electron experiences significant electron-electron repulsion in the compact 2p subshell. In Cl, the added electron enters the larger 3p subshell, where repulsions are less significant. Consequently, Cl has a more negative electron gain enthalpy than F.
Comparing P and S: S (Group 16) has a more negative electron gain enthalpy than P (Group 15). Adding an electron to S (3p⁴) gives a stable 3p⁵ configuration. Adding an electron to P (3p³) disrupts the stable half-filled configuration, making it less favorable, leading to a less negative or even positive electron gain enthalpy for P.
Generally, electron gain enthalpy order for these elements is approximately: P < S < F < Cl (from least negative to most negative). Therefore, Cl has the most negative electron gain enthalpy among the given options.
While halogens generally have the most negative electron gain enthalpies, the electron gain enthalpy of chlorine is more negative than that of fluorine. Values (kJ/mol): F = -328, Cl = -349, S = -200, P = -72.

70. Which one among the following is the correct arrangement in increasing

Which one among the following is the correct arrangement in increasing order of ionic radii of O²⁻, F⁻, Na⁺, Mg²⁺ ?

[amp_mcq option1=”Mg²⁺ < Na⁺ < F⁻ < O²⁻" option2="O²⁻ < F⁻ < Na⁺ < Mg²⁺" option3="Mg²⁺ < Na⁺ < O²⁻ < F⁻" option4="O²⁻ < F⁻ < Mg²⁺ < Na⁺" correct="option1"]

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The correct arrangement in increasing order of ionic radii is Mg²⁺ < Na⁺ < F⁻ < O²⁻.
The given ions O²⁻, F⁻, Na⁺, and Mg²⁺ are isoelectronic species, meaning they all have the same number of electrons (10 electrons, with the electronic configuration $1s^2 2s^2 2p^6$).
For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases. A higher nuclear charge exerts a stronger attraction on the same number of electrons, pulling them closer to the nucleus and reducing the ionic size.
The atomic numbers are: Oxygen (O) = 8, Fluorine (F) = 9, Sodium (Na) = 11, Magnesium (Mg) = 12.
The effective nuclear charges are approximately +8 for O²⁻, +9 for F⁻, +11 for Na⁺, and +12 for Mg²⁺.
As the nuclear charge increases (from +8 to +12), the attraction on the 10 electrons increases, leading to a decrease in ionic size.
Order of increasing nuclear charge: O²⁻ (+8) < F⁻ (+9) < Na⁺ (+11) < Mg²⁺ (+12). Order of decreasing ionic radii: O²⁻ > F⁻ > Na⁺ > Mg²⁺.
Order of increasing ionic radii: Mg²⁺ < Na⁺ < F⁻ < O²⁻.
Cations are generally smaller than their parent atoms because they lose valence electrons and the remaining electrons are held more tightly by the nucleus. Anions are generally larger than their parent atoms because they gain electrons, increasing electron-electron repulsion and expanding the electron cloud. Among isoelectronic species, negative ions are larger than positive ions, and within negative ions, the size increases with increasing negative charge. Within positive ions, the size decreases with increasing positive charge.
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