Question 1

A transition metal M gives a volatile chloride whose vapour density is 94.8. If this chloride contains 74.75% chlorine by mass, what is the formula of the metal chloride?

Accepted Answer

MCl₄. MW = 2 x 94.8 = 189.6; Cl mass = 0.7475 x 189.6 = 141.7 g -> 141.7/35.5 = 4 Cl atoms, so MCl4. Stored MCl2 is wrong.

Question 2

What is the ratio of the magnetic moments of Fe(III) and Co(II)?

Accepted Answer

√7: √3. The magnetic moment is determined by the number of unpaired electrons in the d-orbitals of the transition metals. Fe(III) has 5 unpaired electrons, leading to a magnetic moment of √35, while Co(II) has 4 unpaired electrons, resulting in a magnetic moment of √24. The ratio of their magnetic moments simplifies to √7: √3.

Question 3

Which of the following species contains the greatest number of unpaired electrons?

Accepted Answer

Fe2+. Mg2+ (0), Ti3+ 3d1 (1), V3+ 3d2 (2), Fe2+ 3d6 (4 unpaired). Fe2+ has the most unpaired electrons, so the answer is Fe2+, not V3+.

Question 4

Which of the following species contains an atom in the +6 oxidation state?

Accepted Answer

CrO2Cl2. In chromyl chloride CrO2Cl2, Cr is +6 (2 O at -2 and 2 Cl at -1). The stored Cr2O3 has Cr in +3, and MnO4- is +7, [Cr(CN)6]3- is +3.

Question 5

Which of the following pairs contains compounds in which the metal in each case is present in its maximum possible oxidation state?

Accepted Answer

CrO2Cl2, MnO4−. CrO2Cl2 (Cr +6) and MnO4- (Mn +7) both have the metal at its maximum oxidation state. The stored pair includes MnO2 (Mn +4), which is not the maximum.

Question 6

Among the following metal oxides, which one exhibits antiferromagnetic behavior?

Accepted Answer

MnO2. MnO2 exhibits antiferromagnetic behavior due to its crystal structure and the presence of manganese ions, which have unpaired electrons that align oppositely in neighboring ions, resulting in a net magnetic moment of zero.

Question 7

Which oxide exhibits metallic-type electrical behavior?

Accepted Answer

Chromium dioxide. Chromium dioxide is a transition metal oxide that exhibits metallic conductivity due to its partially filled d-orbitals, allowing for the movement of electrons, unlike the other options which are insulators or semiconductors.

Question 8

Bronze is an alloy made by combining which pair of metals?

Accepted Answer

Copper and tin. Bronze is an alloy of copper and tin. (Copper + zinc is brass; lead + tin is solder.)

Question 9

Which formula represents yellow ammonium sulphide?

Accepted Answer

(NH4)2Sx. Yellow ammonium sulphide is an ammonium polysulphide, written generally as (NH4)2Sx, where the polysulphide chain (x sulphur atoms) gives the yellow colour. Plain (NH4)2S is colourless.

Question 10

Among the elements having the following outer electronic configurations, which one can show the greatest number of oxidation states?

Accepted Answer

3d⁵ 4s². The element with the outer electronic configuration 3d⁵ 4s² can exhibit multiple oxidation states due to the half-filled d subshell, which provides stability and allows for the loss of varying numbers of electrons from both the s and d orbitals.

Question 11

The catalytic action shown by transition metals and their compounds is chiefly attributed to their:

Accepted Answer

capacity to exhibit multiple oxidation states. Transition metals and their compounds act as catalysts chiefly because they exhibit multiple (variable) oxidation states, allowing them to form intermediate complexes and provide alternative low-energy reaction paths.

Question 12

What is observed when an excess of dilute nitric acid is added to a solution of potassium chromate?

Accepted Answer

Dichromate ions and water are produced.. When dilute nitric acid is added to potassium chromate, the acidic environment facilitates the conversion of chromate ions to dichromate ions, which is a common reaction in acidic solutions, resulting in the formation of dichromate ions and water.

Question 13

Using the following standard reduction potentials:
E°(Sc³+/Sc) = −0.37, E°(Mn³+/Mn²+) = +1.57,
E°(Cr²+/Cr) = −0.90, E°(Cu²+/Cu) = +0.34,
which statement is not correct?

Accepted Answer

Mn³+ is more stable than Mn²+.. E(Mn3+/Mn2+) = +1.57 V is large and positive, so Mn3+ is readily reduced to Mn2+; the half-filled d5 Mn2+ is the more stable species. Hence the statement that Mn3+ is more stable than Mn2+ is the incorrect one.

Question 14

In which of the following species is the observed colour not due to a d–d electronic transition?

Accepted Answer

All of these. In Cr2O7^2-, CrO4^2-, and CrO2Cl2 chromium is in the +6 state (d0), so there are no d electrons for d-d transitions. Their colour arises from ligand-to-metal charge transfer, so 'All of these' is correct.

Question 15

Among the following lanthanides, which element shows the greatest stability in the +2 oxidation state?

Accepted Answer

Eu. Europium (Eu) is known for its stable +2 oxidation state due to its electron configuration, which allows it to easily lose two electrons and achieve a stable half-filled f-orbital. This stability is less pronounced in the other lanthanides listed.

Question 16

On strong heating, which of the following nitrates decomposes to give the metal as the residue?

Accepted Answer

Silver nitrate. Nitrates of very unreactive (noble) metals decompose to the metal itself: 2AgNO3 -> 2Ag + 2NO2 + O2. Cu, Mn and Fe nitrates give the metal oxide, not the metal.

Question 17

Which of the following sequences does not show the correct increasing order of the property mentioned alongside it?

Accepted Answer

V2+ < Cr2+ < Mn2+ < Fe2+: paramagnetic nature. Paramagnetism follows unpaired electrons: V2+(d3)=3, Cr2+(d4)=4, Mn2+(d5)=5, Fe2+(d6)=4. So the order V2+ < Cr2+ < Mn2+ < Fe2+ is wrong because Fe2+ (4) is less paramagnetic than Mn2+ (5). The other three orders (ionic radius, +3 aqueous stability, number of oxidation states) are correct.

Question 18

Which of the following is considered the principal reason for lanthanoid contraction?

Accepted Answer

Poor shielding of one 4f electron by another within the same subshell. Lanthanoid contraction arises because 4f electrons shield one another from the nuclear charge very poorly (imperfect/poor shielding of one 4f electron by another). The effective nuclear charge therefore increases across the series, steadily contracting the atomic/ionic radii.

Question 19

Which of the following elements exhibits the greatest variety of oxidation states in its compounds?

Accepted Answer

Am. Americium (Am) is an actinide and shows a wide range of oxidation states (+2, +3, +4, +5, +6), far more than the lanthanides Eu (+2,+3) or Gd (+3). The listed option 'Ld' is not a real element, so Am is the answer.

Question 20

When concentrated sulfuric acid is added to potassium permanganate, an explosive compound is produced. Which product is formed in this reaction?

Accepted Answer

Mn2O7. The reaction between concentrated sulfuric acid and potassium permanganate produces manganese heptoxide (Mn2O7), which is known for its explosive properties. This compound forms due to the oxidation of manganese in the permanganate ion under acidic conditions.

Question 21

Which of the following electronic configurations has a spin-only magnetic moment of 2.82 BM?

Accepted Answer

d4 in a strong ligand field. The d4 configuration in a strong ligand field results in a low-spin state with two unpaired electrons, leading to a calculated spin-only magnetic moment of 2.82 BM, which corresponds to the presence of two unpaired spins.

Question 22

A red-colored solid does not dissolve in water, but it dissolves when a little KI is added to the water. On heating this red solid in a test tube, violet vapours are evolved and tiny droplets of a metal deposit on the cooler part of the tube. The red solid is

Accepted Answer

HgI2. The red solid is HgI2 (mercury(II) iodide), which is known to be insoluble in water but dissolves in potassium iodide (KI) due to the formation of a complex ion. Upon heating, HgI2 decomposes, releasing violet iodine vapors and leaving behind metallic mercury as droplets.

Question 23

Among four consecutive elements of the first transition series given below, which one has a positive standard reduction potential for the M2+/M couple (E°M2+/M)?

Accepted Answer

Cu (Z = 29). Among Fe, Co, Ni, Cu, only copper has a positive standard reduction potential for M2+/M (E = +0.34 V); the others (Fe -0.44, Co -0.28, Ni -0.25 V) are all negative. This is why Cu does not displace H2 from acids.

Question 24

What is the magnetic moment of the Gd3+ ion (Z = 64)?

Accepted Answer

7.9 BM. Gd3+ has the 4f7 configuration with 7 unpaired electrons, giving spin-only moment = sqrt(7(7+2)) = sqrt(63) = 7.9 BM.

Question 25

An aqueous solution of a metal ion (M) gives a yellow precipitate (B) with ammonium sulfide, (NH4)2S, and the precipitate remains insoluble in ammonium polysulfide, (NH4)2S2. Which cation is present in (M)?

Accepted Answer

Cd2+. The formation of a yellow precipitate with ammonium sulfide indicates the presence of cadmium ions (Cd2+), as cadmium sulfide (CdS) is known for its characteristic yellow color. Additionally, the insolubility of the precipitate in ammonium polysulfide further confirms that the metal ion is cadmium, as it does not dissolve in this reagent.

Question 26

The major components in “Gun Metal” are:

Accepted Answer

Cu, Sn and Zn. Gun metal is primarily an alloy of copper, tin, and zinc, which gives it strength and corrosion resistance, making it suitable for various applications.

Question 27

Which commercially available form of iron is the most highly purified?

Accepted Answer

Wrought iron. Wrought iron is the most highly purified form of iron, characterized by its low carbon content and the absence of impurities, making it malleable and ductile, unlike pig iron and cast iron which contain higher levels of carbon and other elements.

Question 28

Which one of the following nitrates will leave behind a metal on strong heating?

Accepted Answer

Silver nitrate. Silver nitrate decomposes upon strong heating to produce silver metal, nitrogen dioxide, and oxygen, leaving behind solid silver. This property distinguishes it from the other nitrates listed, which do not yield a metal upon decomposition.

Question 29

Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect?

Accepted Answer

The +4 oxidation state of cerium is not known in solutions. The statement is incorrect because cerium can exist in the +4 oxidation state in solutions, although it is less stable compared to the +3 state. This means that while +4 cerium can be present, it is not as commonly found in solution as the +3 state.

Question 30

The lanthanide contraction is responsible for the fact that

Accepted Answer

Zr and Hf have about the same radius. The lanthanide contraction leads to a decrease in the size of the lanthanide ions, which in turn affects the size of the elements in the same period. As a result, zirconium (Zr) and hafnium (Hf), which are in the same group, have similar atomic radii despite being in different periods.

Question 31

The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct one is

Accepted Answer

d⁴ (in strong ligand fields). The 'spin only' magnetic moment of 2.84 BM corresponds to a d⁴ configuration in a strong ligand field, where the electrons are paired up in the lower energy orbitals, resulting in a specific number of unpaired electrons that contribute to the magnetic moment.

Question 32

Which of the following factors may be regarded as the main cause of lanthanide contraction?

Accepted Answer

Poor shielding of one of the 4f electrons by another in the subshell. The correct option highlights that the poor shielding of 4f electrons from each other leads to a reduced effective nuclear charge felt by the outer electrons, resulting in a greater contraction of the lanthanide series as the atomic number increases.

Question 33

Lanthanoid contraction is caused due to

Accepted Answer

the imperfect shielding on outer electrons by 4f electrons from the nuclear charge. Across the lanthanoids the 4f electrons shield the outer electrons poorly from the increasing nuclear charge. This imperfect (weak) shielding lets the effective nuclear charge rise steadily, pulling electron shells inward and causing the gradual decrease in size known as the lanthanoid contraction.

Question 34

The "spin-only" magnetic moment [in units of Bohr magneton, (μB)] of Ni2+ in aqueous solution would be (At. No. Ni = 28)

Accepted Answer

2.84. The spin-only magnetic moment for transition metal ions can be calculated using the formula μₛ = √(n(n+2)), where n is the number of unpaired electrons. For Ni2+, which has 2 unpaired electrons, the calculation yields a magnetic moment of approximately 2.84 μB.

Question 35

Identify the incorrect statement among the following:

Accepted Answer

4f and 5f orbitals are equally shielded.. The incorrect statement is that 4f and 5f orbitals are equally shielded: 5f electrons are more diffuse and poorly shielded compared with 4f. The other statements are correct - La (5d^1, f^0) and Lu (5d^1, f^14) both have a partially filled d-orbital and no partially filled f-orbital, lanthanoid chemistry is very similar, and d-block elements show irregular properties.

Question 36

The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because

Accepted Answer

the 5f orbitals extend further from the nucleus than the 4f orbitals. The correct option is right because the 5f orbitals, being higher in energy and extending further from the nucleus compared to the 4f orbitals, allow for greater involvement in bonding and electron exchange, leading to a wider range of oxidation states in actinoids.

Question 37

Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being

Accepted Answer

lesser energy difference between 5f and 6d than between 4f and 5d. In actinoids the 5f, 6d and 7s orbitals are close in energy (small 5f-6d gap), so more electrons take part in bonding, giving a larger range of oxidation states than the lanthanoids where the 4f-5d gap is larger.

Question 38

Knowing that the chemistry of lanthanoids(Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect?

Accepted Answer

Ln (III) compounds are generally colourless.. Ln (III) compounds often exhibit vibrant colors due to f-f electronic transitions, contrary to the statement that they are generally colorless. This characteristic is a result of the partially filled 4f orbitals, which can absorb visible light.

Question 39

Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect?

Accepted Answer

Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.. Ferrous compounds tend to be more easily hydrolyzed than ferric compounds due to their lower charge density, which allows them to interact more readily with water molecules, leading to hydrolysis.

Question 40

Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E° M3+/M2+ value ?

Accepted Answer

Co(Z = 27). Cobalt has a higher E° M3+/M2+ value compared to the other transition metals listed due to its favorable half-cell reduction potential, which is influenced by its electronic configuration and the stability of its oxidation states.

Question 41

Which series of reactions correctly represents chemical reactions related to iron and its compound?

Accepted Answer

Fe O2, heat→ Fe3O4 CO, 600 °C→ FeO CO, 700 °C→ Fe. This option accurately depicts the reduction of iron oxides through a series of reactions involving heat and carbon monoxide, illustrating the transformation of iron from its oxide forms back to elemental iron.

Question 42

Match the catalysts to the correct processes:
Catalyst Process
(A) TiCl4 (i) Wacker process
(B) PdCl2 (ii) Ziegler - Natta polymerization
(C) CuCl2 (iii) Contact process
(D) V2O5 (iv) Deacon's process

Accepted Answer

(A)-(ii), (B)-(i), (C)-(iv), (D)-(iii). TiCl4 is used in Ziegler-Natta polymerization to produce polymers, PdCl2 is a catalyst in the Wacker process for converting alkenes to carbonyl compounds, CuCl2 is involved in the Deacon's process for producing chlorine, and V2O5 is a catalyst in the Contact process for synthesizing sulfuric acid.

Question 43

The colour of KMnO4 is due to:

Accepted Answer

L → M charge transfer transition. The color of KMnO4 arises from the L → M charge transfer transition, where electrons are excited from ligand orbitals (L) to metal orbitals (M), resulting in the absorption of specific wavelengths of light that give the compound its characteristic color.

Question 44

Which of the following compounds is metallic and ferromagnetic?

Accepted Answer

CrO2. CrO2 is a metallic compound that exhibits ferromagnetism due to the presence of chromium in a specific oxidation state, allowing for unpaired electrons that contribute to magnetic ordering.

Question 45

Which of the following compounds is not colored yellow ?

Accepted Answer

Zn2[Fe(CN)6]. (NH4)3[As(Mo3O10)4] (ammonium arsenomolybdate), BaCrO4 and K3[Co(NO2)6] are all yellow. Zn2[Fe(CN)6] (zinc ferrocyanide) is a white precipitate, so it is the one not coloured yellow.

Question 46

Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E°3+/2+ value?
(1) Fe (Z = 26)
(2) Co (Z = 27)
(3) Cr (Z = 24)
(4) Mn (Z = 25)

Accepted Answer

Co (Z = 27). Cobalt (Co) has a higher E°3+/2+ value compared to the other elements listed because it has a more favorable electron configuration and a stronger tendency to stabilize its +2 oxidation state, making it more efficient in accepting electrons.

Question 47

Which of the following arrangements does not represent the correct order of the property stated against it?
(1) Co3+ < Fe3+ < Cr3+ < Sc3+: stability in aqueous solution
(2) Sc < Ti < Cr < Mn: number of oxidation states
(3) V2+ < Cr2+ < Mn2+ < Fe2+: paramagnetic behaviour
(4) Ni2+ < Co2+ <

Accepted Answer

V2+ < Cr2+ < Mn2+ < Fe2+: paramagnetic behaviour. The correct option is right because the paramagnetic behavior of transition metal ions is influenced by the number of unpaired electrons, and in this case, the order of increasing paramagnetism should reflect the increasing number of unpaired electrons, which is not accurately represented in the given arrangement.

Question 48

Match the catalysts to the correct processes:
Catalyst
(A) TiCl3
(B) PdCl2
(C) CuCl2
(D) V2O5
Process
(i) Wacker process
(ii) Ziegler-Natta Polymerization
(iii) Contact process
(iv) Deacon's process

Accepted Answer

(A) - (ii), (B) - (i), (C) - (iv), (D) - (iii). The correct option matches each catalyst with its respective process based on their known roles in chemical reactions: TiCl3 is used in Ziegler-Natta polymerization for producing polymers, PdCl2 is essential in the Wacker process for converting alkenes to carbonyl compounds, CuCl2 is involved in Deacon's process for producing chlorine, and V2O5 is a catalyst in the Contact process for sulfuric acid production.

Question 49

When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3, a dark green product is formed which disproportionate in acidic solution to afford a dark purple solution, X is - [JEE-Main On line-2018]

Accepted Answer

Mn. Fusing MnO2 with KOH and an oxidizer (KNO3) gives green potassium manganate K2MnO4 (Mn +6), which disproportionates in acid to purple permanganate KMnO4. So X = Mn.

Question 50

The incorrect statement out of the following is - [JEE-Main On line-2018]

Accepted Answer

Cu2+ and Ni2+ ions give black precipitate with H2S in presence of HCl solution. With H2S in presence of HCl, only group-II cations like Cu2+ precipitate (black CuS). Ni2+ needs an ammoniacal/basic medium (group IV) and does not give a precipitate in acidic solution, so the statement that both Cu2+ and Ni2+ give a black precipitate with H2S in HCl is incorrect.

JEE Main Chemistry: The d- and f-Block Elements questions with solutions

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