Question 1

The count of d-electrons in Fe2+ differs from the number of:

Accepted Answer

p-electrons in a Chlorine atom. The number of d-electrons in Fe2+ is 6, which differs from the number of p-electrons in a Chlorine atom, making it the correct comparison.

Question 2

Which sequence of reactions accurately depicts the chemical processes involving iron and its associated compounds?

Accepted Answer

Fe reacts with dilute H2SO4 to form FeSO4, which further reacts with H2SO4 and O2 to produce Fe2(SO4)3, and upon heating, reverts to Fe.. The correct sequence involves iron reacting with dilute sulfuric acid to form iron(II) sulfate, which then reacts with sulfuric acid and oxygen to produce iron(III) sulfate, and upon heating, it reverts back to iron.

Question 3

When H2S interacts with an acidic solution of K2Cr2O7, which of the following compounds is not produced?

Accepted Answer

CrSO4. In acidic medium K2Cr2O7 oxidizes H2S to elemental sulfur while Cr2O7^2- is reduced to Cr^3+, forming Cr2(SO4)3 and K2SO4. Chromium is reduced only to the +3 state, never +2, so CrSO4 (Cr^2+) is not produced. The compound not formed is CrSO4 (index 0); the stored answer K2SO4 is in fact produced.

Question 4

When an excess amount of sodium thiosulfate is combined with copper sulfate in water, the resulting product is:

Accepted Answer

Sodium dithiosulfatocuprate(I). The reaction between excess sodium thiosulfate and copper sulfate in water results in the formation of sodium dithiosulfatocuprate(I), which is a complex ion.

Question 5

Among the given lanthanoid ions, which one exhibits diamagnetic behavior? (Atomic numbers: Ce = 58, Sm = 62, Eu = 63, Yb = 70)

Accepted Answer

Yb2+. Ytterbium(II) ion exhibits diamagnetic behavior due to its completely filled or empty orbitals, resulting in no unpaired electrons.

Question 6

Which of these statements is accurate?

Accepted Answer

Both Hg2+ and Hg22+ ions display divalent behavior.. Both Hg2+ and Hg22+ ions indeed display divalent behavior, as they can both form compounds with a +2 oxidation state.

Question 7

The shift in color from green to purple occurs because of:

Accepted Answer

the transformation of Mn⁶⁺ into Mn⁷⁺. The green colour is MnO4^2- (manganate, Mn in +6); on disproportionation/oxidation it becomes purple MnO4^- (permanganate, Mn in +7). So the green-to-purple shift is the conversion of Mn6+ to Mn7+, option (a).

Question 8

In the given reaction: 3K₂MnO₄ + 2H₂O + 4CO₂ → 2KMnO₄ + MnO₂ + 4KHCO₃, what role does CO₂ play?

Accepted Answer

To acidify the solution by producing KHCO₃. CO₂ plays a role in acidifying the solution by reacting with potassium ions to produce KHCO₃, which in turn affects the pH of the solution and influences the reaction.

Question 9

A clear aqueous solution contains nitrates of two different metals, X and Y. When this solution is mixed with aqueous NaCl, a white solid forms. This solid partially dissolves in hot water, leaving behind a residue P and producing a solution Q. Residue P dissolves in aqueous ammonia and al

Accepted Answer

Silver and Lead. The white precipitate formed is a mixture of silver chloride (AgCl) and lead chloride (PbCl₂). PbCl₂ partially dissolves in hot water, leaving AgCl as residue P, which dissolves in ammonia and sodium thiosulfate. The hot solution Q contains Pb²⁺ ions, which react with KI to form yellow lead iodide (PbI₂).

Question 10

Which of the following is an INCORRECT statement about potassium permanganate (KMnO4) and potassium dichromate (K2Cr2O7)?

Accepted Answer

Both are paramagnetic and exhibit colour. Both Mn in KMnO4 (+7, d⁰) and Cr in K2Cr2O7 (+6, d⁰) have no unpaired electrons, making both compounds diamagnetic, not paramagnetic. All other options are correct: both are strong oxidising agents in acid, O2 is involved in their industrial preparation, and the tetrahedral anions MnO4⁻ and Cr2O7²- involve sp³ (sometimes described as d³s) hybridisation.

Question 11

When excess aqueous KI is added to HgCl2 solution, a precipitate X forms first, which then dissolves in excess KI to give a solution Y. Which of the following statements about X and Y is correct?

Accepted Answer

The Hg atom in compound Y has sp³ hybridisation. HgCl2 + 2KI gives HgI2 (X, scarlet-red precipitate, not green), which dissolves in excess KI to form K2[HgI4] (Y). In [HgI4]²-, Hg is tetra-coordinated in a tetrahedral geometry, giving sp³ hybridisation. Nessler's reagent (Y) is alkaline, not neutral, so option D is wrong.

Question 12

Arrange the following ions in decreasing order of their magnetic moments: (P) V⁴+ (Q) Mn⁴+ (R) Fe³+ (S) Ni²+

Accepted Answer

R > Q > S > P. Fe³+ has 5 unpaired electrons (mu ≈ 5.92 BM), Mn⁴+ has 3 (mu ≈ 3.87 BM), Ni²+ has 2 (mu ≈ 2.83 BM), and V⁴+ has 1 (mu ≈ 1.73 BM). Hence the decreasing order is R > Q > S > P.

Question 13

Which of the following pairs of lanthanide ions both represent stable (commonly observed) oxidation states?

Accepted Answer

Eu2+ and Tb4+. Eu2+ ([Xe]4f7, half-filled 4f) and Tb4+ ([Xe]4f7, half-filled 4f) are both stabilised by the extra stability of the half-filled 4f configuration, making this the correct pair of stable oxidation states.

Question 14

Which of the following metals does NOT form an amalgam with mercury?

Accepted Answer

Fe. Iron (Fe) does not form an amalgam with mercury because mercury cannot wet iron. This is why mercury can be stored in iron containers. Zinc, gold, and most other metals readily amalgamate.

Question 15

Identify the INCORRECT statement(s) among the following:

Accepted Answer

Both sodium dichromate and potassium dichromate can be used as primary standards in volumetric estimations. In K2MnO4, Mn is +6; in CrO2Cl2 (chromyl chloride), Cr is also +6 - so statement A is correct. K2Cr2O7 on strong heating gives 4K2Cr2O7 -> 4K2CrO4 + 2Cr2O3 + 3O2 (green Cr2O3 formed, O2 evolved) - statement C is correct. KMnO4 + KOH(s) on heating gives K2MnO4 + MnO2 + H2O (black MnO2 or green manganate, not O2) - statement D is incorrect. Na2Cr2O7 is hygroscopic and NOT a primary standard - statement B is incorrect.

Question 16

In a laboratory experiment, an orange-colored solid compound A decomposes on heating to give a green solid residue B, a colorless gas C, and water vapor. The dry gas C, when passed over heated magnesium, produces a white solid D. D reacts with water to give a gas E, which forms dense white

Accepted Answer

Ammonium dichromate ((NH4)2Cr2O7). Ammonium dichromate (orange) decomposes on heating as: (NH4)2Cr2O7 -> Cr2O3 (green) + N2(g) + 4H2O. N2 over heated Mg gives Mg3N2 (white solid D). Mg3N2 + H2O -> Mg(OH)2 + NH3 (gas E). NH3 + HCl -> NH4Cl (dense white fumes).

Question 17

Among the following pairs of ions, identify the pair in which BOTH ions possess a half-filled 4f subshell and therefore exhibit a stable oxidation state.

Accepted Answer

Eu2+ and Tb4+. Eu2+ has configuration [Xe]4f⁷ and Tb4+ also has [Xe]4f⁷, both having a half-filled f subshell which imparts extra stability. All other pairs lack both ions with half-filled f shells simultaneously.

Question 18

Which of the following statements about periodic trends and properties of elements are INCORRECT? (A) Cr2+(g) has a greater magnetic moment than Co3+(g). (B) The atomic size of the group 13 elements follows the order B < Ga < Al < Tl. (C) Lanthanide contraction is responsible for the ionis

Accepted Answer

B and C only. Statement A: Both Cr2+ and Co3+ are d⁴ in gas phase (high-spin), so they have equal magnetic moments -- A is INCORRECT as a statement claiming Cr2+ has greater moment. Statement B: Due to poor d-electron shielding, Ga is actually smaller than Al, so the correct order is B < Al < Ga < Tl; the given order is INCORRECT. Statement C: Lanthanide contraction affects 5d transition metals, not 6p elements like Pb directly; Pb vs Sn IE difference is due to relativistic effects and inert pair effect, not lanthanide contraction. C is INCORRECT. Statement D: Large jump between IE3 (849) and

Question 19

When MnO2 undergoes alkaline oxidative fusion, it is converted into a manganate salt. What is the oxidation number of Mn in the product formed during this process?

Accepted Answer

+6. Alkaline oxidative fusion converts MnO2 to potassium manganate (K2MnO4). In K2MnO4, the oxidation number of Mn is +6, since 2(+1) + x + 4(-2) = 0 gives x = +6.

Question 20

In the periodic table, volatile metals are found in which group?

Accepted Answer

Group 12. Group 12 elements (Zn, Cd, Hg) are known as volatile metals because they have low boiling points compared to typical transition metals; mercury is even liquid at room temperature.

Question 21

Consider the disproportionation reaction: 3K2MnO4 + 2H2O + 4CO2 → 2X + Y + 4KHCO3. If the oxidation state of Mn in product X is n1 and in product Y is n2, find the value of (n1 + n2):

Accepted Answer

11. K2MnO4 contains Mn in +6 state. On reaction with CO2 and water (acidic environment), Mn(+6) disproportionates: 2 moles form KMnO4 [Mn = +7] and 1 mole forms MnO2 [Mn = +4]. So X = KMnO4 (n1 = 7), Y = MnO2 (n2 = 4), n1 + n2 = 11. (The original options 1-4 appear to be a scanning artifact; correct answer is 11.)

Question 22

From the list Mn2O7, CrO3, Cr2O3, CrO, MnO2, V2O4, which oxides are amphoteric?

Accepted Answer

MnO2 and Cr2O3. In transition metal chemistry: Mn2O7 (+7 Mn) is strongly acidic. CrO3 (+6 Cr) is acidic. CrO (+2 Cr) is basic. Cr2O3 (+3 Cr) is amphoteric - it dissolves in both acid and base. MnO2 (+4 Mn) is amphoteric. V2O4 (+4 V) is also amphoteric. Among the given options, MnO2 and Cr2O3 together represent amphoteric oxides.

Question 23

An ion X^(n+) of element with atomic number 26 has a magnetic moment of sqrt(24) B.M. Find the number of unpaired electrons and the value of n respectively.

Accepted Answer

4 and 2. Magnetic moment = sqrt(n*(n+2)) = sqrt(24). So n*(n+2) = 24 => n² + 2n - 24 = 0 => (n+6)(n-4) = 0 => n = 4 unpaired electrons. Element Z = 26 is Fe: [Ar] 3d⁶ 4s². Fe²+ (n=2): remove 2 electrons from 4s -> [Ar] 3d⁶. In 3d⁶: 4 unpaired (one pair + 4 singles by Hund's rule in 5 d-orbitals: 2 orbitals share 1 pair, 4 have 1 electron each - actually 3d⁶ gives 4 unpaired: t2g has 4 and eg has 2 in high-spin, giving 4 unpaired). So Fe²+ has 4 unpaired electrons. n = 2. Answer: 4 unpaired electrons and n = 2.

Question 24

Among the lanthanide ions La3+, Ce3+, Eu3+, and Lu3+, which has the greatest tendency to form coordination complexes?

Accepted Answer

Lu3+. All four ions carry a 3+ charge. Due to lanthanide contraction, the ionic radius decreases steadily from La3+ (largest) to Lu3+ (smallest). A smaller ionic radius gives a higher charge-to-radius ratio, making the ion a stronger Lewis acid and thus increasing its tendency to accept electron pairs from ligands. Therefore Lu3+ has the maximum complex-forming tendency.

Question 25

Chromite ore is fused with K2CO3 in the presence of air. The product gives a yellow solution Y. On acidification, Y gives an orange solution X. X is used as a powerful oxidizing agent. What are the oxidation states of the metal in the anion of X and Y respectively?

Accepted Answer

+6, +6. The process: 4FeCr2O4 + 8K2CO3 + 7O2 -> 8K2CrO4 (Y, yellow solution, chromate). 2K2CrO4 + H2SO4 -> K2Cr2O7 (X, orange, dichromate) + K2SO4 + H2O. In K2CrO4: oxidation state of Cr = +6. In K2Cr2O7: oxidation state of Cr = +6. So both X and Y have Cr in +6 state. Answer: +6, +6.

Question 26

Regarding CrO5 (chromium pentoxide), which of the following statements is/are correct? a. The oxidation number of Cr is +6. b. Four oxygen atoms in the molecule are involved in peroxide linkage. c. Only one oxygen atom has oxidation number -2.

Accepted Answer

All are correct. CrO5 (chromium pentoxide or chromium peroxide): structure has one Cr=O group (oxygen ON = -2) and two peroxide groups (-O-O-) where each O has ON = -1. So there are 4 peroxide oxygen atoms (from two O-O groups) and 1 terminal oxygen. Total oxygen count: 5 = 4+1 (correct). ON check: Cr + 1*(-2) + 4*(-1) = 0 => Cr = +6. So: a is correct (Cr = +6); b is correct (4 oxygens in peroxide linkage); c is correct (only 1 oxygen has ON = -2). All statements correct.

Question 27

A vanadium chloride compound has a magnetic moment of 1.73 BM. What is the formula of this vanadium chloride?

Accepted Answer

VCl4. mu = sqrt(n(n+2)) = 1.73 BM -> n(n+2) = 3 -> n=1. V in VCl2 is V²+ with config [Ar]3d³ (3 unpaired). V in VCl3 is V³+ with [Ar]3d² (2 unpaired). V in VCl4 is V⁴+ with [Ar]3d¹ (1 unpaired). V in VCl5 is V⁵+ with [Ar]3d⁰ (0 unpaired). Only VCl4 gives 1 unpaired electron and mu = sqrt(3) = 1.732 BM.

Question 28

In which of the following oxoanions are the M-O bond lengths NOT all identical?

Accepted Answer

Cr2O7²-. MnO4⁻: permanganate, tetrahedral, Mn in +7, all four Mn-O bonds are equivalent by symmetry. MnO4²-: manganate, also tetrahedral, all Mn-O bonds equivalent. CrO4²-: chromate, tetrahedral, all Cr-O bonds equivalent. Cr2O7²-: dichromate, two CrO4 tetrahedra sharing one oxygen vertex. The bridging O (Cr-O-Cr) has a longer, weaker bond than the terminal Cr=O bonds. Therefore in Cr2O7²-, all Cr-O bond lengths are NOT identical — the bridging Cr-O bond is longer than the terminal Cr-O (or Cr=O) bonds.

Question 29

Identify the pair in which BOTH ions produce a pink colour in their aqueous solutions.

Accepted Answer

(Mn²+, Co²+). Mn²+(aq) = pale pink. Co²+(aq) = pink. Fe²+ = pale green (eliminates A). Ni²+ = green (eliminates C). Co³+ is a strong oxidizer, not stable in aqueous solution (eliminates C). Cr³+ = violet/green (eliminates D). Only Mn²+ and Co²+ are both pink.

Question 30

Which of the following cations is NOT precipitated by the reagent combination NH4Cl + NH4OH?

Accepted Answer

Fe²+. In qualitative analysis (Group III A), NH4Cl suppresses OH⁻ concentration. Al(OH)3, Cr(OH)3, and Fe(OH)3 (very low Ksp) precipitate. Fe(OH)2 (higher Ksp) does not precipitate here; Fe²+ is detected later as FeS in Group III B.

Question 31

In the permanganate ion MnO4²- (manganate ion), how many d-pi to p-pi (dp-pp) bonds are present? (Consider that Mn uses d-orbitals to form pi bonds with oxygen's p-orbitals.)

Accepted Answer

4. In MnO4²- (manganate ion), manganese is in the +6 oxidation state with electron configuration [Ar] 3d¹. The ion is tetrahedral with 4 oxygen atoms bonded to Mn. Each Mn-O bond involves: (1) one sigma bond from O lone pair to Mn, and (2) one d-pi to p-pi back bond where the filled p-orbital of oxygen donates into empty/partially filled d-orbital of Mn. With 4 Mn-O bonds, there are 4 d-pi to p-pi bonds in MnO4²-.

Question 32

A very important laboratory reagent X is prepared from the naturally occurring ore pyrolusite. When pyrolusite is fused with alkali in the presence of O2, a green compound Y is produced. Y is converted to X by electrolysis or by treatment with ozone. Let a = oxidation state of the central

Accepted Answer

0. Pyrolusite = MnO2. Fusion: MnO2 + 2KOH + (1/2)O2 -> K2MnO4 (potassium manganate) + H2O. Y = MnO4²- (manganate ion, green color). Oxidation state of Mn in MnO4²-: x + 4(-2) = -2 => x = +6. So a = 6. X = MnO4⁻ (permanganate), produced by electrolytic oxidation or ozone. The permanganate ion MnO4⁻ has tetrahedral geometry; due to resonance, all four Mn-O bonds are equivalent. So b = 4. Therefore a - b = 6 - 4 = 2.

Question 33

Consider the reaction: 2 Cu²+ + 4 X⁻ -> Cu2X2 (solid) + X2. Which of the following can be X⁻?

Accepted Answer

I⁻. In this reaction, X⁻ is oxidized (X⁻ -> (1/2)X2 + e⁻) and Cu²+ is reduced to Cu⁺ (in Cu2X2). For this reaction to occur spontaneously, the reducing power of X⁻ must be sufficient to reduce Cu²+. Among the halogens: F⁻ is too stable (very poor reducing agent). Cl⁻ cannot reduce Cu²+ under normal conditions. Br⁻ also cannot reduce Cu²+ to Cu⁺. I⁻ is the only halide that spontaneously reduces Cu²+ to Cu⁺ giving the insoluble precipitate Cu2I2. This is a classic disproportionation/redox reaction used in iodometric titrations: 2Cu²+ + 4I⁻ -> Cu2I2 + I2.

Question 34

The sulphate salt X of a 3d-series element A, on heating, produces a mixture of two gases both containing sulphur. Adding a few drops of HNO3 to a solution of X, followed by KSCN, gives a blood-red colour; excess sodium oxalate removes the red colour by forming complex Y. Which of the foll

Accepted Answer

(B) Decomposition of X gives oxide of A in +3 state. A is iron (Fe). X is FeSO4 (ferrous sulphate). Heating: 4FeSO4 -> 2Fe2O3 + 4SO2 + O2, or at higher temperature SO3 also forms; two S-containing gases are SO2 and SO3. HNO3 oxidises Fe²+ to Fe³+; KSCN gives blood red [Fe(SCN)]²+. Excess oxalate forms [Fe(ox)3]³- (Y), displacing SCN-. A: Fe + dil HCl -> FeCl2 (+2 state, NOT +3) -> FALSE. B: Fe2O3 formed (+3 state) -> TRUE. C: SO2 and SO3 -> TRUE. D: Fe is paramagnetic -> FALSE. E: [Fe(ox)3]³- has Fe³+ (d⁵ high-spin, 5 unpaired) -> paramagnetic -> TRUE. Correct statements: B, C, E.

Question 35

Match the following species with their spin-only magnetic moments: List-I (Species): (P) Cu+ (Q) Fe³+ (R) Mn⁴+ (S) Sc³+ List-II (Magnetic moment in units of mu_B): (1) 0 (2) sqrt(35) (3) sqrt(3) (4) sqrt(15) (5) sqrt(8) Select the correct matching.

Accepted Answer

P -> 1; Q -> 2; R -> 4; S -> 1. P) Cu+: [Ar]3d¹⁰, 0 unpaired electrons. mu = sqrt(0*2) = 0. -> (1). Q) Fe³+: [Ar]3d⁵, 5 unpaired electrons (high spin). mu = sqrt(5*7) = sqrt(35). -> (2). R) Mn⁴+: [Ar]3d³, 3 unpaired electrons. mu = sqrt(3*5) = sqrt(15). -> (4). S) Sc³+: [Ar]3d⁰, 0 unpaired electrons. mu = 0. -> (1). Correct matching: P->1, Q->2, R->4, S->1.

Question 36

Assertion (A): The decrease in atomic radius per element in the actinide series is larger than the corresponding decrease in the lanthanide series. Reason (R): The 5f electrons in actinides are poor at shielding the nuclear charge from one element to the next within the series.

Accepted Answer

Both A and R are true and R is the correct explanation of A.. Actinide contraction (per element step) is indeed greater than lanthanide contraction because 5f electrons shield nuclear charge even more poorly than 4f electrons. Poor 5f shielding means each added proton has a greater net pull on the outer electrons, causing a larger reduction in radius. Thus R correctly explains A.

Question 37

How many of the following reagent combinations produce a pink-colored solution, precipitate, or product? (i) MnCl2 + H2S(g) + HCl (ii) Mn2+ + S2O8²- (persulfate) (iii) Mn2+ + NaBiO3 + HNO3 (iv) Mn2+ + PbO2 + HNO3 (v) K2MnO4 + Cl2 (vi) ZnCl2 + H2S + NH4OH(aq)

Accepted Answer

2. (i) MnCl2 + H2S in acidic (HCl) medium: in acidic conditions MnS does NOT precipitate (Ksp too large for acidic medium). No pink. (ii) Mn2+ + S2O8²-: persulfate oxidizes Mn2+ to MnO4- (purple/permanganate), which could be considered purple-pink. This gives pink/purple color. (iii) Mn2+ + NaBiO3 + HNO3: NaBiO3 is a strong oxidizer that converts Mn2+ to MnO4- (permanganate, purple-pink). PINK. (iv) Mn2+ + PbO2 + HNO3: PbO2 also oxidizes Mn2+ to MnO4-. PINK/purple. (v) K2MnO4 + Cl2: Cl2 oxidizes MnO4²- (green) to MnO4- (purple-pink). Could be counted as pink. (vi) ZnCl2 + H2S + NH4OH: gives wh

Question 38

In the laboratory, an aqueous solution of manganese(II) ions is oxidised by ________ to form ________. Choose the option that correctly fills both blanks.

Accepted Answer

Peroxydisulphate; Permanganate. In analytical chemistry, Mn²+(aq) is oxidised to the intensely violet permanganate ion MnO4⁻ using peroxydisulphate (persulphate) S2O8²- as the oxidising agent, often with an Ag⁺ catalyst. This is used as a qualitative test for Mn²+. Thiosulphate is a reducing agent, not an oxidiser. Sulphate has no oxidising power toward Mn²+ under lab conditions.

Question 39

A metal cation M²+ forms a black precipitate with acidified H2S that dissolves in hot concentrated HNO3. An aqueous solution of M²+ gives a brown precipitate on treatment with potassium ferrocyanide (K4[Fe(CN)6]). Which of the following statements are correct?

Accepted Answer

M²+(aq) reacts with excess KCN to release a gas.. M is copper (Cu). CuS is black, dissolves in hot conc. HNO3. Cu²+ gives reddish-brown precipitate with K4[Fe(CN)6]. A: Cu²+ + KCN(excess) → [Cu(CN)4]³- + (CN)2 gas — correct. B: Cu is extracted from copper pyrites by auto-reduction (2Cu2O + Cu2S → 6Cu + SO2) — correct. C: E0(Cu²+/Cu) = +0.34 V > 0 — correct. D: Deep blue colour with excess NH3 (not Cl⁻) — incorrect.

Question 40

Arrange the following ions in decreasing order of their magnetic moments: (A) V^+4, (B) Mn^+4, (C) Fe^+3, (D) Ni^+2.

Accepted Answer

C > B > D > A. V^+4: [Ar]3d¹ — 1 unpaired e, mu = sqrt(3) ≈ 1.73 BM. Mn^+4: [Ar]3d³ — 3 unpaired e, mu = sqrt(15) ≈ 3.87 BM. Fe^+3: [Ar]3d⁵ — 5 unpaired e (high spin), mu = sqrt(35) ≈ 5.92 BM. Ni^+2: [Ar]3d⁸ — 2 unpaired e, mu = sqrt(8) ≈ 2.83 BM. Order: Fe³+ > Mn⁴+ > Ni²+ > V⁴+, i.e., C > B > D > A.

Question 41

Cr²+ acts as a reducing agent primarily because:

Accepted Answer

Its configuration changes from d⁴ to d³, achieving a half-filled t2g level in octahedral field.. Cr²+ has [Ar]3d⁴ configuration. It readily loses one electron to form Cr³+ ([Ar]3d³). In an octahedral crystal field, d³ corresponds to t2g³ eg⁰ — a half-filled t2g set, which is extra stable due to exchange energy. This thermodynamic driving force makes Cr²+ a good reducing agent.

Question 42

From the following ions, how many can be precipitated as sulfides from an acidic solution by passing H2S gas? Cd²+, Co²+, Cu²+, Ca²+, Mn²+, Mg²+, As³+, Al³+, Cr³+, Bi³+

Accepted Answer

4. In acidic solution H2S provides a low [S²-] sufficient only to exceed Ksp of the most insoluble sulfides (Group II). From the given list: Cu²+ (CuS — very insoluble), Cd²+ (CdS — very insoluble), As³+ (As2S3 — very insoluble), Bi³+ (Bi2S3 — very insoluble) precipitate in acidic medium. Co²+ (CoS), Mn²+ (MnS) require near-neutral conditions (Group III). Ca²+, Mg²+, Al³+, Cr³+ do not form sulfide precipitates under these conditions. Total = 4.

Question 43

The mineral hemimorphite has the formula Znₓ(OH)₂ Si₄ O₁₄ (where x is an integer). Determine the value of x.

Accepted Answer

4. Hemimorphite is a zinc silicate mineral. Its formula is Zn₄(OH)₂ Si₄ O₁₄ (sometimes written as Zn₄(Si₂O₇)(OH)₂ * H2O after factoring). Charge balance in Znₓ(OH)₂ Si₄ O₁₄: each Zn²+ contributes +2x. OH(-) contributes -2. Si₄O₁₄: Si is +4 each, O is -2 each. Charge from Si₄O₁₄ = 4*(+4) + 14*(-2) = 16 - 28 = -12. Total charge = 2x + (-2) + (-12) = 0. 2x - 14 = 0 => x = 7. Hmm that gives 7, not an option. Let me try Si₄O₁₁ for the silicate group. Or reconsider: in Znₓ(OH)₂Si₄O₁₁: 2x - 2 + 4*4 + 11*(-2) = 0 => 2x-2+16-22=0 => 2x-8=0 => x=4. The formula might be Znₓ(OH)₂ Si₄ O₁₁ not O₁₄. But the

Question 44

Chromite ore is fused with Na2CO3 in excess air to prepare K2Cr2O7. Which of the following statements about this process is correct?

Accepted Answer

The oxidation number of Cr in the fusion product is +6.. Chromite ore = FeCr2O4, where Fe is in +2 state (option A says +2, which is actually correct for Fe). But the statement says iron in +2, which is correct for chromite. However, option B about Cr being +6 in fusion product is the most definitively correct statement. In Na2CrO4, Cr is +6. CrO4²- is tetrahedral with all equivalent Cr-O bonds (option C is also correct). Cr(VI) is d⁰, so no unpaired electrons, yet it shows colour due to charge transfer (option D - this statement claims no unpaired electron AND it is coloured, which is true bu

Question 45

Solid sodium chloride is mixed with solid potassium dichromate and concentrated sulfuric acid is added. A reddish-brown gaseous product X is formed. How many d-orbitals participate in the hybridization of X?

Accepted Answer

1. The reaction is: K2Cr2O7 + 4NaCl + 6H2SO4 → 2CrO2Cl2 (chromyl chloride) +... Chromyl chloride (CrO2Cl2) is the reddish-brown fuming gas. Chromium in CrO2Cl2 is in the +6 oxidation state. The structure has Cr bonded to 2 oxygen atoms (double bonds) and 2 chlorine atoms, with no lone pairs on Cr, giving 4 electron domains. The hybridization is sp3 (some sources describe it as involving one d orbital giving sp2d or dsp2, but the tetrahedral geometry of Cr with 4 sigma bonds and pi bonds via d-orbitals means one d orbital is used in hybridization). Cr: [Ar] 3d5 4s1 in ground state; in CrO2Cl2 t

Question 46

Which of the following statements is/are incorrect? (A) Zr (Z = 40) belongs to the f-block elements (B) IE(Be) > IE(B) (C) Zn is a transition element (D) Gd (Z = 64) is an inner-transition element

Accepted Answer

(A) Zr (Z = 40) belongs to the f-block elements. Zr (Z=40) is a d-block element (group 4), not an f-block element (statement A is incorrect). Zn (Z=30) has fully filled d-orbitals in all oxidation states and is not a transition element by strict definition (statement C is also incorrect). Statements B and D are correct.

Question 47

How many of the following pairs of ions can be separated by passing H2S gas in dilute HCl medium? Pairs: Bi³+ and Sn⁴+; Al³+ and Hg²+; Cd²+ and Zn²+; Fe³+ and Cu²+; As³+ and Sb³+

Accepted Answer

3. In dilute HCl with H2S, only Group II sulfides precipitate. Pair 2 (Al³+ stays in solution; Hg²+ precipitates), Pair 3 (Zn²+ stays; Cd²+ precipitates), and Pair 4 (Fe³+ stays; Cu²+ precipitates) can be separated. Pairs 1 and 5 both precipitate together. Total = 3 pairs.

Question 48

In the context of lanthanoids (elements La through Lu), which of the following statements are correct? (Select all that apply.)

Accepted Answer

Because of similar properties, the separation of lanthanoids is not easy.. A: TRUE — lanthanoids have very similar chemical properties due to similar outer electronic configurations, making their separation by conventional methods difficult (ion-exchange chromatography is needed). B: FALSE — not all lanthanoids form +4 compounds; only Ce, Pr, and Tb commonly do. C: TRUE — lanthanoid contraction: successive addition to 4f subshell provides poor shielding, leading to gradual decrease in ionic/atomic radius across the series. D: TRUE — all lanthanoids exhibit the +3 oxidation state as their chara

Question 49

Identify the incorrect statement from the following:

Accepted Answer

The red colour of ruby is due to the presence of Co3+ ions.. Ruby is a form of corundum (Al2O3) in which some Al3+ ions are replaced by Cr3+ ions; the d-d transitions of Cr3+ in the crystal field produce the characteristic deep red colour. Co3+ is not present in ruby. The other statements are correct: VOSO4 contains V4+ which can be oxidised to V5+ or it acts as oxidant by being reduced; Cr2O3 is indeed amphoteric (dissolves in both acid and base); RuO4 is a powerful oxidizing agent used in organic synthesis.

Question 50

Match each element in Column-I with its correct properties in Column-II. Column-I: (A) Ba (B) Pb (C) Ag (D) Ca Column-II: (P) Its cation in solution gives a brick-red precipitate with CrO4²- ions. (Q) Its cation in solution gives a yellow precipitate with CrO4²- ions. (R) Its salt gives an

Accepted Answer

(A)-(Q,R,T); (B)-(P,Q,S); (C)-(Q,T); (D)-(S). Ag2CrO4 is brick red (P -> C/Ag); BaCrO4 and PbCrO4 are yellow (Q -> A/Ba and B/Pb); Ba gives apple green flame (R -> A/Ba); Ca gives brick red flame (S -> D/Ca); CaCrO4 is soluble so no ppt (T -> D/Ca). The best match among the given options is option 1.

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

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