If the uncertainty in velocity and position of a minute particle in space are, 2.4 × 10-26(ms-1) and 10-7(m) respectively. The mass of the particle in g is __________. (Given: h = 6.626 × 10-34g)

Δ V = 2 . 4 × 1 0 − 2 6 m / s e c And Δ x = 1 0 − 7 m According to Heisenberg’s uncertainty Δ v × Δ x = h 4 π × m ∴ m = h 4 π × Δ v × Δ x = 6 . 6 2 6 × 1 0 − 3 4 4 × 3 . 1 4 × 2 . 4 × 1 0 − 2 6 × 1 0 − 7 ∴ m = 0.02198 Kg Or 21.98 gm Or 22 gm in the nearest integer.

Identify the incorrect statement for PCl5 from the following

Structure PCl5 is trigonal bipyramidal, Hybridization of P is sp3d Equatorial bonds lie in a plane Axial bonds are longer than equatorial bonds so axial bonds are weaker than equatorial bonds.

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27°C to 42°C. Its energy of activation in J/mol is (Take in 5 = 1.6094; R = 8.314 J mol⁻¹ K⁻¹)

k = Ae? Ea/RT ln (K? /K? ) = (Ea/R) [1/T? - 1/T? ] ln (5) = (Ea/8.314) [1/300 - 1/315] 1.6094 = (Ea/8.314) [15 / (300 × 315)] Ea = 84297 J

Consider the following equations : 2Fe²⁺ + H₂O₂ → xA + yB (in basic medium) 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → x'C + y'D + z'E (in acidic medium) The sum of the stoichiometric coefficients x, y, x', y' and z' for products A, B, C, D and E, respectively, is

(i) 2Fe²? + H? O? → 2Fe³? + 2OH? (ii) 2MnO? + 5H? O? + 6H? → 2Mn²? + 5O? + 8H? O So sum of (x + y + x¹ + y¹ + z¹) = 2 + 2 + 2 + 5 + 8 = 19

The number of chiral centres present in threonine is

Threonine has two chiral carbon atoms.

The molecule in which hybrid MOs involve only one d-orbital of the central atom is :

[Ni(CN)₄]²⁻

[CrF₆]³⁻

The mechanism of action of "Terfenadine" (Seldane) is :

Inhibits the action of histamine receptor

Inhibits the secretion of histamine

Helps in the secretion of histamine

Activates the histamine receptor

The incorrect statement(s) among (a) - (c) is (are): (a) W(VI) is more stable than Cr(VI). (b) in the presence of HCl, permanganate titrations provide satisfactory results. (c) some lanthanoid oxides can be used as phosphors.

(b) only

(a) only

(a) and (b) only

Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :

Cᵥ(T₂ - T₁)

-RTV₂/V₁

The process that is NOT endothermic in nature is

H(g) + e⁻ → H⁻(g)

O⁻(g) + e⁻ → O²⁻(g)

Na(g) → Na⁺(g) + e⁻

Ar(g) + e⁻ → Ar⁻(g)

The processes of calcination and roasting in metallurgical industries, respectively, can lead to:

Global warming and acid rain

Photochemical smog and global warming

Global warming and photochemical smog

Photochemical smog and ozone layer depletion

The one that can exhibit highest paramagnetic behaviour among the following is: gly = glycinato ; bpy = 2,2 '-bipyridine

[Co(OX)₂(OH)₂]⁻ (Δ₀ > P)

[Fe(en)(bpy)(NH₃)₂]²⁺

[Pd(gly)₂]

[Ti(NH₃)₆]³⁺

An alkaline earth metal ' M' readily forms water soluble sulphate and water insoluble hydroxide. Its oxide MO is very stable to heat and does not have rock-salt structure. M is:

Sr

The Crystal Field Stabilization Energy (CFSE) of [CoF 3 (H 2 O) 3 ](Δo < P) is :

−0.4Δo

−0.4Δo + P

−0.8Δo + 2P

The reaction in which hybridization of the underlined atom is affected is:

H₂SO₄ + NaCl →

H₃PO₂ →

NH₃ + H⁺ →

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1M AgNO3 and 0.1M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be : (E°Ag+/Ag = 0.8 V, E°Au+/Au = 1.69 V)

Only gold

Silver and gold in equal mass proportion

Only silver

Silver and gold in proportion to their atomic weights

A sample of red ink (a colloidal suspension) is prepared by mixing eosine dye, egg white, HCHO and water. The component which ensures stability of the ink sample is :

HCHO

Eosin dye

Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight: Overview, Questions, Preparation

Q: Metal deficiency defect in shown by Fe0.93O. In the crystal, some Fe2+ cations are missing and loss of positive charge is compensated by the presence of Fe3+ ions. The percentage of Fe2+ ions in the Fe0.93O crystals is___________.

In Fe0.93O Let us suppose that fraction of Fe2+ is x and fraction of Fe3+ is (1 – x) x × ( + 2 ) + ( 1 − x ) × ( + 3 ) = O.S. of Fe atom o r 2 x + 3 − 3 x = + 2 0 0 9 3 ∴ % o f F e 2 + = 8 5 % a n d % o f F e 3 + = 1 5 %

Q: 2g of a non-volatile non-electrolyte solute is dissolved in 200g of two different solvents A and B whose ebullioscopic constant are in the ratio of 1:8. The elevation in boiling points of A and B are in the ratio x y ( x : y ) . The value of y is __________. (Nearest integer)

Δ T b = i × K b × m or Δ T b = K b × m (i = 1 for non-electrolyte) o r ( Δ T b ) A ( Δ T b ) B = ( K b ) A ( K b ) B o r x y = 1 8 | ∴ y = 8 |

Q: 65. Match the catalysts given in Column I with the processes given in Column II.

65. (i) - (c); (ii)- (d); (iii)- (b); (iv)- (e); (v)- (a)

Q: 1. Identify A to E and explain reaction involves:

1. CuCO3 CuO + CO2 (D) Ca (OH)2 + CO2 → CaCO3 + H2O (E) CaCO3 + CO2 + H2O → Ca (HCO3)2 clear sol. CuO + CuS 3Cu + SO2 (A) Cu + 4HNO3 → Cu (NO3)2 + 2NO2 + 2H2O (conc.) (B) Cu (NO3)2 + 4NH3 → [Cu (NH3)4] (NO3)2 (C) blue solution

Q: 2. When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with ???, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

2. (A) = FeCr2O4 (B) = Na2CrO4 (C) = Na2Cr2O7 (D) = K2Cr2O7 4FeCrO4 + 8NaCO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 (A) (B) 2NaCrO4 + 2H + → Na2Cr2O7 + 2Na + + H2O Na2Cr2O7 + KCl→ K2Cr2O7 + 2NaCl (C) (D)

Q: The gas produced by treating an aqueous solution of ammonium chloride with sodium nitride is

N H 4 C l ( a q ) + N a N O 2 ( a q ) → N 2 ( g ) + 2 H 2 O ( l ) + N a C l ( a q ) Hence, N2 gas is produced.

Q: Arrange the following coordination compounds in the increasing order of magnetic moments (Atomic numbers: Mn = 25; Fe = 26) (A) [ F e F 6 ] 3 − (B) [ F e ( C N ) 6 ] 3 − (C) [ M n C l 6 ] 3 − (high spin) (D) [ M n ( C N ) 6 ] 3 − Choose the correct answer from the options given below

A. [ F e F 6 ] 3 − Fe3+ - 3d54s0 F- is weak field ligand, so no pairing of electrons : Number of unpaired electrons, n = 5 B. [ F e ( C N ) 6 ] 3 − F e 3 + − 3 d 5 4 s 0 CN- is strong field ligand, so pairing of electrons takes place. Number of unpaired electrons, n = 1 C. [ M n C l 6 ] 3 − M n 3 + − 3 d 3 4 s 0 Number of unpaired electrons, n = 3 D. [ M n ( C N ) 6 ] 3 − M n 3 + − 3 d 3 4 s 0 Number of unpaired electrons, n = 3 Higher the number of unpaired electrons. Then higher be the magnetic moment as; Magnetic moment, μ = n ( n + 2 ) B . M Hence, order of magnetic moment is ; B < D = C < A

Updated on Aug 21, 2025 11:27 IST

By Raj Pandey

Table of content

The D- And F- Block Element Questions and Answers
JEE Main 27th June 2022 (Second Shift)
JEE Mains Solutions 2022,27th june ,chemistry , first
JEE Mains 2020

The D- And F- Block Element Questions and Answers

1. Identify A to E and explain reaction involves:

Ans:

CuCO₃ CuO + CO₂

(D)

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

(E)

CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂

clear sol.

CuO + CuS 3Cu + SO2

(A)

Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

(conc.) (B)

Cu(NO₃)₂ + 4NH₃ → [Cu(NH₃)₄](NO₃)₂

2. When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with ��, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

Ans: (A) = FeCr₂O₄

(B) = Na₂CrO₄

(D) = K₂Cr₂O₇

4FeCrO₄ + 8NaCO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

(A) (B)

2NaCrO₄ + 2H + → Na₂Cr₂O₇ + 2Na + + H₂O

Na₂Cr₂O₇ + KCl→ K₂Cr₂O₇ + 2NaCl

3. When an oxide of manganese (A) is fused with KOH in the presence of an oxidizing agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionate in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.

Ans: (A) = MnO₂

(B) = K₂MnO₄

(D) = KIO₃

2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

(A) (B)

3MnO₄^2- + 4H⁺ → 2MnO₄^- + MnO₂ + 2H₂O

(C)

2MnO₄^2- + 2H₂O + KI→ 2MnO₂ + 2SO₄^- + KIO₃

(A) (D)

4. On the basis of lanthanoid contraction, explain the following:

(a) Nature of bonding in La₂O₃and Lu₂O₃.

Ans: As per fajan’s rule, smaller the size of ion, greater is its tendency to make covalent bonds as the positively charged cation strongly attract the negatively charged electron cloud.

On moving from La to Lu in the lanthanoid series, the atomic size decreases so the covalent character increases. Hence, La₂O₃is ionic and Lu₂O₃ is covalent.

(b) Trends in the stability of oxo-salts of lanthanoids from La to Lu.

Ans: The stability of Oxo salts is directly proportional to the size of atom, as we move from La to Lu, the size of the atom decreases and hence the stability of oxo salts also decreases.

(c) Stability of the complexes of lanthanides.

Ans: As we move along the lanthanide series, the atomic size decreases. As a result, the charge/size ratio increases and the stability of the complexes also increases.

(d) Radii of 4d- and 5d-block elements.

Ans: As an effect of lanthanide contraction, zirconium and hafnium have similar radius of 160pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii.

(e) Trends in acidic character of lanthanide oxides.

Ans: Acidic or basic character of oxides is related to the nature of oxides. As we move along the lanthanide series from La to Lu, the covalent character increases.

Therefore, their basic character decreases or acidic character increases.

5. (A) Answer the following questions:

(i) Which element of the first transition series has the highest second ionization enthalpy?

Ans: Out of all the elements of the first transition series copper has the highest second ionisation enthalpy.

Electronic configuration of Copper is: 3d¹⁰4s¹

After the Loss of first electron from the 4s copper acquires 3d¹⁰ configuration which is stable. Therefore, removal of second electron from the field 3-D orbital is very difficult and requires high amount of energy.

6. (ii) Which element of the first transition series has highest third ionization enthalpy?

Ans: Among the elements of first transition series zinc has the highest third ionisation enthalpy. Electronic configuration of zinc is: 3d¹⁰4s²

After the loss of two electrons from 4s orbital, Z and +2 Ion acquires 3d¹⁰ fully filled configuration which is highly stable therefore removal of third electron from 3d 10orbital will be more difficult and requires a large amount of energy.

(iii) Which element of the first transition series has lowest enthalpy of atomization?

Ans: Zinc is the element of the first transition series that has the lowest enthalpy of atomisation. The electronic configuration of zinc is: 3d¹⁰4s².This is because it has filled 3d-subshell and filled 4s-subshell so no unpaid electron is available for metallic bonding.

(b) Identify the metal and justify your answer:

(I) CarbonylmM(CO)₅

Ans: Fe(CO)₅as per EAN Rule.

EAN Rule: Effective atomic number (EAN), number that represents the total number of electrons surrounding the nucleus of a metal atom in a metal complex.

(II) MO₃F

Ans: The M in the given complex is in +7 O.S.

Manganese is the only element in the first transition series that shows +7 oxidation state, therefore this compound is MnO₃F.

6. Mention the type of compounds formed when small atoms like H get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.

Ans: Interstitial compounds are formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals.

Characteristics:

They are hard and rigid in nature having high melting points.

Like pure metals only, they are conductors. They can gain chemical inertness.

8. (A) Transition metals can act as catalysts because these can change their oxidation state.

How does Fe(III) catalyse the reaction between iodide and persulphate ions?

Ans: 2I^- + S₂O₈^2- →Fe(III) I₂ + 2SO₄^2-

role of fe ions:

2Fe³⁺+ 2I^- → 2Fe²⁺ + I₂

2Fe²⁺ + S₂O₈^2- → 2Fe³⁺ + 2SO₄^2-

(B) Mention any three processes where transition metals act as catalysts.

Ans: 1. Fine powdered state of Nickel is used in the hydrogenation of oils into fats.

Iron is used in the formation of ammonia in Haber’s process.
In contact process in manufacture of H₂SO₄, vanadium is used in its oxide form- V₂O₅.

9. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H₂SO₄and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

Ans: KMnO₄ K₂MnO₄ + MnO₂ + O₂

(A) (B) (C)

MnO₂ + KOH → 2K₂MnO₄ + 2H₂O

MnO₂ + 4NaCl + 4H₂SO₄→ MnCl₂ + 2NaHSO₄+2H₂O + Cl₂

(D)

(A) = KMnO₄

(B) = K₂MnO₄

(D) = MnCl₂

Commonly asked questions

65. Match the catalysts given in Column I with the processes given in Column II.

1. Identify A to E and explain reaction involves:

2. When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with ???, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

JEE Main 27th June 2022 (Second Shift)

Commonly asked questions

Identify the incorrect statement for PCl₅ from the following

The gas produced by treating an aqueous solution of ammonium chloride with sodium nitride is

Arrange the following coordination compounds in the increasing order of magnetic moments (Atomic numbers: Mn = 25; Fe = 26)

(A) ${[F e F_{6}]}^{3 -}$ (B) ${[F e {(C N)}_{6}]}^{3 -}$ (C) ${[M n C l_{6}]}^{3 -}$ (high spin) (D) ${[M n {(C N)}_{6}]}^{3 -}$

Choose the correct answer from the options given below

JEE Mains Solutions 2022,27th june ,chemistry , first

Commonly asked questions

Metal deficiency defect in shown by Fe_0.93O. In the crystal, some Fe²⁺ cations are missing and loss of positive charge is compensated by the presence of Fe³⁺ ions. The percentage of Fe²⁺ ions in the Fe_0.93O crystals is___________.

If the uncertainty in velocity and position of a minute particle in space are, 2.4 × 10^-²⁶(ms^-¹) and 10^-⁷(m) respectively. The mass of the particle in g is __________.

(Given: h = 6.626 × 10^-³⁴g)

2g of a non-volatile non-electrolyte solute is dissolved in 200g of two different solvents A and B whose ebullioscopic constant are in the ratio of 1:8. The elevation in boiling points of A and B are in the ratio $\frac{x}{y} (x : y) .$ The value of y is __________. (Nearest integer)

Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight Exam

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