NEET Previous Year Questions (2014-2024): The d & f-Block Elements | Chemistry Class 12 PDF Download

2024

Q1: 'Spin only' magnetic moment is same for which of the following ions?
A. Ti³⁺
B. Cr²⁺
C. Mn²⁺
D. Fe²⁺
E. Sc³⁺
Choose the most appropriate answer from the options given below.
(a) B and D only
(b) A and E only
(c) B and C only
(d) A and D only      (NEET 2024)
Ans: (a)
Spin only magnetic moment is given by
∴ Cr²⁺ and  Fe²⁺ will have same spin only magnetic moment.

Q2: Given below are two statements :
Statement I: Both [Co(NH₃)₆]³⁺ and [CoF₆]³⁻ complexes are octahedral but differ in their magnetic behaviour.
Statement II: [Co(NH₃)₆]³⁺ is diamagnetic whereas [CoF₆]³⁻ is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
(a) Both Statement I and Statement II are true
(b) Both Statement I and Statement II are false
(c) Statement I is true but Statement II is false
(d) Statement I is false but Statement II is true      (NEET 2024)
Ans: (a)

Q3: The E^∘ value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of
(a) d⁵ to d⁴ configuration
(b) d⁵ to d² configuration
(c) d⁴ to d⁵ configuration
(d) d³ to d⁵ configuration       (NEET 2024)
Ans: (c)

Electronic configuration of
Electronic configuration of
Electronic configuration of
Electronic configuration of
As Mn³⁺ from d⁴ configuration goes to more stable d⁵ configuration (Half filled), due to more exchange energy in d⁵ configuration.

Q4: The pair of lanthanoid ions which are diamagnetic is
(a) Ce⁴⁺ and Yb2+
(b) Ce³⁺ and Eu²⁺
(c) Gd³⁺ and Eu³⁺
(d) Pm³⁺ and Sm³⁺        (NEET 2024)
Ans: (a)
Magnetic moment
n ↣ number of unpaired electron

Hence Ce⁴⁺ and Yb²⁺ are only diamagnetic.

Q5: During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe²⁺ ion?
(a) dilute hydrochloric acid
(b) concentrated sulphuric acid
(c) dilute nitric acid
(d) dilute sulphuric acid     (NEET 2024)
Ans: (d)
Mohr's salt is the ammonium iron(II) sulfate, with the chemical formula (NH4)₂Fe(SO₄)₂⋅6H₂O. It is known for its stability compared to the other iron(II) salts which tend to readily oxidize to iron(III) salts when exposed to the air. To prevent the oxidation and hydrolysis of the Fe²⁺ ion during the preparation of Mohr's salt, an acid is added. This acid serves several purposes: it maintains the acidic environment necessary to prevent hydrolysis, aids in the solubilization of iron(II) sulfate, and minimizes the oxidation of Fe²⁺ ions to Fe³⁺.
The options provided give four different types of acids, from which one needs to be selected for the preparation of this salt. We can evaluate them based on their appropriateness:

• Dilute Hydrochloric Acid: While capable of maintaining a low pH to prevent oxidation, HCl can potentially introduce chloride ions (Cl⁻), which can form complexes with iron, changing the composition of the solution.
• Concentrated Sulphuric Acid: This choice is too strong and can lead to excessive acidity as well as potential safety issues during handling and dilution. It is unnecessary for this application.
• Dilute Nitric Acid: Nitric acid is an oxidizing agent and can promote the oxidation of Fe²⁺ to Fe³⁺, which is undesirable in the preparation of Mohr's salt.
• Dilute Sulphuric Acid: This is a very common choice since it helps maintain the stability of the Fe²⁺ ion without contributing extraneous ions that could form undesired complexes or products. Moreover, it keeps the solution acidic, helping prevent oxidation and hydrolysis, while not introducing any oxidizing characteristics.

Given these considerations, the best choice for preventing hydrolysis of the Fe²⁺ ion during the preparation of Mohr's salt is Dilute Sulphuric Acid (Option D). This is because it maintains the suitable acidic conditions needed for stabilizing the ferrous ion and does not interfere with the redox stability of the iron by providing an oxidative chemical environment.

2023

Cu⁺² is more stable than Cu⁺ because released hydration energy is more in case of Cu⁺² than Cu⁺.


Q2: Which of the following statements are INCORRECT?   (NEET 2023)
A. All the transition metals except scandium form MO oxides which are ionic.
B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc₂O₃ to Mn₂O₇.
C. Basic character increases from V₂O₃ to V₂O₃ to V₂O₅.
D. V₂O₄ dissolves in acids to give VO₄³⁻ salts.
E. CrO is basic but Cr₂O₃ is amphoteric.
Choose the correct answer from the options given below:
(a) A and E only
(b) B and D only
(c) C and D only
(d) B and C only
Ans: (c)

D→V₂O₅ dissolve in acid to give VO₄⁻³ salts. This doesn't shown by V₂O₄

₂₀₂₂

Q1: Given below are two statements :           (NEET 2022)
Statement I : Cr²⁺ is oxidising and Mn³⁺ is reducing in nature.
Statement II : Sc³⁺ compounds are repelled by the applied magnetic field.
In the light of the above statements,
choose the most appropriate answer from the options given below :
(a) Statement I is incorrect but Statement II is correct
(b) Both Statement I and Statement II are correct
(c) Both Statement I and Statement II are incorrect
(d) Statement I is correct but Statement II is incorrect
Ans: (a)
Statement I : Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.
Statement II : Sc³⁺ has zero unpaired electron, so magnetic moment is also zero. Hence, Sc³⁺ will repelled by the applied magnetic field.

2021

Q1: Zr (Z=40) and Hf (Z=72) have similar atomic and ionic radii because of:       (NEET 2021)
(a) lanthanoid contraction
(b) having similar chemical properties
(c) belonging to same group
(d) diagonal relationship
Ans: (a)

• Hf is a post lanthanoid element. Due to presence of4f-orbitals which have poor shielding effect, the effective nuclear charge on valence shell electrons is more which result in the decrease of the size of Hf. This effect is known as lanthanoid contraction.
• The almost identical radii of Zr (160 pm) and Hf (159 pm) is a consequence of the lanthanoid contraction.

Q2: The incorrect statement among the following is :      (NEET 2021)
(a) Lanthanoids are good conductors of heat and electricity.
(b) Actinoids are highly reactive metals, especially when finely divided.
(c) Actinoid contraction is greater for element to element than Lanthanoid contraction.
(d) Most of the trivalent Lanthanoid ions are colorless in the solid state.
Ans: (d)

• The surface area increases when actinoids are finely divided which results in exposure of more reactant molecules to react. Hence, rate increases and so, actinoids are highly reactive metals when finely divided.
• The shielding effect of 5f-orbitals in actinoids is poor than the shielding effect of4f-orbitals. So, the effective nuclear charge on valence electrons is more in actinoids. Hence, actinoid contraction is greater than lanthanoid contraction.
• Trivalent lanthanoid ions are coloured in the solid state due to presence of f-electrons.
• Lanthanoids are inner transition metals. So, they are good conductors of heat and electricity.

Q3: Which of the following reactions is the metal displacement reaction? Choose the right option.      (NEET 2021)
(a)

(b)

(c)

(d)

Ans: (c)

• Both reactions (a) and (b) are examples of decomposition reactions.
• Reactions (c) and (d), both are examples of displacement reactions, while reaction (c) is an example of metal displacement reaction n in which a more reactive metal displaces/replaces the less reactive metal.

2020

Q1: Identify the incorrect statement.     ( NEET 2020)
(a) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
(b) The oxidation states of chromium in CrO²_4- and Cr₂O₇^2- are not the same.
(c) Cr²⁺(d⁴) is a stronger reducing agent than Fe²⁺(d⁶) in water.
(d) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
Ans: (b)
Oxidation state of Cr in CrO₄^2- and Cr₂O₇^2- is + 6. 

2018

Q1: Which one of the following ions exhibits d-d transition and paramagnetism as well?
(a) CrO₄^2-
(b) Cr₂O₇^2–
(c) MnO₄^-
(d) MnO₄^2-       (NEET 2018)
Ans: (d)
In CrO₄^2-       Cr⁺⁶ (n = 0)       diamagnetic
In Cr₂O₇^2–       Cr⁺⁶ (n = 0)       diamagnetic
In MnO₄^-       Mn⁺⁷ (n = 0)       diamagnetic
In MnO₄^2-       Mn⁺⁶ (n = 1)       paramagnetic
In MnO₄^2-, one unpaired electron(n) is present in d-orbital so, d-d transition is possible. 

Q2: Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :
              (NEET 2018)

(a)

(b)

(c)

(d)

Ans: (a)

2017

Q1: Name the gas that can readily decolourise acidified KMnO₄ solution :    (NEET 2017)
(a) SO₂
(b) NO₂
(c) P₂O₅
(d) CO₂
Ans: (a)
Acidified KMnO₄ is a strong oxidizing agent thus, among the given option which readily undergoes oxidation with KMnO₄ will decolourise it. CO₂, NO₂ and P₂O₅ are already in their highest oxidation state while SO₂ can further oxidize with KMnO₄ to give sulphate ions.
2MnO₄^– + 5SO₂ + 2H₂O $\to$ 2Mn²⁺ + 5SO₄^2– + 4H⁺ 


Q2: HgCl₂ and I₂ both when dissolved in water containing I^– ions the pair of species formed is :    (NEET 2017)
(a) HgI₂, I^–
(b)
(c) Hg₂I₂, I^–
(d) HgI₂, I^¯₃
Ans: (b)
HgCl₂ + 4I^– (aq) $\to$ HgI₄^2– (aq) + 2Cl^–(aq)
I₂(s) + I^– (aq) $\to$ I₃^–(aq) 


Q3: The reason for greater range of oxidation states in actinoids is attributed to :-    (NEET 2017)
(a) actinoid contraction
(b) 5f, 6d and 7s levels having comparable energies
(c) 4f and 5d levels being close in energies
(d) the radioactive nature of actinoids
Ans: (b)
Minimum or comparable energy gap between 5f, 6d and 7s subshell makes electron excitation easier, hence there is a greater range of oxidation states in actinoids.

2016

Q1: Which one of the following statements is correct when SO₂ is passed through acidified K₂Cr₂O₇ solution?    (NEET 2016)
(a) Green Cr₂(SO₄)₃ is formed.
(b) The solution turns blue
(c) The solution is decolourized.
(d) SO₂ is reduced.
Ans: (a)
K₂Cr₂O₇ + H₂SO₄ + 3SO₂  →  K₂SO₄ + Cr₂(SO₄)₃ (Green) + H₂O 

Q2: The electronic configurations of Eu (Atomic No. 63) Gd (Atomic No. 64) and Tb (Atomic No. 65) are :     (NEET 2016)
(a) [Xe]4f⁷ 6s², [Xe]4f⁷5d16s² and [Xe]4f⁹6s²
(b) [Xe]4f⁷ 6s², [Xe]4f⁸ 6s² and [Xe]4f⁸ 5d¹6s²
(c) [Xe]4f⁶ 5d¹6s², [Xe]4f⁷ 5d¹6s² and [Xe]4f⁹ 5d¹6s²
(d) [Xe]4f⁶ 5d¹6s², [Xe]4f⁷ 5d¹6s² and [Xe]4f⁸ 5d¹6s²
Ans: (a)
Eu (63) = [Xe] 4f⁷ 6s²
Gd (64) = [Xe] 4f⁷ 5d¹ 6s²
Tb (65) = [Xe] 4f⁹ 6s² 

Q3: Which one of the following statements related to lanthanoids is incorrect.
(a) Europium shows + 2 oxidation state.
(b) The basicity decreases as the ionic radius decreases from Pr to Lu.
(c) All the lanthanons are much more reactive than aluminium.
(d) Ce(+4) solutions are widely used as oxidizing agent in volumetric analysis.    (NEET 2016)
Ans: (c)
The first few members of the lanthanoid series are quite reactive, almost like calcium. However, with increasing atomic number, their behavior becomes similar to that of aluminium.

2015

Q1: Which of the following processes does not involve oxidation of iron ?    (AIPMT 2015 Cancelled Paper )
(a) Liberation of H₂ from steam by iron at high temperature
(b) Rusting of iron sheets
(c) Decolourization of blue CuSO₄ solution by iron
(d) Formation of Fe(CO)₅ from Fe
Ans: (d)


Q2:  Because of lanthanoid contraction, which of the following pairs of elements have nearly same atomic radii ? (Numbers in the parenthesis are atomic numbers).    (AIPMT 2015 Cancelled Paper )
(a) Zr (40) and Ta (73)
(b) Ti (22) and Zr (40)
(c) Zr (40) and Nb (41)
(d) Zr (40) and Hf (72)
Ans: (d)

Q3: Magnetic moment 2.84 B.M. is given by (At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27)

(a) Cr²⁺
(b) Co²⁺
(c) Ni²⁺
(d) Ti³⁺   (AIPMT 2015 Cancelled Paper )
Ans: (c)

Cr²⁺ – [Ar]3d⁴4s⁰ , 4 unpaired electrons
Co²⁺ – [Ar]3d⁷4s⁰ , 3 unpaired electrons
Ni²⁺ – [Ar]3d⁸ 4s⁰ , 2 unpaired electrons
Ti³⁺ – [Ar]3d¹4s⁰ , 1 unpaired electron 

Q4: Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium
(a) [Xe]4f⁹5s¹
(b) B[Xe]4f⁷5d¹6s²
(c) [Xe]4f⁶5d²6s²
(d) [Xe]4f⁸6d²        (NEET / AIPMT 2015)
Ans: (b)
Gd(64) = [Xe]4f⁷5d¹6s² 

Q5: Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified KMnO₄ for complete oxidation
(a) FeSO₃
(b) FeC₂O₄
(c) Fe(NO₂)₂
(d) FeSO₄         (NEET / AIPMT 2015)
Ans: (d)
FeSO₄ will require the least amount of acidified KMnO₄ for complete oxidation.

2014

Q1: Reason of lanthanoid contraction is :    (NEET 2014)
(a) Decreasing nuclear charge
(b) Decreasing screening effect
(c) Negligible screening effect of 'f ' orbitals
(d) Increasing nuclear charge
Ans: (c)
The shape of f-orbitals is very much diffused and they have poor shielding effect. The effective nuclear charge increases which causes the contraction in the size of electron charge cloud. This contraction in size is quite regular and known as lanthanoid contraction.

Q2: The reaction of aqueous KMnO₄ with H₂O₂ in acidic conditions gives :    (NEET 2014)
(a) Mn²⁺ and O₃
(b) Mn⁴⁺ and MnO₂
(c) Mn⁴⁺ and O₂
(d) Mn²⁺ and O₂
Ans: (d)
Hydrogen peroxide is oxidised to H₂O and O₂.
2KMnO₄ + 3H₂SO₄ + 5H₂O₂ $\to$ K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
Thus, Mn²⁺ and O₂ are produced.