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

2023

Q1: The stability of Cu2+ is more than Cu+ salts in aqueous solution due to   (NEET 2023)
(a) First ionisation enthalpy
(b) Enthalpy of atomization
(c) Hydration energy
(d) Second ionisation enthalpy
Ans:
(c)

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

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. V2Odissolves 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)

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

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 t2g 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) NEET Previous Year Questions (2014-2024): The d & f-Block Elements | Chemistry Class 12

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

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

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

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 CrO24- and Cr2O72- 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 CrO42- and Cr2O72- is + 6. 

2018

Q1: Which one of the following ions exhibits d-d transition and paramagnetism as well?
(a) CrO42-
(b) Cr2O72–
(c) MnO4-
(d) MnO42-       (NEET 2018)
Ans: 
(d)
In CrO42-       Cr+6 (n = 0)       diamagnetic
In Cr2O72–       Cr+6 (n = 0)       diamagnetic
In MnO4-       Mn+7 (n = 0)       diamagnetic
In MnO42-       Mn+6 (n = 1)       paramagnetic
In MnO42-, 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)

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

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

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

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

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

Ans: (a)

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


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 + 2H2 2Mn2+ + 5SO42– + 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)NEET Previous Year Questions (2014-2024): The d & f-Block Elements | Chemistry Class 12
(c) Hg2I2, I
(d) HgI2, I¯3
Ans: 
(b)
HgCl2 + 4I (aq)  HgI42– (aq) + 2Cl(aq)
I2(s) + I (aq)  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]4f6s2, [Xe]4f75d16s2 and [Xe]4f96s2
(b) [Xe]4f6s2, [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)
NEET Previous Year Questions (2014-2024): The d & f-Block Elements | Chemistry Class 12

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


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)

NEET Previous Year Questions (2014-2024): The d & f-Block Elements | Chemistry Class 12
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)
FeSOwill require the least amount of acidified KMnOfor 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  K2SO4 + 2MnSO4 + 8H2O + 5O2
Thus, Mn2+ and O2 are produced. 

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FAQs on NEET Previous Year Questions (2014-2024): The d & f-Block Elements - Chemistry Class 12

1. What are the key properties of d-block elements?
Ans. D-block elements are characterized by their variable oxidation states, formation of colored compounds, and catalytic properties. They also exhibit magnetic behavior and can form complex compounds.
2. How do d-block elements differ from f-block elements?
Ans. D-block elements are transition metals that fill the d-orbitals, while f-block elements are inner transition metals that fill the f-orbitals. D-block elements are typically harder, have higher melting and boiling points, and exhibit variable oxidation states compared to f-block elements.
3. Why do transition metals exhibit variable oxidation states?
Ans. Transition metals have incompletely filled d-orbitals, which allows them to lose different numbers of electrons and exhibit multiple oxidation states. This flexibility in oxidation states is due to the ease with which electrons can be added or removed from the d-orbitals.
4. How do transition metals form colored compounds?
Ans. Transition metals form colored compounds due to the absorption of certain wavelengths of light by the d-electrons in their partially filled d-orbitals. The energy difference between the d-orbitals corresponds to the energy of visible light, causing the compounds to appear colored.
5. What role do d-block elements play in catalysis?
Ans. D-block elements are known for their catalytic properties due to their ability to easily change oxidation states and form temporary bonds with reactants. They can act as catalysts in various chemical reactions by providing an alternative pathway with lower activation energy, thus increasing the rate of the reaction.
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