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Chemistry – Class XII – d & f Block Elements - NEET & JEE Test Worksheet – 45 Multiple Choice Questions – Very Useful for Students appearing in Engineering & Other Entrance Examinations CA$4.24   Add to cart
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Chemistry – Class XII – d & f Block Elements - NEET & JEE Test Worksheet – 45 Multiple Choice Questions – Very Useful for Students appearing in Engineering & Other Entrance Examinations

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Chemistry – Class XII – d & f Block Elements - NEET & JEE Test Worksheet – 45 Multiple Choice Questions – Very Useful for Students appearing in Engineering & Other Entrance Examinations

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  • August 22, 2024
  • 5
  • 2024/2025
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  • chemistry
  • isc
  • class12
  • classxii
  • dfblockelements
  • neet
  • jee
  • afmc
  • ism
  • mcqs
  • multiplechoicequestions
  • enggmedicalentranceexamination

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d and f Block Elements
NEET TEST
Marking Scheme: Q.6 Excess of KI reacts with CuSO4 solution and
(i) Each question is allotted 4 (four) marks for then Na2S2O3 solution is added to it.Which of
each correct response. the statements is incorrect for this reaction –
(ii) ¼ (one fourth) marks will be deducted for (1) Evolved I2 is reduced (2) CuI2 is formed.
indicating incorrect response of each question. No (3) Na2S2O3 is oxidised (4) Cu2I2 is formed.
deduction from the total score will be made if no
response is indicated for an item in the answer Q.7 Lanthanoid contraction is caused due to -
sheet. (1) the same effective nuclear charge from Ce
to Lu.
---------------------------------------------------------------
(2) the imperfect shielding on outer electrons
Q.1 Arrange Ce3+, La3+, Pm3+, and Yb3+ in by 4f electrons from the nuclear charge.
increasing order of their ionic radius - (3) the appreciable shielding on outer




JI
(1) Yb3+ < Pm3+ < Ce3+ < La3+ electrons by 4f electrons from the nuclear
(2) Ce3+ > Yb3+ < Pm3+ < La3+ charge.
(4) the appreciable shielding on outer
(3) Yb3+ > Pm3+ < La3+ < Ce3+
electrons by 5d electrons from the nuclear
(4) Pm3+ < La3+ < Ce3+ < Yb3+ charge.
Q.2 What would happen when a solution of Q.8 Identify the incorrect statement among the
potassium chromate is treated with an excess following
of dilute nitric acid - (1) d-block elements show irregular and
LA
(1) Cr3+ and Cr2O72– are formed erratic chemical properties among
(2) Cr2O72– and H2O are formed themselves.
(3) Cr2O72– is reduced to +3 state of Cr (2) La and Lu have partially filled d-orbitals
and no other partially filled orbitals.
(4) Cr2O72– is oxidised to +7 state of Cr (3) The chemistry of various lanthanoids is
Q.3 The atomic number of vanadium (V), very similar.
chromium (Cr), manganese (Mn) and iron (Fe) (4) 4f and 5f-orbitals are equally shielded.
are 23, 24, 25 and 26 respectively. Which one Q.9 In context with the transition elements, which
of these may be expected to have the highest of the following statements is incorrect ?
second ionisation enthalpy - (1) In the highest oxidation states, the
A

(1) V (2) Cr transition metal show basic character and
(3) Mn (4) Fe form cationic complexes.
Q.4 The radius of La3+ is 1.06 Å, which of the (2) In the highest oxidation states of the first
following given values will be closest to the five transition elements (Sc to Mn), all the
radius of Lu3+. (At no. of Lu = 71, La = 57) 4s electrons are used for bonding.
(1) 1.6 Å (2) 1.4 Å (3) Once the d5 configuration is exceeded, the
B


(3) 1.06 Å (4) 0.85 Å tendency to involve all the 3d electrons in
Q.5 Cerium (Z = 58) is an important member of the bonding decreases.
lanthanoids. Which of the following statement (4) In addition to the normal oxidation states,
about cerium is incorrect the zero oxidation state is also shown by
(1) Cerium (IV) acts as an oxidising agent. these elements in complexes.
(2) The +3 oxidation state of cerium is more Q.10 The number of 3d-electrons remained in Fe2+
stable than the +4 oxidation state. (At. no. of Fe = 26) is –
(3) The +4 oxidation state of cerium is not (1) 4 (2) 5
known in solutions. (3) 6 (4) 3
(4) The common oxidation states of cerium are
+3 and +4.




1

CLASSES BY ANKUR SIR 7983744732

, Q.11 The correct order of E values with Q.17 Preparation of looking mirrors involves the use
M 2 /M of
negative sign for the four successive elements (1) Red lead
Cr, Mn, Fe and Co is (2) Ammoniacal silver nitrate
(1) Mn > Cr > Fe > Co (3) Ammoniacal AgNO3 + red lead
(2) Cr > Fe > Mn > Co (4) Ammoniacal AgNO3 + red lead + HCHO
(3) Fe > Mn > Cr > Co
(4) Cr > Mn > Fe > Co Q.18 Why is HCl not used to make the medium
Q.12 In context of the lanthanoids, which of the acidic in oxidation reactions of KMnO4 in
following statement is not correct ? acidic medium ?
(1) There is a gradual decrease in the radii of (1) Both HCl and KMnO4 act as oxidising
the members with increasing atomic agents.




JI
number in the series. (2) KMnO4 oxidises HCl into Cl2 which is
(2) All the member exhibit +3 oxidation state. also an oxidising agent.
(3) Because of similar properties the (3) KMnO4 is a weaker oxidising agent than
separation of lanthanoids is not easy.
HCl.
(4) Availability of 4f electrons results in the
(4) KMnO4 acts as a reducing agent in the
formation of compounds in +4 state for all
the members of the series. presence of HCl.
Q.13 Which of the following arrangements does not Q.19 The magnetic nature of elements depends on
LA
represent the correct order of the property the presence of unpaired electrons. Identify the
stated against it ? configuration of transition elements, which
(1) Number of oxidation states : shows highest magnetic moment
Sc < Ti < Cr < Mn (1) 3d7 (2) 3d5
(2) Paramagnetic behaviour : (3) 3d8 (4) 3d2
V2+ < Cr2+ < Mn2+ < Fe2+ Q.20 The correct order of number of unpaired
(3) Ionic size : Ni2+ < Co2+ < Fe2+ < Mn2+ electrons is
(4) Stability in aqueous solution : (1) Cu2+ > Ni2+ > Cr3+ > Fe3+
Co3+ < Fe3+ < Cr3+ < Sc3+ (2) Ni2+ > Cu2+ > Fe3+ > Cr3+
Q.14 The salts of Cu in +1 oxidation state are (3) Fe3+ > Cr3+ > Ni2+ > Cu2+
A

unstable because (4) Cr3+ > Fe3+ > > Ni2+ > Cu2+
(1) Cu+ has 3d10 configuration Q.21 Highest oxidation state of manganese in
(2)Cu+ disproportionates easily to Cu(0)& Cu2+ fluoride is +4 (MnF4) but highest oxidation
(3)Cu+ disproportionates easily to Cu2+ & Cu3+ state in oxides is +7 (Mn2O7) because
(4) Cu+ is easily reduced to Cu2+ (1) Fluorine is more electronegative than
Q.15 If KMnO4 is reduced by oxalic acid in an acidic oxygen.
B


medium, then oxidation number of Mn changes (2) Fluorine does not possess d-orbitals.
from (3) Fluorine stabilises lower oxidation state.
(1) 4 to 2 (2) 6 to 4 (4) In covalent compounds, fluorine can form
(3) 7 to 2 (4) 7 to 4 single bond only while oxygen forms
Q.16 The trend of basicity of lanthanoid hydroxides double bond.
(1) Increases across the lanthanoid series Q.22 Native silver metal forms a water soluble
(2) Decreases across the lanthanoid series complex with a dilute aqueous solution of
(3) First increases and then decreases NaCN in the presence of
(4) First decreases and then increases. (1) Nitrogen (2) Oxygen
(3) Carbon (4) Argon




2



CLASSES BY ANKUR SIR 7983744732

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