COMEDK 2025 Evening Shift | d and f Block Elements Question 2 | Chemistry

A.	$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$	P.	is predominantly a basic compound.
B.	$\mathrm{FeCr}_2 \mathrm{O}_4$	Q.	is predominantly a acidic compound
C.	$\mathrm{CrO}_3$	R.	forms yellow chromate on reaction with $\mathrm{Na}_2 \mathrm{CO}_3$ in air.
D.	$\mathrm{CrO}$	S.	is interconvertible with $\mathrm{MO}_4^{2-}$ depending on the pH of the medium.

$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$

$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$

$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{Q} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{P}$

$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

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Choose the statements which are incorrect in the case of Lanthanoids.

A. $\mathrm{Ce}^{4+}$ is diamagnetic while $\mathrm{Sm}^{3+}$ is paramagnetic.

B. The atomic size of the transition metals having atomic number greater than 71 are very close to that of the elements above them.

C. Lanthanoids react with hot water forming water soluble $\mathrm{Ln}(\mathrm{OH})_3$ with the liberation of $\mathrm{O}_2$.

D. The general electronic configuration of Lanthanoids is $(n-2) \mathrm{f}^{1-14} 5 \mathrm{~d}^0 6 \mathrm{~s}^2$, where $n=6$.

C & D

B & D

A & B

A & D

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

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Larger number of oxidation states are exhibited by the actinoids than those of lanthanoids. The reason is:

Lesser energy difference between 5 f and 6 d than between 4 f and 5 d orbitals

More energy difference between 5 f and 6 d than between 4 f and 5 d orbitals

4 f orbitals are more diffused than 5 f orbitals

Highly reactive nature of the actinoids

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