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NEET CHEMISTRYThe d-and f-Block ElementsMedium

Question

In acidic medium, H2O2\text{H}_2\text{O}_2 changes Cr2O72−\text{Cr}_2\text{O}_7^{2-} to CrO5\text{CrO}_5 which has two (–O – O–) bonds. The oxidation state of Cr\text{Cr} in CrO5\text{CrO}_5 is:

A

+5+5

B

+3+3

C

+6+6

D

−10-10

Step-by-Step Solution

In CrO5\text{CrO}_5, the central chromium atom is bonded to one oxygen atom via a double bond (oxide linkage) and four oxygen atoms via single bonds forming two peroxide linkages (–O–O–). Let the oxidation state of Cr\text{Cr} be xx. The oxidation state of oxygen in an oxide linkage is −2-2 . The oxidation state of oxygen in a peroxide linkage is −1-1 . So, the sum of oxidation states is: x+1(−2)+4(−1)=0  ⟹  x−6=0  ⟹  x=+6x + 1(-2) + 4(-1) = 0 \implies x - 6 = 0 \implies x = +6. Thus, the oxidation state of Cr\text{Cr} in CrO5\text{CrO}_5 is +6+6.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from The d-and f-Block Elements. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThe d-and f-Block Elementsacidicmediumtexthtextochangestextcrtexto

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