Single Choice

The correct order in which the $$O-O$$ bond length increases as :

A$$H_{2}O_2 < O_{2} < O_{3}$$
B$$O_{3} < H_{2}O_{2} < O_{2}$$
C$$O_{2} < O_{3} < H_{2}O_{2}$$
Correct Answer
D$$O_{2} < H_{2}O_{2} < O_{3}$$

Solution

Bond length is inversely proportional to bond strength and bond strength is directly proportional to bond dissociation energy and $$H_2O_2$$ has single bond , $$O_3$$ has partial double bond because of resonance and $$O_2$$ has double bond so correct order is $$ O_2 < O_3 < H_2O_2$$ . Bond length of $$O-O$$ in $$O_2= 1.21A^0$$ $$O_3= 1.273A^0$$ and $$H_2O_2= 1.49A^0$$

SIMILAR QUESTIONS

Chemical Bonding

Which of the following facts regarding change in bond length is correct?

Chemical Bonding

Assuming that all the four valencies of carbon atom in propane pointing towards the corners of a regular tetrahedron, If the distance between the terminal carbon atoms in propane is $$X$$ $$\mathring { A } $$. Find $$100X$$. Given, $$C-C$$ bond length is $$1.54\mathring { A } $$, (sin $$109.5$$ is $$0.94$$)

Chemical Bonding

The correct order in which the $$O- O$$ bond length increases in the following:

Chemical Bonding

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Chemical Bonding

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Chemical Bonding

Which of the following is not correct:

Chemical Bonding

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Chemical Bonding

The correct order of decreasing bond lengths of CO, $$CO_2$$ and $$CO^{2-}_3$$ is:

Chemical Bonding

Which of the following statements are correct?

Chemical Bonding

$$(a)$$ Find the angle $$\theta $$ between adjacent nearest-neighbor bonds in the silicon lattice. Recall that each silicon atom is bonded to four of its nearest neighbors.The four neighbors form a regular tetrahedron—a pyramid whose sides and base are equilateral triangles. $$(b)$$ Find the bond length, given that the atoms at the corners of the tetrahedron are $$388$$ pm apart.

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