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Is copper more reactive than iron? Give a reaction in support of your answer.In gaseous phase, the angle between the planes of the \[{H_2}{O_2}\] molecule is:
A.\[{101.5^0}\]
B.\[{90^0}\]
C.\[{111.5^0}\]
D.\[{109^0}{28'}\]

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Last updated date: 24th Jul 2024
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Answer
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Hint: Hydrogen peroxide is a chemical alloy made up of hydrogen and water molecules. \[{H_2}{O_2}\] is the chemical formula for it. Hydrogen peroxide is commonly seen as a transparent liquid with a slight pale blue coloration when it is in its purest form.

Complete answer:
In the gas phase, the H-O-O dihedral angle (angle between the planes comprising the H-O-O group) is \[{111.5^0}\]. So Option C is correct.
In general, we know that in the crystalline phase, there is more hydrogen bonding than in the vapour phase. On the oxygens, we realise there are lone pairs, and lone-pair/lone-pair repulsion can probably be reduced. We'd also like to keep eclipsed and gauche experiences to a minimum.
Actually, for HOOH, it appears that there is only one Newman projection possible: the lone-pair/lone-pair interaction, as well as the lone-pair/hydrogen interaction. We can only reduce these interactions; hence, the energy-minimized conformer is unlikely to be the one with all of them eclipsed.
First, don't consider that the gas state differs from the solid state. Instead, consider that the solid state differs from the gas state. There is very little intermolecular interaction in the gaseous state (very, very few species do not follow the ideal gas law to many significant digits).
The gas state geometry is similar to that of \[s{p^3}\] tetrahedral geometry and can be considered natural. The torsional rotation energies were not easy to describe.
Due to hydrogen bonding, bond lengths and angles in liquid and solid phases are slightly altered. In the crystalline state, the bond angle between the two planes decreases to \[{90.2^0}\].

Note:
The valence bond principle can also be used to describe the structure of $H_2O_2$, since both oxygen atoms are sp-hybridized. Lone pairs of electrons occupy two of these hybrid orbitals on each oxygen. The third hybrid orbital forms an O-H sigma bond with the s-orbital of a hydrogen atom, while the fourth forms a sigma bond with the second oxygen atom's half-filled hybrid orbital.