
A typical sesquioxide of $Rb/R{{b}_{4}}{{O}_{6}}$ can be represented as:
A. $[R{{b}_{4}}{{({{O}_{2}})}^{2-}}{{({{O}_{2}})}_{2}}^{2}]$
B. $R{{b}_{4}}{{({{O}^{2-}})}_{6}}$
C. $Rb{{O}_{2}},{{(R{{b}_{2}}{{O}_{2}})}_{2}}$
D. $R{{b}_{2}}O,{{(R{{b}_{2}}{{O}_{2}})}_{2}}$
Answer
481.5k+ views
Hint: Sesquioxide can be defined as an oxide which contains three atoms of oxygen with two atoms or radicals of another element. Examples of sesquioxide are aluminium oxide which is represented as $A{{l}_{2}}{{O}_{3}}$.
Complete step by step answer:
Generally all alkali metal sesquioxides are expected as in case of sesquioxides it contains peroxide or superoxide but in case of alkali metals sesquioxide contain peroxide and superoxide both. Peroxide is represented by ${{O}_{2}}^{2-}$ while superoxide is represented by ${{O}_{2}}^{-}$ ions. Out of the various alkali oxide the sesquioxides $R{{b}_{4}}{{O}_{6}}$ and $C{{s}_{4}}{{O}_{6}}$ are of special interest. With the help of electronic structure calculations by using the local spin-density approximation predicted that $R{{b}_{4}}{{O}_{6}}$ should be a half-metallic ferromagnet which was later contradicted when an experimental investigation of the temperature dependent magnetization of $R{{b}_{4}}{{O}_{6}}$ showed a low-temperature magnetic transition and differences between ZFC means zero field cooled and FC i.e. field cooled measurements. This proves that $R{{b}_{4}}{{O}_{6}}$ and $C{{s}_{4}}{{O}_{6}}$ comprises two different types of dioxygen anions known as hyperoxide and the peroxide anions. The nonmagnetic peroxide anions do not contain unpaired electrons while the superoxide anions contain unpaired electrons in antibonding orbitals.
And a typical sesquioxide of $Rb/R{{b}_{4}}{{O}_{6}}$ is represented as $[R{{b}_{4}}{{({{O}_{2}})}^{2-}}{{({{O}_{2}})}_{2}}^{2}]$
Hence we can say that option A is the correct answer.
Note: Many sesquioxides contain the metal in the +3 oxidation state and their crystal structures are verified by using powder x-ray diffraction. The mixed valency of both compounds is confirmed by using Raman spectroscopy and time-dependent magnetization experiments which indicates that $Rb/R{{b}_{4}}{{O}_{6}}$ show magnetic frustration.
Complete step by step answer:
Generally all alkali metal sesquioxides are expected as in case of sesquioxides it contains peroxide or superoxide but in case of alkali metals sesquioxide contain peroxide and superoxide both. Peroxide is represented by ${{O}_{2}}^{2-}$ while superoxide is represented by ${{O}_{2}}^{-}$ ions. Out of the various alkali oxide the sesquioxides $R{{b}_{4}}{{O}_{6}}$ and $C{{s}_{4}}{{O}_{6}}$ are of special interest. With the help of electronic structure calculations by using the local spin-density approximation predicted that $R{{b}_{4}}{{O}_{6}}$ should be a half-metallic ferromagnet which was later contradicted when an experimental investigation of the temperature dependent magnetization of $R{{b}_{4}}{{O}_{6}}$ showed a low-temperature magnetic transition and differences between ZFC means zero field cooled and FC i.e. field cooled measurements. This proves that $R{{b}_{4}}{{O}_{6}}$ and $C{{s}_{4}}{{O}_{6}}$ comprises two different types of dioxygen anions known as hyperoxide and the peroxide anions. The nonmagnetic peroxide anions do not contain unpaired electrons while the superoxide anions contain unpaired electrons in antibonding orbitals.
And a typical sesquioxide of $Rb/R{{b}_{4}}{{O}_{6}}$ is represented as $[R{{b}_{4}}{{({{O}_{2}})}^{2-}}{{({{O}_{2}})}_{2}}^{2}]$
Hence we can say that option A is the correct answer.
Note: Many sesquioxides contain the metal in the +3 oxidation state and their crystal structures are verified by using powder x-ray diffraction. The mixed valency of both compounds is confirmed by using Raman spectroscopy and time-dependent magnetization experiments which indicates that $Rb/R{{b}_{4}}{{O}_{6}}$ show magnetic frustration.
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