The bond angles in the nitrite ion , nitrogen dioxide and the Nitronium ion are 115°,134°,180° respectively .
In the nitrite ion , N-atom has sp2 hybridization so it has trigonal planner geometry and bent shaped and has bond angle of 115° because of there is a lone pair on nitrogen atom due to which there is a repulsion occurs between the bond pair and lone pair . so the bond angle slightly deviate from 120° (general bond angle in case of sp2 hybridization ) and form a angle of 115° .
according to the Lewis structure nitrogen of nitrite ion contains 6 valance electron of oxygen and 5 valence electron of nitrogen . from which one oxygen and nitrogen bonded by double bond by sharing- pairing 4 electrons and other oxygen by single bond .
In case of Nitronium ion , N-atom has sp-hybridization , so it has linear geometry and the bond angle is 180° . also there is a positive charge on the nitrogen atom . which cause the valence electron of nitrogen becomes 4 . and the valance electron of oxygen is 6 . so according to Lewis structure ,the two oxygen bonded to the nitrogen atom by double bond . hence the shape will be linear and bond angle will be 180° .
In case of nitrogen dioxide , it involves the sp2 hybridization . according to Lewis dot structure the nitrogen dioxide have one 2 sigma and 1 lone pair electron . it has bent geometry .due to the lone pair there is a greater repulsion occurs due to which bond angle becomes 134° .
<h3>What is Lewis structure ?</h3>Lewis structure is also known as Lewis dot structure , Lewis dot diagram , electron dot structure . it shows the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule .
<h3>Hybridization :</h3>it is the intermixing of atomic orbitals to make new hybrid orbital .
Learn more about hybridization here :
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Answer:
1387.2g of Al2O3
Explanation:
The number of moles of aluminum oxide required to give the actual yield is obtained from the given mass of aluminum produced by the reaction. The percentage yield is used to obtain the theoretical yield of the aluminum. The number of moles produced theoretically can be used to calculate the amount of aluminum oxide required in moles and when we multiply the value obtained by the molar mass of aluminum oxide, we obtain the mass of aluminum oxide required.
