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Nimfa-mama [501]
1 year ago
6

The bond angles in the nitrite ion, nitrogen dioxide, and the nitronium ion (NO₂⁺) are 115°, 134°, and 180°, respectively. Expla

in these values using Lewis structures and VSEPR theory.
Chemistry
1 answer:
rewona [7]1 year ago
3 0

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 :

brainly.com/question/22765530

#SPJ4

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How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
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Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

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