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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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disa [49]

Explanation:

b. What useful functions do oxidation numbers  serve?

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1 mole = 6.022 * 10^23 molecules

c. What is the name given to the number of  molecules in 1 mole?

Avogadro's Number of molecules

21. a. What is the molar mass of an element?

This is the mass of an element divided by the number of moles.

Molar mass = Mass / Number of moles

b. Write the molar mass rounded to two  decimal places of carbon, neon, iron and  uranium.

amu = Atomic Mass Unit

Carbon = 12.01 amu

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3 years ago
Express the following quantity in scientific notation. The answer needs to have the correct number of significant figures. 21,30
melisa1 [442]
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4 0
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Which of the following materials is a substance, silver, air ,gasoline or staintless steel?
Irina-Kira [14]
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4 0
3 years ago
Could someone please care to explain what the answer is
polet [3.4K]

Calculate the percentage composition of Ca(MnO4)2.

<u>Firstly, we see how many atoms are there for each element in the formula</u>.

Ca= 1 atom

Mn= 2 atoms

O= 8 atoms

<u>Next, we are going to consult our periodic table for the atomic mass of each element.</u>

Ca= 40

Mn= 55

O= 16

Then, we have to find the molar mass for the compound..

Here is the formula for calculating molar mass of an element:

Molar Mass= ( no. of atoms of the element × atomic mass of the element)

Now, we have to calculate the atomic mass of the compound. So using the molar mass formula for an element, we calculate the molar mass for each element then we sum up their molar masses to get the compounds molar mass.

Molar mass (Ca)= 1× 40

(Ca)= 40

Molar mass (Mn)=2×55

(Mn)= 110

Molar mass (O)= 8×16

(O)= 128

Now: Molar mass( compound)= (Ca)+(Mn)+(O)

= 40+ 110 128

= 278

This is everything we need to calculate our percentage composition for each element..

* The example says to find the percentage composition for Ca. So we only find for Ca, Which is already done using the formula and the answer is 14.39%.

To prove that your answer is correct, find the percentage composition for Mn and O as well. Then you add up their percentage compositions.

If you do and you get 100 as your answer, then your percentage compositions are correct.

Why don't you try finding the percentage composition for Mn and O, then add up all the three percentage compositions. If you 100 as their sum, then your percentage composition for each of the elements are correct.

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Strike441 [17]
Hi!
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