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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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Vikki [24]
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6 0
3 years ago
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Can someone balance this chemical equation for me please : )
Gnoma [55]

Answer:

2

Explanation:

Tried out 1, but couldn't get whole numbers on the right side then, so went up to 2, worked

all four numbrrs: 2, 7, 4, 6

The 7 for oxygen got adjusted in the last step of the thinking, because it's the simplest to adjust.

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3 0
3 years ago
An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t
Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

5 0
2 years ago
I'm sorry for asking so many questions this assignment is past due and I need it done NOW
marishachu [46]

Answer:

7N to the right

Explanation:

7 0
2 years ago
Given that Delta.G for the reaction below is –957.9 kJ, what is Delta.Gf of H2O?
Alexxandr [17]

Answer:

6ΔG°(f) H₂O = -229 Kj/mol

Explanation:

                    4NH₃(g)          +      5O₂(g)       =>        4NO(g)           +     6H₂O(g)

ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O

Hess's Law

ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants

-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]

-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj

ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj =  -228.56 Kj ≅ -228.6 Kj*

*Verified with Standard Heat of Formation Table

8 0
3 years ago
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