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 .
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Complete Question
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices
Answer:
Explanation:
From the question we are told that:
Wavelength
Generally the equation for Atom Transition is mathematically given by
Where
Rydberg constant
Therefore
0.84 milligrams of a 2. 000 mg sample remain after 6. 55 years, according to radioactive decay.
Given data,
amount of sample = 2.000mg initially = 0.002grams
According to radioactive decay,
( )λ =
=
= 0.133
According to radioactive decay,
- λt
= ln0.002 - (0.133×6.55)
= -6.21 - 0.87 = -7.08 = 0.00084g = 0.84mg
Therefore, 0.84 milligrams of a 2. 000 mg sample remain after 6. 55 years.
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