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tankabanditka [31]
3 years ago
14

In the Lewis structure of AB4 where B is more electronegative than A. Both are main group elements where A has 8 valence electro

ns and each B has 7 valence electrons.
Required:
a. What is the total number of valence electrons?
b. How many lone pairs are in the molecule?
Chemistry
1 answer:
Ivenika [448]3 years ago
5 0

Answer:

1. 36

2. Two

Explanation:

The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.

For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.

There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.

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Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.
lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

<u>E = 2.18 x 10^-18 J</u>

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

<u>E = 13.6 eV</u>

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Hope this helps!!
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Read 2 more answers
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