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Oliga [24]
1 year ago
8

Give the hybridization for the br in bro4⁻.

Chemistry
1 answer:
elixir [45]1 year ago
8 0

The hybridization for the Br in BrO4⁻ is sp^{3}. So, the correct option is (e).

In chemistry, the idea of combining atomic orbitals to create new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) is known as orbital hybridisation (or hybridization). These new hybrid orbitals are suitable for the pairing of electrons to form chemical bonds in valence bond theory.

Because more directional hybridised orbitals result in higher overlap when creating bonds, stronger bonds are formed, which favours the hybridization of orbitals. When hybridization takes place, this leads to more stable molecules.

One s orbital and three p orbitals combine to form four sp^{3} orbitals, each of which has a 25% s character and a 75% p character. This process is known as sp^{3} hybridization. Anytime an atom is surrounded by four groups of electrons, this kind of hybridization is necessary.

Learn more about hybridization here:

brainly.com/question/12207339

#SPJ4

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How many millimeters (mm) is the length of a standard table if it is
matrenka [14]

Answer:

1,500 mm

Explanation:

if 1 meter = 1000 mm, 0.5 meters is 500 mm, so 1.50 meters is 1,500 mm

5 0
3 years ago
How would having too much sample in the melting point tube most likely affect the melting point measurement? Select the correct
oksano4ka [1.4K]

Answer:

2-4 mm height of capillary tube.

Explanation:

Sample should be around 2-4 mm in height.

It should be packed well so that it does not have air packets that caues the lowering of melting point.

If you take greater amount, then there will be needed more heat, resulting a wide range of melting point.

7 0
3 years ago
How many grams of Hydrogen (H) would need to react with 143 grams of Oxygen (O) to make 235 grams of Water (H20)? 235
drek231 [11]

Answer:

45 grams

Explanation:

7 0
2 years ago
Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0
2 years ago
Name four products of incomplete combustion<br><br> 1 - ____<br> 2 -____<br> 3 -____<br> 4 -_____
faltersainse [42]

Answer:

Carbon Monoxide / Carbon Dioxide / Sulfur  and Nitrogen Dioxide

Explanation:

3 0
3 years ago
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