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lara [203]
3 years ago
14

Which element can expand its valence shell to accommodate more than eight electrons? which element can expand its valence shell

to accommodate more than eight electrons? o he n br?
Chemistry
2 answers:
amm18123 years ago
4 0

Answer:

Br can expand its valence shell to accommodate more than eight electrons.

Explanation:

In a bond, elements tends to accommodate their valence electrons to reach the configuration of a noble gas (The Octet rule), so they try to complete their valence shell with 8 electrons, and they can do that only if they use the <u>orbitals s</u> (may contain only 2 electrons) <u>and p</u> (may contain only 6 electrons).

Sometimes, there are elements that need to share more than 8 valence electrons to use all the available electrons, but this is possible only if the atom has <u>orbitals d</u> (may contain 10 electrons) <u>or f</u> (may contain 14 electrons).

That's why <u>only the elements that has a number 3 of period or more can expand its valence shell</u>, because level 3 atoms has a <em>d subshell</em>.

In this case, He atom is in period number 1, N atom is in period number 2 and Br atom is in period number 3, that's why Br atom can expand its valence shell, and He and N atoms can't.

Anvisha [2.4K]3 years ago
3 0
<span>According to octet rule,  atoms with an atomic number less than 20 tend to combine  with other atom such that both of these atoms have eight electrons in their valence shells, which gives them the same electronic configuration as that of noble gas.

However, there are few compound that donot obey octel rule. Among the elements mentioned above i.e. oxygen and helium obeys octet rule.

In case of nitrogen, oxide of nitrogen (like NO and NO2) have incomplete octet. 

While there are few compounds of Br wherein Br has expanded octet. For example, in BrF5, Br has 12 electrons in valence shell. </span>
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When adding the measurements 42.1014 g + 190.5 g, the answer has ___ significant figures.
andrezito [222]

Answer: The answer has 7 significant figures

Explanation:

 The addition of  190.5  and 42.1014 will give 232.6014.  Counting the digits will give 7 significant figures.

4 0
4 years ago
29. Ammonia is formed with one nitrogen and three hydrogen atoms. What is the electron configuration of nitrogen? How
masya89 [10]

Answer:

<u>One lone-Pair is present in Ammonia</u>

<u></u>

Explanation:

The number of valence electron in N = 5

The number of Valence electron in H = 1

The formula of ammonia = NH3

Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8

The lewis structure suggest that :

Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.

3 electron of Nitrogen are involved in sharing with Hydrogen

So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair

So, there is only 1 lone pair in the ammonia molecule .

The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.

Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.

5 0
4 years ago
A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car- bon dioxide is lib
MrRissso [65]

Answer:

53.9%

Explanation:

1 mole of BaCO₃ yields  1 mole of CO₂,

1 mole of SrCO₃ yields    1 mole of CO₂

m₁ = mass of BaCO₃

m₂ =  mass SrCO₃

molar mass of SrCO₃  = 147.63 g/mol

molar mass of  BaCO₃ = 197.34 g/mol

molar mass of CO₂ = 44.01 g/mol

mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479

mole of BaCO₃ = m₁ / 197.34

mole of SrCO₃  = m₂ / 147.63

mole of BaCO₃ + mole of SrCO₃  = 0.00479

(m₁ / 197.34) + (m₂ / 147.63) = 0.00479

147.63 m₁ + 197.34 m₂ = 139.55

m₁ + m₂ = 0.8

m₁ = 0.8 - m₂

147.63 (0.8 - m₂) + 197.34 m₂ = 139.55

118.104 - 147.63 m₂ + 197.34 m₂ = 139.55

49.71 m₂ = 139.55 - 118.104 = 21.446

m₂ = 21.446 / 49.71 = 0.431

the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%

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