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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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aluminum nitrite and ammonium cloride react to form aluminium chloride. nitrogen and water. what mass of alluminum chloride is p
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Answer:

35.7 g of NH₄Cl

Explanation:

The reaction is:

Al(NO₂)₃ +  3NH₄Cl → AlCl₃ + 3N₂ + 6H₂O

Ratio is 3:1. 3 mol of ammoium chloride react to produce 1 mol of chloride

We convert the mass of ammoium chloride to moles:

43 g . 1 mol/ 53.45 g = 0.804 moles

(0.804 . 1) / 3 = 0.268 moles of AlCl₃ are formed

Let's convert the moles to mass → 0.268 mol . 133.33 g/1mol = 35.7 g

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3 years ago
What is greatest velocity? How do you get greatest velocity?
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Explanation:

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A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.51 kJ/mol at 25 °C.
SVEN [57.7K]

Answer:

(a) [A] = 0.13 M, [B]= 0.23 M and [C] = 0.17 M.

(b) Option B.

Explanation:

The reaction given:

                     A(aq)  +  B(aq)  ⇄  C(aq)      (1)

<u>Initial:</u>            0.30M   0.40M      0M         (2)                

<u>Equilibrium:</u>  0.3 - x     0.4 - x       x           (3)

The equation of Gibbs free energy of the reaction (1) is the following:

\Delta G^{\circ} = - RTLn(K_{eq})    (4)

<em>where ΔG°: is the Gibbs free energy change at standard conditions, R: is the gas constant, T: is the temperature and K_{eq}: is the equilibrium constant </em>

(a) To calculate the concentrations of A, B, and C at equilibrium, we need first determinate the equilibrium constant using equation (4), with ΔG°=-4.51x10³J/mol, T=25 + 273 = 298 K, R=8.314 J/K.mol:

K_{eq} = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{-\frac{-4.51\cdot 10^{3} J/mol}{8.314 J/K.mol \cdot 298 K}} = 6.17      (5)

Now, we can calculate the concentrations of A, B, and C at equilibrium using the equilibrium constant calculated (5):

K_{eq} = \frac{[C]}{[A][B]} = \frac{x}{(0.3 - x)(0.4 - x)}     (6)    

Solving equation (6) for x, we have two solutions x₁=0.69 and x₂=0.17, and by introducing the solution x₂ into equation (3) we can get the concentrations of A, B, and C at equilibrium:                          

[A] = 0.3 - x_{2} = 0.3 - 0.17 = 0.13 M

[B] = 0.4 - x_{2} = 0.4 - 0.17 = 0.23 M

[C] = x = 0.17 M

<u>Notice that the solution x₁=0.69 would have given negative values of the A and B concentrations.</u>  

(b) If the reaction had a standard free-energy change of +4.51x10³J/mol, the equilibrium constant would be:

K_{eq} = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{-\frac{4.51\cdot 10^{3} J/mol}{8.314 J/K.mol \cdot 298 K}} = 0.16

By solving the equation (6) for x, with the equilibrium constant calculated, we can get again two solutions x₁ = 6.9 and x₂= 0.017, and by introducing the solution x₂ into equation (3) we can get the concentrations of A, B, and C at equilibrium:

[A] = 0.3 - x_{2} = 0.3 - 0.017 = 0.28 M

[B] = 0.4 - x_{2} = 0.4 - 0.017 = 0.38 M

[C] = x = 0.017 M        

<u>Again, the solution x₁=6.9 would have given negative values of the A and B concentrations.</u>

Hence, the correct answer is option B: there would be more A and B but less C.

I hope it helps you!

7 0
3 years ago
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