All categories

3 years ago

Why doesn’t neon readily form an ionic bond?

2 answers:

6 0

By definition of noble gases, neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

<h3>Noble gases</h3>

Noble gases are not very reactive, that is, they practically do not form chemical compounds. This means that they do not react with other substances, nor do they even react between atoms of the same gas, as is the case with diatomic gases such as oxygen (O₂).

The chemical stability of the noble gases and therefore the absence of spontaneous evolution towards any other chemical form, implies that they are already in a state of maximum stability.

All chemical transformations involve valence electrons, they are involved in the process of covalent bond formation and the formation of ions. Therefore, the practically null reactivity of the noble gases is due to the fact that they have a complete valence shell, which gives them a low tendency to capture or release electrons.

Since the noble gases do not react with the other elements, they are also called inert gases.

Neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

Learn more about noble gases:

brainly.com/question/8361108

brainly.com/question/11960526

brainly.com/question/19024000

Oliga [24]3 years ago

6 0

Atoms of neon do not lose or gain electrons of their outermost orbit easily. So they do not form ionic bonds.Neon do not form ionic bonds is that atoms of neon already have a stable octet in their outer shell.

<h3>Hope this helps!!</h3>

You might be interested in

Which equation agrees with the ideal gas law?

photoshop1234 [79]

<span> $\frac{(P1)(V1)}{T1}=\frac{(P2)(V2)}{T2}$ is directly related to the ideal gas law.

The ideal gas law states that: $PV=nRT$ . Rewriting this gives: $\frac{PV}{T}=nR$ . R is the universal gas constant, so its value is fixed. For a given sample, n is the number of moles of the gas, so its value would be fixed too. This means that the value of $\frac{PV}{T}$ would be a fixed constant as well. Therefore, whatever the initial value $\frac{(P1)(V1)}{T1}$ is, it should be equal to the final value $\frac{(P2)(V2)}{T2}$ .</span>

6 0

3 years ago

Read 2 more answers

Adenosine triphosphate is the:

OlgaM077 [116]

Answer:

B. Energy Molecule

Explanation:

Adenosine triphosphate, or ATP, is the primary carrier of energy in cells. The water-mediated reaction known as hydrolysis releases energy from the chemical bonds in ATP to fuel cellular processes. Adenosine triphosphate (ATP), energy-carrying molecule found in the cells of all living things.

5 0

3 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

Read 2 more answers

Name a common greenhouse gas with only hydrogen and oxygen

AVprozaik [17]

I'm guessing water vapour. i can't think of any other gas limited to only hydrogen and oxygen

hope it helps :)

3 0

3 years ago

Write a resonance form for pyrrole in which nitrogen has a formal charge of 1. Are comparable resonance forms possible for pyrid

Svetach [21]

Answer:

See explanation and image attached

Explanation:

Here attached are resonance forms of pyrrole and pyridine. The images were obtained from quora and researchgate respectively.

Now, we can see that in the resonance forms of pyrrole, the nitrogen atom in the heterocycle has a formal charge of +1. However, in the six membered pyridine hetrocycle, the nitrogen atom may have a formal charge of -1 or +1 as shown in the canonical structures attached. The structures in which nitrogen has a +1 formal charge in pyridine are comparable to structures obtained from pyrrole. These structures have less contribution to the structure of pyridine.

3 0

3 years ago