Bond Strength And Electronegativity Flashcards

1
Q

What is bond dissociation enthalpy and what is it always?

A

It is a measure of the strength of a chemical bond

It is always endothermic and always in gaseous state

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2
Q

Why can’t exact bond dissociation enthalpy changes only be used for molecules greater than 2 atoms?

A

Because in the case of water, the two successive dissociations have different enthalpies

Averages have to be used instead

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3
Q

What relation does the bond dissociation enthalpy and enthalpy of atomisation of water have in common?

A

The average bond dissociation enthalpy is = to half the enthalpy of atomisation - 924 / 2 = 462

(Splitting water into its constituent atoms - gaseous)

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4
Q

Define the enthalpy of atomisation

A

Refer to the enthalpy change that accompanies the separation of 1 mole of compound into its constituent atoms in the gas phase

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5
Q

How do you find the number of bonds per atom in a compound / lattice?

A

Divide the number of bonds by the number of atoms it links.

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6
Q

What are the bonding trends down groups in the p-block?

A

Bond distance increases down groups because atomic radii increases

Bond enthalpy decreases down the groups because of decreasing orbital overlap - weaker bonds

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7
Q

What experimental data can be used to see the leap in atomic radii between periods 2 and 3?

A

A graph showing bond enthalpies of group 14 elements - all with diamond-type structure.

There’s a large distance between C and Si which reflects this leap in atomic radii between periods 2 and 3 - bonds are weaker with less overlap

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8
Q

Why does metallic character of p-block element increase down the groups?

A

Because valence electrons are less tightly held

Hence bonding becomes less directional and electrons more delocalised

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9
Q

Why does pi bonding mainly occur between atoms of the first row of the p-block?

A

Because the atoms are sufficiently small so that their p-orbitals can overlap effectively

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10
Q

Why are O-O and N-N single bonds considerably weaker than a C-C bond?

A

Because the small atoms bring their lone pairs in close contact - so repel each other away

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11
Q

What’s a consequence of the considerably N-N and O-O single bonds?

A

A consequence of this is that compounds containing these bonds are particularly reactive

Example - hydrogen peroxide
H - O — O - H which is prone to decombustion

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12
Q

What drives the decombustion of hydrogen peroxide?

A

The formation of the O=O double bond - being more than twice as strong as a O-O single bond

Hence exothermic

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13
Q

Why is the formation of ethylene exothermic?

A

Because a C=C double bond is less than twice as strong as a C-C single bond

Hence the polymerisation of ethylene is exothermic

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14
Q

Why is nitrogen a gas but phosphorus a solid at ambient conditions?

A

Because nitrogen forms very strong N=N triple bonds - hence nitrogen consists of small non-polar N2 molecules

In contrast, pi bonds in phosphorus is weaker due to less pi overlap between the larger atoms - hence phosphorus atoms tend to make 3 single bonds to other P atoms

This results in a polymeric solid.

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15
Q

What does the enthalpy of reaction of HCl being exothermic suggest?

A

It indicates that the polar HCl bond must be stronger than the average strength of the non-polar H2 and Cl2 bonds.

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16
Q

How do heteronuclear bonds compare to homonuclear bonds?

A

A heteronuclear bond A-B tends to be stronger than the average strength of the corresponding homonuclear bonds A-A and B-B

17
Q

What does a greater difference in electronegativity cause to a bond?

A

A greater difference means the bond is more ionic in character

Smaller difference = more covalent

18
Q

What’s the equation for electronegativity of a bond?

A

∆ = [x(A) - x(B)]^2 eV

19
Q

What is 1 eV equal to?

A

1 eV = 96.5 KJ/mol

20
Q

How can the strength of hydrogen bonding be estimated?

A

From the enthalpy of evaporation = 42KJ/mol

So D(H-bond) ≈ 1/2 x 42 = 21KJ/mol

21
Q

What is on the x and y axis of the Ketelaar triangle?

A

X axis = x(average) = 1/2 (x(A) + x(B))

Y axis = ∆x = x(A) - x(B)
- the difference