Linus Pauling noticed the the energy of a polar shortcut is often greater than expected. He attributed the better bond energy to

a coulombic attraction between atoms with partially positive and negative charges. The better bond lengths that the heteronuclear bonds. Among the countless unexplainable phenomena that researchers encounter. The capability of heteronuclear varieties to form twin and triple bonds. The greater variety of valence electrons found in heteronuclear molecules. Answer

a.a coulombic attraction in between atoms v partially positive and negative charges.

You are watching: As bond order increases bond length


Exercise (PageIndex2)

In molecules, together bond order increases,

both link length and also bond energy increase. Both bond length and also bond power decrease. Bond length increases and also bond energy is unchanged. Bond length is unchanged and also bond power increases. Bond size decreases and also bond power increases. Answer

e.bond length decreases and bond power increases.


Exercise (PageIndex3)

Determine the shortcut order the NO+?

a. 2 b. 1.5 c. 1 d. 2.5 e. 3

Answer

e. 3


Exercise (PageIndex4)

Use Lewis structures to guess the link order because that a sulfur-oxygen link in SO32-?

a. 1 b. 4/3 c. 5/2 d. 2 e. 3/2

Answer

a. 1


Exercise (PageIndex5)

Use Lewis frameworks to predict the shortcut order for a sulfur-oxygen shortcut in the sulfite ion?

a. 1 b. 3/2 c. 4/3 d. 2 e. 5/2

Answer

c. 4/3


Bond Energy


Exercise (PageIndex6)

Based ~ above the following data, what is the F-F shortcut energy?

H2(g) + F2(g) → HF(g); ΔrH = –272.5 kJ/mol-rxn

Bond Bond power (kJ/mol)

H-H 435

H-F 565

a. 500 kJ/molb. 150 kJ/molc. –150 kJ/mold. –695 kJ/mole. 695 kJ/mol

Answer

b.150 kJ/mol

Cleavage of one H-H bond, (Delta H_1= 435 kJ/mol;)

Cleavage that one H-F bond, (Delta H_2= 565 kJ/mol;)

Formation of two F-F bonds, (Delta H_3= ?)

ΔHf of the reaction is = -272.5 kJ/mol = -545 kJ/2 mol

(Delta H_f=(Delta H_1+Delta H_3)-(2*Delta H_2))

(-545 = (435+Delta H_3)-(2*565))

(-545 = (435+Delta H_3)-1130) (585 = 435+Delta H_3)

(153 kJ = Delta H_3)


Exercise (PageIndex7)

Using bond-energy data, what is ΔrH because that the following reaction?

CH4(g) + 2Cl2(g) → CCl4(g) + 2H2(g)

Bond Bond energy (kJ/mol)

C-H 413

H-H 432

Cl-Cl 242

C-Cl 328

a. –40 kJ/mol-rxnb. –150 kJ/mol-rxnc. 40 kJ/mol-rxnd. 1415 kJ/mol-rxne. 150 kJ/mol-rxn

Answer

a. –40 kJ/mol-rxn


Exercise (PageIndex8)

Using the adhering to data reactions:

ΔH° (kJ/mol-rxn)

H2(g) + Cl2(g) → 2HCl(g) –184

H2(g) → 2H(g) 432

Cl2(g) → 2Cl(g) 239

calculate the energy of an H-Cl bond.

a. 92 kJ/molb. 855 kJ/molc. 487 kJ/mold. 428 kJ/mole. 244 kJ/mol

Answer

d. 428 kJ/mol


Exercise (PageIndex9)

Which of the following types has the shortest carbon−nitrogen bond?

a.CH3CNb. CH3N2c. CH2NHd. CH3CONH2e. CH3NH3+

Answer

a.CH3CN


Exercise (PageIndex10)

Use the link energies detailed to finish the following statement.

________ when all of the bonds in acetic mountain (CH3COOH) are broken.

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Bond Bond power (kJ/mol)

C-H 413

C-O 358

O-H 463

C=O 745

C-C 348

C=C 614

3153 kJ/mol of energy is spend 3153 kJ/mol of energy is released 2805 kJ/mol of power is released 2805 kJ/mol of energy is spend 2766 kJ/mol of power is consumed Answer

a.3153 kJ/mol of power is consumed


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