Consider the following ionization energies for aluminum. Al( g ) → Al + (g)+e – I 1 =580 kJ/mol Al + ( g ) → Al 2+ ( g )+e – I 2 =1815 kJ/mol Al 2+ ( g ) → Al 3+ ( g )+e – I 3 =2740 kJ/mol Al 3+ ( g ) → Al 4+ ( g )+e – I 4 =11,600 kJ/mol a. Account for the increasing trend in the values of theionization energies. b. Explain the large increase between I 3 and I 4 . c. Which one of the four ions has the greatest electronaffinity? Explain. d. List the four aluminum ions given in the precedingreactions in order of increasing size, and explainyour ordering. ( Hint: Remember that most of thesize of an atom or ion is due to its electrons.)
Consider the following ionization energies for aluminum. Al( g ) → Al + (g)+e – I 1 =580 kJ/mol Al + ( g ) → Al 2+ ( g )+e – I 2 =1815 kJ/mol Al 2+ ( g ) → Al 3+ ( g )+e – I 3 =2740 kJ/mol Al 3+ ( g ) → Al 4+ ( g )+e – I 4 =11,600 kJ/mol a. Account for the increasing trend in the values of theionization energies. b. Explain the large increase between I 3 and I 4 . c. Which one of the four ions has the greatest electronaffinity? Explain. d. List the four aluminum ions given in the precedingreactions in order of increasing size, and explainyour ordering. ( Hint: Remember that most of thesize of an atom or ion is due to its electrons.)
Solution Summary: The author explains how the ionization energy of Al is determined by how effectively valence electron is held by the nucleus.
Consider the following ionization energies for aluminum.
Al(
g
)
→
Al
+
(g)+e
–
I
1
=580 kJ/mol
Al
+
(
g
)
→
Al
2+
(
g
)+e
–
I
2
=1815 kJ/mol
Al
2+
(
g
)
→
Al
3+
(
g
)+e
–
I
3
=2740 kJ/mol
Al
3+
(
g
)
→
Al
4+
(
g
)+e
–
I
4
=11,600 kJ/mol
a. Account for the increasing trend in the values of theionization energies. b. Explain the large increase between
I
3
and
I
4
. c. Which one of the four ions has the greatest electronaffinity? Explain. d. List the four aluminum ions given in the precedingreactions in order of increasing size, and explainyour ordering. (Hint: Remember that most of thesize of an atom or ion is due to its electrons.)
Lab Data
The distance entered is out of the expected range.
Check your calculations and conversion factors.
Verify your distance. Will the gas cloud be closer to the cotton ball with HCI or NH3?
Did you report your data to the correct number of significant figures?
- X
Experimental Set-up
HCI-NH3
NH3-HCI
Longer Tube
Time elapsed (min)
5 (exact)
5 (exact)
Distance between cotton balls (cm)
24.30
24.40
Distance to cloud (cm)
9.70
14.16
Distance traveled by HCI (cm)
9.70
9.80
Distance traveled by NH3 (cm)
14.60
14.50
Diffusion rate of HCI (cm/hr)
116
118
Diffusion rate of NH3 (cm/hr)
175.2
175.2
How to measure distance and calculate rate
For the titration of a divalent metal ion (M2+) with EDTA, the stoichiometry of the reaction is typically:
1:1 (one mole of EDTA per mole of metal ion)
2:1 (two moles of EDTA per mole of metal ion)
1:2 (one mole of EDTA per two moles of metal ion)
None of the above
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell