🗊Презентация Atomic structure

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Recitation
Atomic Structure
Описание слайда:
Recitation Atomic Structure

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Q1
Select the best choice :
Most energy necessary to remove one electron:   
             Cu	             Cu+		  Cu2+      
Highest electron affinity	                          
         Cl		    Br		    I             
Greatest volume                                       
              S2-                      Ar                 Ca2+
Описание слайда:
Q1 Select the best choice : Most energy necessary to remove one electron: Cu Cu+ Cu2+ Highest electron affinity Cl Br I Greatest volume S2- Ar Ca2+

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R1
Select the best choice :
Most energy necessary to remove one electron:   
             Cu	             Cu+		  Cu2+      
   Has the greatest surplus of protons; smallest the most difficult to attract electrons from
Highest electron affinity	                          
         Cl		    Br		    I               
EA is the greatest for the smallest halogen; strongest attraction between nucleus and outermost electrons), so Cl has the highest value
Greatest volume                                       
              S2-                      Ar                 Ca2+
Описание слайда:
R1 Select the best choice : Most energy necessary to remove one electron: Cu Cu+ Cu2+ Has the greatest surplus of protons; smallest the most difficult to attract electrons from Highest electron affinity Cl Br I EA is the greatest for the smallest halogen; strongest attraction between nucleus and outermost electrons), so Cl has the highest value Greatest volume S2- Ar Ca2+

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Q2
Explain why Ag+ is the most common ion for silver. 				                                    
Which is the more likely configuration for Mn2+ : [Ar]4S23d3 or [Ar]3d5  .
Описание слайда:
Q2 Explain why Ag+ is the most common ion for silver. Which is the more likely configuration for Mn2+ : [Ar]4S23d3 or [Ar]3d5 .

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R2
Ag+ is the most common ion for silver because it has  [Kr]4d10  . With filled 4d subshells	                                    (0.25pts)
Описание слайда:
R2 Ag+ is the most common ion for silver because it has [Kr]4d10 . With filled 4d subshells (0.25pts)

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R1C
The preferred configuration 
of Mn2+ is [Ar]3d5
The 3d orbital are lower in energy than the 4s. In addition, the configuration minimizes electron-electron repulsions (because each d electron is in a separate space) and maximizes the  stabilizing effect of electrons with parallel spin e)
Описание слайда:
R1C The preferred configuration of Mn2+ is [Ar]3d5 The 3d orbital are lower in energy than the 4s. In addition, the configuration minimizes electron-electron repulsions (because each d electron is in a separate space) and maximizes the stabilizing effect of electrons with parallel spin e)

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3.2 Units
A) Electromagnetic Radiation
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3.2 Units A) Electromagnetic Radiation

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Spectrum
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Spectrum

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ELECTROMAGNETIC  RADIATION
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ELECTROMAGNETIC RADIATION

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3.2 EM Radiation
A) Electromagnetic Radiation
The frequency of radiation used in a typical microwave oven is 1.00  1011 Hz. What is the energy of a mole of microwave photons with this frequency
Описание слайда:
3.2 EM Radiation A) Electromagnetic Radiation The frequency of radiation used in a typical microwave oven is 1.00  1011 Hz. What is the energy of a mole of microwave photons with this frequency

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3.2 Units
A) Electromagnetic Radiation
The frequency of radiation used in a typical microwave oven is 1.00  1011 Hz. What is the energy of a mole of microwave photons with this frequency
Описание слайда:
3.2 Units A) Electromagnetic Radiation The frequency of radiation used in a typical microwave oven is 1.00  1011 Hz. What is the energy of a mole of microwave photons with this frequency

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Solution 1
E = h, 
Multiply this value by the Avogadro constant to find the energy of a mole of photons (where the Avogadro constant, NA, is the number of entities in a mole.
E = h, to calculate the energy of one photon where
h, the Planck constant = 6.626  10–34 J s
  =  1.00  1011 Hz (s–1) 
E =  (6.626  10–34 J s)  (1.00  1011 s–1) E =  
   6.63  10–23 J
The value for a single photon can then be converted into the energy for one mole of photons by multiplying by the Avogadro constant, 6.022  1023 mol–1
Описание слайда:
Solution 1 E = h, Multiply this value by the Avogadro constant to find the energy of a mole of photons (where the Avogadro constant, NA, is the number of entities in a mole. E = h, to calculate the energy of one photon where h, the Planck constant = 6.626  10–34 J s  = 1.00  1011 Hz (s–1) E = (6.626  10–34 J s)  (1.00  1011 s–1) E = 6.63  10–23 J The value for a single photon can then be converted into the energy for one mole of photons by multiplying by the Avogadro constant, 6.022  1023 mol–1

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Solution
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Solution

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3.2 Atomic Spectra
A) What is the ionization energy (kJ/mol) for an excited state of hydrogen in which the electron has already been promoted to n = 2 level?
Описание слайда:
3.2 Atomic Spectra A) What is the ionization energy (kJ/mol) for an excited state of hydrogen in which the electron has already been promoted to n = 2 level?

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3.2 Response
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3.2 Response

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3.2 Response
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3.2 Response

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Exercise
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Exercise

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Useful Table 3.4.  The atomic spectrum of hydrogen
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Useful Table 3.4. The atomic spectrum of hydrogen

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Exercise
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Exercise

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Exercise
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Exercise

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Exercise
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Exercise

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3.2   Light Interference
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3.2 Light Interference

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1.2   Light Interference
A set of maxima and minima in an interference pattern suggests a totally different effect. As shown in Figures 3.7 and 3.8 (p.114), this experiment is demonstrating the wave like properties of light, where the interference pattern is generated from individual waves adding together (in phase) or subtracting from one another (out of phase).
Описание слайда:
1.2 Light Interference A set of maxima and minima in an interference pattern suggests a totally different effect. As shown in Figures 3.7 and 3.8 (p.114), this experiment is demonstrating the wave like properties of light, where the interference pattern is generated from individual waves adding together (in phase) or subtracting from one another (out of phase).

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Q4
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Q4

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Atomic structure, слайд №25
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Radial Distance
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Radial Distance

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Exercise
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Exercise

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Radial nodes for S = n-1 Radial nodes for p = n-2
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Radial nodes for S = n-1 Radial nodes for p = n-2

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Particle in a Box
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Particle in a Box

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Atomic structure, слайд №30
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Atomic structure, слайд №34
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Atomic structure, слайд №36
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 Show that substituting the value of r given in question C into =Asinrx and applying the normalization requirement gives  : 
 Show that substituting the value of r given in question C into =Asinrx and applying the normalization requirement gives  :
Описание слайда:
Show that substituting the value of r given in question C into =Asinrx and applying the normalization requirement gives : Show that substituting the value of r given in question C into =Asinrx and applying the normalization requirement gives :

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Atomic structure, слайд №38
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Integration
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Integration

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Schrodinger Equation
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Schrodinger Equation

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Atomic structure, слайд №41
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Answer-4
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Answer-4

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Use Slater’s rules to determine the relative sizes of N, O and F atoms.
Use Slater’s rules to determine the relative sizes of N, O and F atoms.
Описание слайда:
Use Slater’s rules to determine the relative sizes of N, O and F atoms. Use Slater’s rules to determine the relative sizes of N, O and F atoms.

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Step-1
Step-1
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Step-1 Step-1

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Use Slater’s rules to determine the relative sizes of N, O and F atoms.
Use Slater’s rules to determine the relative sizes of N, O and F atoms.
Описание слайда:
Use Slater’s rules to determine the relative sizes of N, O and F atoms. Use Slater’s rules to determine the relative sizes of N, O and F atoms.

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Atomic structure, слайд №46
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Q3
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Q3

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Atomic structure, слайд №48
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R1c
Explain factors that cause lanthanide contraction. 				                                   (0.25pts)
Explain why Ag+ is the most common ion for silver. 				                                    (0.25pts)
Which is the more likely configuration for Mn2+ : [Ar]4S23d3 or [Ar]3d5  . 	                                         (0.25pts)
Описание слайда:
R1c Explain factors that cause lanthanide contraction. (0.25pts) Explain why Ag+ is the most common ion for silver. (0.25pts) Which is the more likely configuration for Mn2+ : [Ar]4S23d3 or [Ar]3d5 . (0.25pts)

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Atomic structure, слайд №50
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Atomic structure, слайд №51
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