Download Chem 220 In Class Socrative Qs: atomic orbitals 28/09/2016 1

Chem 220 In Class Socrative Qs: atomic orbitals
28/09/2016
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Which type of force determines the potential energy of an electron in an atom?
We’re first going to review atomic structure and properties. What properties of atoms do you think are most useful to help us understand the properties of molecules? Submit as many answers as you like.
a)
b)
c)
d)
e)
We’ll revisit this question in a couple of weeks.
The electrostatic force of attraction between the (–) electron and (+) protons in the nucleus generates the potential energy of the electron.
gravity
nuclear
friction
electrostatic
centripetal
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As the electron kinetic energy increases, what happens to the number of nodes in a wavefunction Ψ?
Relative to the nucleus, where is the point described by
(r,θ,φ) = (1 pm, 90°, 180°)?
a)
b)
c)
d)
a)
b)
c)
d)
decreases
stays the same
increases
e– KE and wavefunction nodes are unrelated
The KE of a wave depends on its curvature, how quickly the amplitude of Ψ changes in space. Tighter curvature leads to more phase changes and more nodes.
1 pm away along the negative x‐axis 1 pm away along the positive y‐axis 1 pm away along the negative y‐axis 1 pm away along the negative z‐axis r = distance from nucleus
θ = angle downward from +z
φ = counter‐clockwise angle
from +x in xy plane
z
θ = 90°
y
φ = 180°
x
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Chem 220 In Class Socrative Qs: atomic orbitals
How many of these statements help explain why an H atom 4s
orbital electron is higher energy than a 2s?
•
•
•
•
The 4s orbital has a higher value of .
The 4s orbital is larger.
The 4s orbital has more nodes.
The 4s orbital has more regions of positive amplitude.
• The 4s orbital has higher overall electron density. a) 1 b) 2 c) 3 d) 4 e) 5
Orbital energy is related to wave curvature: n ↑ = # nodes ↑ = higher energy
Orbital energy is related to e– distance and orbital size:
n ↑ = r ↑ = higher energy
28/09/2016
Is the most likely place to find the electron in an H atom at the nucleus (r = 0) or at 1 Bohr radius distance (r = 53 pm) from the nucleus?
a) At the nucleus, because Ψ2 for a 1s orbital is a maximum when r = 0
b) 53 pm from the nucleus, because 4πr2Ψ2 for a 1s orbital is a maximum when r = 1a0 (1 Bohr radius = 53 pm)
c) Both – the first answer describes the most likely single point in space, while the second describes the most likely distance away from the nucleus
d) Neither – the electron is best described as a wave with indeterminate position, so the question is nonsensical
“place” is ambiguous: Ψ2 describes probability at a point, 4πr2Ψ2 describes probability at a distance
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a)
b)
c)
d)
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As quantum number n increases the total number of nodes in an orbital increases. (# nodes = n – 1) What kind of orbital is this? To determine the energy of a specific electron in a ground state argon atom, which kind(s) information do you need?
3d
4s
4d
4f
b) Whether the electron occupies an s orbital or a p
orbital
a) The shell of the electron
c) Whether the electron is spin up or spin down
d) More than one of the above, but not all three
# total nodes = 3 = n – 1, so n = 4
# angular nodes = 2 = 
n = 4 and  = 2 means 4d
d orbitals always have two angular nodes,
but gain interior radial nodes as n 
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e) All of a) b) and c)
In multi‐e– atoms, electron energy is determined by n (the shell, so a) and  (the subshell or orbital type, so b). An electron’s spin does not affect its energy (unless a magnetic field is present).
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Chem 220 In Class Socrative Qs: atomic orbitals
28/09/2016
For the 3p orbitals in phosphorus (P), Zeff = 4.89.
Which of the following statements is true?
a) The 3p orbitals of S must have Zeff < 4.89, because S has more electrons to screen the charge than P
b) The 2p orbitals of N must have Zeff > 4.89, because they are lower energy than 3p orbitals
c) The 3s orbitals of P must have Zeff > 4.89, because they penetrate closer to the nucleus than 3p
d) All orbitals of P have Zeff = 4.89, because all P atoms have the same Z value
The 3s is lower‐E than the 3p because of greater penetration and higher Zeff. S has more protons than P, so its 3p e– experience a higher Zeff, not lower. N has far fewer protons, but its e– are lower‐E due to a smaller n (= 2 not 3).
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Which element has a ground‐state configuration of [Ne]3s23p2 ?
a)
b)
c)
d)
e)
Be
C Si Ge none of these
Valence electron configuration is 3s23p2, np2 is Group 14, n = 3 is third row (after Ne): Si
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Compared to the 2p orbitals of nitrogen (N), the 2p
orbitals of oxygen (O) are…
Consider the atomic radii of nitrogen (N) and fluorine (F). Which statement is true?
a) higher in energy, because O has a lower ionization energy than N
b) lower in energy, because O electrons experience a greater Zeff than N electrons
c) higher in energy, because N has a stable half‐filled subshell
d) the same energy, because all 2p orbitals are degenerate
e) lower in energy, because O has a less favourable electron affinity than N
a) Fluorine has a larger radius because it has more electrons.
b) Fluorine has a smaller radius because its electrons experience a greater nuclear charge.
c) Fluorine has a larger radius because its electrons experience greater e–‐e– repulsion.
d) Fluorine has a smaller radius because it is closer to a stable octet configuration.
Across a row, Zeff ↑, orbital E ↓.
r(N) = 74 pm, r(F) = 57 pm.
Across a row, Zeff ↑, orbital E ↓ , so atomic radius ↓.
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Chem 220 In Class Socrative Qs: atomic orbitals
Ionization energy of an atom generally decreases down a column of the periodic table because…
a)
There are far more electrons, so the e–‐e– repulsion increases and removing an electron becomes easier
b) The atom gets larger, decreasing e–‐e– repulsion, so the energy released upon ionization decreases.
c) The valence electrons experience greater screening, so Zeff is lower and the electrons are held less strongly.
d) Down a column, n increases, so orbital energy increases.
Down a column, n ↑ = orbital E ↑ = IE ↓ (easier to remove a higher‐E electron). Note that down a column, Zeff also increases, but this effect is smaller than the increase in n.
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28/09/2016
Of Si and P, which has the more favourable electron affinity?
a) Si, because adding an electron to Si creates a stable half‐
filled p3 configuration.
b) P, because it has lower‐E valence orbitals due to a greater Zeff.
c) Si, because the 4th p electron in P– experiences a pairing energy penalty.
d) P, because P is more electronegative and holds electrons more tightly.
e) Si, because the smaller P atom has increased electron repulsion, so it is less favoured to add an electron.
EA(Si) = –1.38 eV, EA(P) = –0.79 eV: Si is more favourable.
Across row, Zeff ↑ = orbital E ↓ = EA generally more favourable. However, Si adds an unpaired e– (3s2p2 → 3s2p3) while P adds a paired e– (3s2p3 → 3s2p4). The pairing energy penalty for P offsets the expected trend, and the EA of P is less
favoured.
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