Reason behind significantly small radius even though it lies beneath in the group should be suggested. Concept introduction: As one moves from top to bottom the shells expands and size increases, while it decreases as one move along the period because of gradual increases in effective nuclear charge. This in turn increases influence of nucleus on valence electron and shrinks the radius along the horizontal row. Lanthanide contraction is phenomenon that leads to significant reduction in sizes along the period in lanthanides. This is attributed to gradual fill up of 4 f orbitals that are highly diffused in shape. Poor overlap increases influence of nucleus for outermost electron manifolds for lanthanides and hence size reduces considerably.
Reason behind significantly small radius even though it lies beneath in the group should be suggested. Concept introduction: As one moves from top to bottom the shells expands and size increases, while it decreases as one move along the period because of gradual increases in effective nuclear charge. This in turn increases influence of nucleus on valence electron and shrinks the radius along the horizontal row. Lanthanide contraction is phenomenon that leads to significant reduction in sizes along the period in lanthanides. This is attributed to gradual fill up of 4 f orbitals that are highly diffused in shape. Poor overlap increases influence of nucleus for outermost electron manifolds for lanthanides and hence size reduces considerably.
Solution Summary: The author explains that lanthanide contraction leads to significant reduction in sizes along the period in d-block elements.
Definition Definition Elements containing partially filled d-subshell in their ground state configuration. Elements in the d-block of the periodic table receive the last or valence electron in the d-orbital. The groups from IIIB to VIIIB and IB to IIB comprise the d-block elements.
Chapter 12, Problem 109E
Interpretation Introduction
Interpretation:Reason behind significantly small radius even though it lies beneath in the group should be suggested.
Concept introduction: As one moves from top to bottom the shells expands and size increases, while it decreases as one move along the period because of gradual increases in effective nuclear charge. This in turn increases influence of nucleus on valence electron and shrinks the radius along the horizontal row.
Lanthanide contraction is phenomenon that leads to significant reduction in sizes along the period in lanthanides. This is attributed to gradual fill up of 4f orbitals that are highly diffused in shape. Poor overlap increases influence of nucleus for outermost electron manifolds for lanthanides and hence size reduces considerably.
3. Consider the compounds below and determine if they are aromatic, antiaromatic, or
non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I
electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly
drawn and you should be able to tell that the bonding electrons and lone pair electrons
should reside in which hybridized atomic orbital 2. You should consider ring strain-
flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti-
aromaticity)
H H
N
N:
NH2
N
Aromaticity
(Circle)
Aromatic Aromatic Aromatic Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic TT
electrons
Me
H
Me
Aromaticity
(Circle)
Aromatic Aromatic Aromatic
Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic πT
electrons
H
HH…
A chemistry graduate student is studying the rate of this reaction:
2 HI (g) →H2(g) +12(g)
She fills a reaction vessel with HI and measures its concentration as the reaction proceeds:
time
(minutes)
[IH]
0
0.800M
1.0
0.301 M
2.0
0.185 M
3.0
0.134M
4.0
0.105 M
Use this data to answer the following questions.
Write the rate law for this reaction.
rate
= 0
Calculate the value of the rate constant k.
k =
Round your answer to 2 significant digits. Also be
sure your answer has the correct unit symbol.
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