How do alkali metals differ from alkaline each metals in atomic structure and general properties?
Expert Solution & Answer
Interpretation Introduction
Interpretation:
The differences between alkali and alkaline earth metals in terms of their atomic structure and properties needs to be explained.
Concept Introduction:
The alkali and alkaline earth metals are referred as the ‘s-block’ elements and belong to Group IA and IIA respectively in the periodic table
Atomic structure of an element depicts the arrangement of electrons around the nucleus of an atom
Variations in the physical and chemical properties across elements present in the periodic table can be explained on the basis of their atomic structure.
Answer to Problem 1E
The main difference between the alkali and alkaline earth metals lies in their electron configuration with the former having one valence electron while the latter having two. It is this difference which brings about variations in their properties.
Explanation of Solution
The alkali earth metal include: Li, Na, K, Rb, Cs and Fr. They have a valence electronic configuration which can be generally represented as ns1, where ‘n’ is the principle quantum number.
The alkaline earth metals include: Be, Mg, Ca, Sr, Ba and Ra. In contrast to alkali metals they have a fully filled outer s-orbital with a valence configuration represented as ns2.
Here are some of the key differences between the two groups
Alkali metals have one electron in their outermost shell whereas, alkaline earth metals have two electrons
Alkali metals form cations with a charge of +1 unlike +2 cations in the case of alkaline earth metals
Alkali metals are more reactive than the alkaline series
Alkali metals tend to be softer and can be cut with a sharp knife. While alkaline metals are harder and form stronger metallic bonds
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Draw the complete mechanism for the reaction below. Please include appropriate arrows, intermediates, and formal charges.
(c) The following data have been obtained for the hydrolysis of sucrose, C12H22O11, to
glucose, C6H12O6, and fructose C6H12O6, in acidic solution:
C12H22O11 + H2O → C6H12O6 + C6H12O6
[sucrose]/mol dm³
t/min
0
0.316
14
0.300
39
0.274
60
0.256
80
0.238
110
0.211
(i) Graphically prove the order of the reaction and determine the rate constant of the
reaction.
(ii) Determine the half-life, t½ for the hydrolysis of sucrose.
(III) adsorbent
(b) Adsorption of the hexacyanoferrate (III) ion, [Fe(CN)6] ³, on y-Al2O3 from aqueous
solution was examined. The adsorption was modelled using a modified Langmuir
isotherm, yielding the following values of Kat pH = 6.5:
(ii)
T/K
10-10 K
280
2.505
295
1.819
310
1.364
325
1.050
Determine the enthalpy of adsorption, AadsHⓇ.
If the reported value of entropy of adsorption, Aads Se = 146 J K-1 mol-1 under the above
conditions, determine Aads Gº.
Campbell Essential Biology with Physiology (5th Edition)
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