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Interpretation:
The relative hardnesses and melting points of rubidium, vanadium, and cadmium with the use of condensed electron configurations are to be predicted.
Concept introduction:
The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.
Metallic bonds are formed between metals. A metallic bond is formed between positively charged ions in a sea of delocalized electrons. The electrons in metals are delocalized over the entire metal.
The metallic character is characterized by the tendency of metals to lose their outermost valence shell electrons. Greater the ease of electron removal, higher will be the electropositivity of the corresponding elements and vice versa.
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Chapter 9 Solutions
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
- 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.arrow_forwardNonearrow_forwardNonearrow_forward
- Q2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forward
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