Interpretation:
It should be stated that how the given set of properties follows the trend from left to right across the period and top to bottom of the group in the periodic table.
Concept Introduction:
Periodic Table: The available chemical elements are arranged considering their
In periodic table the horizontal rows are called periods and the vertical column are called group.
The elements placed on the left of the table are metals which contain its last electron on s-orbital hence considered as s-block elements and the elements placed on the right side of the table contains its last electron in p-orbital which is regarded as p-block elements.
The s and p block elements together are called as main group elements.
The elements with its last electron in d-orbital are called d-block elements also called as
There are seven periods and 18 groups present in the table and some of those groups are given special name as follows,
Cation: Removal of electron from the atom results to form positively charged ion called cation.
Anion: Addition of electron to atom results to form negatively charged ion called anion.
The net charge present in the element denotes the presence or absence of electrons in the element.
Electronegativity: It is the tendency of the element to attract the pair of electrons towards itself.
Concept Introduction:
Shielding effect: The outermost electrons present in the element gets shielded or screened from the nuclear charge by electron present near to the nucleus that is the electrons that surrounds the nucleus (core electrons).
Effective nuclear charge: It is the total positive charge experienced by the valence electrons from the nucleus.
Atomic radius:
Atomic radius is the distance between the atomic nucleus and outermost electron of an atom. From the atomic radius, the size of atoms can be visualized. But there is no specific distance from nucleus to electron due to electron cloud around the atom does not have well-defined boundary.
Electronegativity: It is the tendency of the element to attract the pair of electrons towards itself.
Concept Introduction:
First ionization energy:
The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.
Electronegativity: It is the tendency of the element to attract the pair of electrons towards itself.
Concept Introduction:
Acidity of oxides:
The more electronegativity the element has then the respective oxide of the element has more acidity.
Electronegativity: It is the tendency of the element to attract the pair of electrons towards itself.

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Chapter 4 Solutions
CHEMISTRY:ATOMS FIRST (LL)>CUSTOM PKG.<
- A concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.10 V at 25 °C. What is the ratio of [Sn2+] (i.e., [Sn2+left-half] / [Sn2+right-half])?arrow_forwardElectrochemical cell potentials can be used to determine equilibrium constants that would be otherwise difficult to determine because concentrations are small. What is Κ for the following balanced reaction if E˚ = +0.0218 V? 3 Zn(s) + 2 Cr3+(aq) → 3 Zn2+(aq) + Cr(s) E˚ = +0.0218 Varrow_forwardConsider the following half-reactions: Hg2+(aq) + 2e– → Hg(l) E°red = +0.854 V Cu2+(aq) + 2e– → Cu(s)E°red = +0.337 V Ni2+(aq) + 2e– → Ni(s) E°red = -0.250 V Fe2+(aq) + 2e– → Fe(s) E°red = -0.440 V Zn2+(aq) + 2e– → Zn(s) E°red = -0.763 V What is the best oxidizing agent shown above (i.e., the substance that is most likely to be reduced)?arrow_forward
- Calculate the equilibrium constant, K, for MnO2(s) + 4 H+(aq) + Zn(s) → Mn2+(aq) + 2 H2O(l) + Zn2+(aq)arrow_forwardIn the drawing area below, draw the condensed structures of formic acid and ethyl formate. You can draw the two molecules in any arrangement you like, so long as they don't touch. Click anywhere to draw the first atom of your structure. A C narrow_forwardWrite the complete common (not IUPAC) name of each molecule below. Note: if a molecule is one of a pair of enantiomers, be sure you start its name with D- or L- so we know which enantiomer it is. molecule Ο C=O common name (not the IUPAC name) H ☐ H3N CH₂OH 0- C=O H NH3 CH₂SH H3N ☐ ☐ X Garrow_forward
- (Part A) Provide structures of the FGI products and missing reagents (dashed box) 1 eq Na* H* H -H B1 B4 R1 H2 (gas) Lindlar's catalyst A1 Br2 MeOH H2 (gas) Lindlar's catalyst MeO. OMe C6H1402 B2 B3 A1 Product carbons' origins Draw a box around product C's that came from A1. Draw a dashed box around product C's that came from B1.arrow_forwardClassify each of the amino acids below. Note for advanced students: none of these amino acids are found in normal proteins. X CH2 H3N-CH-COOH3N-CH-COO- H3N-CH-COO CH2 CH3-C-CH3 CH2 NH3 N NH (Choose one) ▼ (Choose one) S CH2 OH (Choose one) ▼ + H3N-CH-COO¯ CH2 H3N CH COO H3N-CH-COO CH2 오오 CH CH3 CH2 + O C CH3 O= O_ (Choose one) (Choose one) ▼ (Choose one) Garrow_forwardAnother standard reference electrode is the standard calomel electrode: Hg2Cl2(s) (calomel) + 2e2 Hg() +2 Cl(aq) This electrode is usually constructed with saturated KCI to keep the Cl- concentration constant (similar to what we discussed with the Ag-AgCl electrode). Under these conditions the potential of this half-cell is 0.241 V. A measurement was taken by dipping a Cu wire and a saturated calomel electrode into a CuSO4 solution: saturated calomel electrode potentiometer copper wire CuSO4 a) Write the half reaction for the Cu electrode. b) Write the Nernst equation for the Cu electrode, which will include [Cu2+] c) If the voltage on the potentiometer reads 0.068 V, solve for [Cu²+].arrow_forward
- 2. (Part B). Identify a sequence of FGI that prepares the Synthesis Target 2,4-dimethoxy- pentane. All carbons in the Synthesis Target must start as carbons in either ethyne, propyne or methanol. Hint: use your analysis of Product carbons' origins (Part A) to identify possible structure(s) of a precursor that can be converted to the Synthesis Target using one FGI. All carbons in the Synthesis Target must start as carbons in one of the three compounds below. H = -H H = -Me ethyne propyne Synthesis Target 2,4-dimethoxypentane MeOH methanol OMe OMe MeO. OMe C₂H₁₂O₂ Product carbons' origins Draw a box around product C's that came from A1. Draw a dashed box around product C's that came from B1.arrow_forwardDraw the skeletal ("line") structure of the smallest organic molecule that produces potassium 3-hydroxypropanoate when reacted with KOH. Click and drag to start drawing a structure. Sarrow_forwardDraw the skeleatal strucarrow_forward
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