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Interpretation: The binding capacity for group 2A cation towards anion gets increases as we move from bottom to top the reason for that trend should be explained.
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. There are seven periods and 18 groups present in the table and some of those groups are given specific name as follows,
Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the
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.
In periodic table the horizontal rows are called periods and the vertical column are called group.
Electronic configuration: It is used to denote the distribution of electrons placed over orbitals that present in the atom.
The rules followed by the electrons are as follows,
The electrons gets distributed starting from lower energy orbital to higher energy orbital, pairing of electrons starts after all the orbitals are singly filled and finally, no two electrons that have same spin can occupy the same orbital.
To Explain: The reason for the increased binding capacity for group 2A as we move from bottom to top.
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Chapter 4 Solutions
Chemistry: Atoms First
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: 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_forward
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- Nonearrow_forwardQ1: Draw the most stable and the least stable Newman projections about the C2-C3 bond for each of the following isomers (A-C). Are the barriers to rotation identical for enantiomers A and B? How about the diastereomers (A versus C or B versus C)? H Br H Br (S) CH3 (R) CH3 H3C (S) H3C H Br Br H A C enantiomers H Br H Br (R) CH3 H3C (R) (S) CH3 H3C H Br Br H B D identicalarrow_forward2. Histamine (below structure) is a signal molecule involved in immune response and is a neurotransmitter. Histamine features imidazole ring which is an aromatic heterocycle. Please answer the following questions regarding Histamine. b a HN =N C NH2 a. Determine hybridization of each N atom (s, p, sp, sp², sp³, etc.) in histamine N-a hybridization: N-b hybridization: N-c hybridization: b. Determine what atomic orbitals (s, p, sp, sp², sp³, etc.) of the lone pair of each N atom resided in N-a hybridization: N-b hybridization: N-c hybridization:arrow_forward
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