ionization of transition metals

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1 Ionization of Transition Metals K Ca Sr Ti V Cr Mn Fe

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Ionization of Transition Metals. K Ca Sr Ti V Cr Mn Fe. Oxidation Numbers: Examples. CaBr 2 CO CO 2 Mg 3 N 2 P 4 O 10 (NH 4 ) 2 S BeF 2 SO 2. H 2 O CH 4 NH 4 Cl NaH CaH 2 KCl RbNO 3 SrSO 4. CHAPTER 7. Chemical Bonding. Chemical Bonds. - PowerPoint PPT Presentation

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Page 1: Ionization of Transition Metals

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Ionization of Transition Metals

K Ca Sr Ti V Cr Mn Fe

Page 2: Ionization of Transition Metals

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Oxidation Numbers: Examples H2O

CH4

NH4Cl NaH CaH2

KCl RbNO3

SrSO4

CaBr2

CO CO2

Mg3N2

P4O10

(NH4)2S

BeF2

SO2

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CHAPTER 7Chemical Bonding

Page 4: Ionization of Transition Metals

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Chemical Bonds Attractive forces that hold atoms together in

compounds are called chemical bonds. There are two main types of chemical bonds

Ionic bonds – resulting from electrostatic attraction between cations and anions

Covalent bonds – resulting from sharing of one or more electron pairs between two atoms

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Lewis Dot Formulas

Valence electrons Electrons which are involved

in chemical bonding These are usually the

outermost electrons These electrons are most

important chemically

Schematic representations of valence electrons in atoms

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Lewis Dot Formulas: Single Atoms We show only electrons in the

outermost occupied shell An electron pair is represented

as a pair of dots An unpaired electron is

represented as a single dot

Li

Na

F

Cl

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Formation of Ionic Compounds

Consider reaction between metallic sodium and gaseous chlorine Electron configurations of the elements

NaCl

Sodium atom has low ionization energy and easily looses the only 3s electron forming a cation of Na+

Chlorine atom has highly negative electron affinity and readily gains an electron becoming an anion of Cl–

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Formation of Ionic Compounds

Na – e– Na+

Cl + e– Cl–Na + Cl Na+ + Cl–

Na+ cations and Cl– anions are electrostatically attracted to each other resulting in an extended ionic lattice

The high energy of the lattice overcomes all other factors involved in the formation of NaCl from elemental sodium and chlorine

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Formation of Ionic Compounds

Na + Cl Na+ Cl– We can write this equation using

Lewis dot formulas

The complete equation is2 Na + Cl2 2 Na+ Cl–

It can also be written as

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Alkali Metals + Halogens Reaction with halogens leading to the formation

of ionic halides M+X– is a general chemical property of alkali metals

It is also a general chemical property of halogens2M(s) + X2 2MX(s)

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Ionic Bonding Electrostatic interaction Non-directional The central ion attempts to

maximize the number of interactions with the ions of opposite charge

Formation of an ionic compound involves loss of electrons by metal (oxidation) and gain of electrons by nonmetal (reduction)

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Alkali Earth Metals + Halogens

The remainder of the IIA metals and VIIA nonmetals react similarly:

M(s) + X2 M2+ (X–)2 (s)

Ca + F2

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Formation of Ionic Compounds

Li + O2

The remainder of the IA metals and VIA nonmetals react similarly:4M(s) + O2(g) 2(M+

2 O2–) (s)

2M(s) + X(s) M+2

X2– (s)X = S, Se, Te, Po

Page 14: Ionization of Transition Metals

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Formation of Ionic Compounds

Mg + O2

The remainder of the IIA metals and VIA nonmetals react similarly:2M(s) + O2(g) 2(M2+ O2–) (s)

M(s) + X(s) M2+ X2– (s)X = S, Se, Te, Po

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ExampleWrite the reaction between calcium and nitrogen. Show what happens to valence electrons using Lewis dot formulas.

Learn Table 7-2

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Covalent Bonding If the difference in electronegativity

of two elements is not large enough, an electron cannot be transferred completely from one atom to the other

It becomes shared between both atoms and a covalent bond is formed

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Formation of H2 Molecule When two H atoms are indefinitely far

from each other, they do not interact If the separation decreases to a

certain distance, the 1s electron of each H atom is attracted by the nucleus of the other H atom, as well as by its own nucleus

If electrons from different atoms can occupy the same orbital, they will form a covalent bond

Page 18: Ionization of Transition Metals

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H2 Molecule We can use Lewis dot formulas

to show covalent bond formation

The covalently bonded atoms are held at a distance corresponding to the lowest total energy

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Covalent Bond We say that the covalent bond is formed

by the overlap of atomic orbitals The covalently bonded atoms are held

together by a pair of shared electons The distance between their nuclei

corresponds to the lowest total energy Below this equilibrium distance the

nucleus-nucleus and electron-electron repulsions become too large, pushing the nuclei back to the equilibrium distance

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HF Molecule

F is more electronegative than H In this molecule the electron pair

will be shifted towards the F atom

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F2 Molecule

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H2O Molecule

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NH3 Molecule

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NH4+ Ion

Lewis formulas can also be drawn for polyatomic ions

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Bonding & Nonbonding Electrons

Representative elements usually attain stable noble gas electron configurations in most of their compounds

Electrons which are shared among two atoms are called bonding electrons

Unshared electrons are called lone pairs or nonbonding electrons

Lewis dot formulas are based on the octet rule

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The Octet RuleS = N - A

S = total number of electrons shared in bonds

N = total number of electrons needed to achieve a noble gas configuration 8 for representative elements 2 for H atoms

A = total number of electrons available in valence shells of the atoms A is equal to the periodic group

number for each element A-S = number of electrons in lone pairs

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Examples F2

H2O

CH4

CO2

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Covalent Bonding Covalent bonds are formed

when atoms share electrons If the atoms share 2 electrons a

single covalent bond is formed If the atoms share 4 electrons a

double covalent bond is formed If the atoms share 6 electrons a

triple covalent bond is formed

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For ions we must adjust the number of electrons available, A: Add one e- to A for each negative charge Subtract one e- from A for each positive charge

The Octet Rule

Example: NH4+

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Assignments & Reminders Go through the lecture notes Read Sections 7-1 through 7-5 Read Sections 4-5 & 4-6 of Chapter 4 Homework #3 due by Oct. 10 Monday (10/10) and Tuesday (10/11) –

lecture quiz #3 based on Chapters 5&6