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Properties of Ions and Covalent Bonding, Slides of Chemical Principles

An in-depth analysis of ionic and covalent bonding, with a focus on the formation of stable bonds, the concept of noble-gas configurations, and the differences between ionic and polar covalent compounds. It also includes examples of ionic and covalent compounds, as well as calculations of dipole moments and bond energies.

Typology: Slides

2013/2014

Uploaded on 01/31/2014

samiksha
samiksha 🇮🇳

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Download Properties of Ions and Covalent Bonding and more Slides Chemical Principles in PDF only on Docsity! Ionic to Covalent • Outline – Binary Ionic Compounds – Partial Ionic Compounds – Covalent Compounds docsity.com Properties of Ions • When will a stable bond be formed? • When one exams a series of stable compounds, it becomes evident that in the majority of compounds, bonding is achieved such that atoms can achieve a nobel-gas configuration • Example: NaCl versus Na+Cl- Na: [Ne]3s1 Cl: [Ne]3s23p5 Na+: [Ne] Cl-: [Ne]3s23p6 = [Ar] docsity.com Properties of Ions (cont.) • Ions on figure correspond to nobel-gas electron configurations. • To form ionic binary compounds, one simply combines in proportions such that total charge is zero. • This approach is not to be applied to transition metals. docsity.com Properties of Ions (cont.) • Note that size decreases for isoelectronic species. • Mainly a consequence of increased charge of nucleus. docsity.com Magnesium (Mg) forms compounds of the form MgX2. What group is X from? A. Group 1 B. Group 2 C. Group 6 D. Group 7 Mg docsity.com Dipole Moments (Lecture 21) • The dipole moment () is defined as:  = QR Charge magnitude Separation distance + center R docsity.com Dipole Moments (cont.) • Example, the dipole moment of HF is 1.83 D. What would it be if HF formed an ionic bond (bond length = 92 pm)?  = (1.6 x 10-19 C)(9.2 x 10-11 m) = 1.5 x 10-29 C.m x (1D/3.336 x 10-30 C.m) = 4.4 D docsity.com Partial Ionic Compounds (cont.) • We can define the ionic character of bonds as follows: % Ionic Character = x 100% (dipole moment X-Y)experimental (dipole moment X+Y-)calculated docsity.com Covalent Compounds • In covalent bonding, electrons are “shared” between bonding partners. • In ionic bonding, Coulombic interactions resulted in the bonding elements being more stable than the separated atoms. • What about covalent bonds…what is the “driving force”? docsity.com Covalent Compounds (cont.) ¢ Back to H,. Hatom Sufficiently far apart to have no interaction H atom H atom The atoms begin to interact as they move closer together. H, molecule Optimum distance to achieve (a) lowest overall energy of system " Hatom Energy (kJ Cc -458 1 Tnternuclear distance (nam) —H——_______ > (b) (H—H bond length) docsity.com Covalent Compounds (cont.) • The same concept can be envisioned for other covalent compounds: Think of the covalent bond as the electron density existing between the C and H atoms. docsity.com Covalent Compounds (cont.) • Example: It takes 1578 kJ/mol to decompose CH3Cl into its atomic constituents. What is the energy of the C-Cl bond? CH3Cl: 3 C-H bonds and 1 C-Cl bond. 3 (C-H bond energy) + C-Cl bond energy = 1578 kJ/mol 413 kJ/mol 1239 kJ/mol + C-Cl bond energy = 1578 kJ/mol C-Cl bond energy = 339 kJ/mol docsity.com Covalent Compounds (cont.) • We can use these bond energies to determine Hrxn: H = sum of energy required to break bonds (positive….heat into system) plus the sum of energy released when the new bonds are formed (negative….heat out from system). Hrxn  Dbonds broken  Dbonds formed docsity.com Covalent Compounds (cont.) • Example: Calculate H for the following reaction using the bond enthalpy method. CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g) Go to Table 13.6: C-H 413 O=O 495 O-H 467 C=O 745 4 x 2 x 4 x 2 x docsity.com
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