Presentation on theme: "Ionic Bonding Electron Transfer e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e-"— Presentation transcript:
Ionic Bonding Electron Transfer e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e-
Ionic eh? Suppose ions are involved? You guessed it! Ionic bonding is the result of a transfer of electrons between two (or more) atoms. The result of the electron transfer are ions (charged atoms). Recall that a cation is an ions with a positive charge; while an anion is an atom with a negative charge.
When is bonding ionic? The general rule of thumb is that a metal and a non-metal make an ionic compound. Δ This is true more often than not, but to be sure we should use the electronegativity difference (ΔEN). Δ To calculate ΔEN we take the greater electronegativity value (atom 1) and subtract the lesser (atom 2). Δ ΔENBonding Character > 1.7Ionic 0.4 – 1.7Polar Covalent < 0.4Non-Polar Covalent 0Purely Covalent Our focus for today.
Testing Δ EN Will sodium chloride (NaCl) be ionic or covalent? Metal and non-metal should be ionic. Δ ΔEN = 3.16 (Cl) – 0.93 (Na) = 2.23 Δ ΔEN > 1.7, therefore NaCl is ionic! Will beryllium (BeCl 2 ) be ionic or covalent? Metal and non-metal should be ionic. Δ ΔEN = 3.16 (Cl) – 1.57 (Be) = 1.59 Δ 0.4 < ΔEN < 1.7, therefore BeCl 2 is polar covalent!!! From now on, check the Δ ΔΔ Δ EN to be sure of the type of bonding!
So how does it work? Ionic bonding arises from transfer of electrons from metals to non-metals in an attempt by each species to gain noble gas configuration. HUH?!?!? “Every atom wants 8 electrons!” (the octet rule) Atoms from Groups 1 – 13 will give up their electrons to make cations and achieve noble gas configuration. Atoms from Groups 15 – 17 will gain electrons and achieve noble gas configuration by making anions.
So how does it work? The ionic bonding 3 step process (consider NaCl): 1)Na loses an electron (after I.E. provided) to make Na + + 1e - 2)Cl picks up the electron (and releases E.A.) from sodium to make Cl - 3)Na + and Cl - come together to make NaCl (and release lattice energy_ Stable octets!!! (noble gas configurations)
An Animated Look The ionic bonding 3 step process 1)2Na + IE 2Na + + 2e - 2)Cl 2 + 2e - 2Cl - + EA 3)2Na + Cl 2 2Na + Cl - (2NaCl) + LE
An Electronic Look Consider the formation of NaCl Na + IE Na + + 1e - 3s 1 2p 6 Cl + 1e- Cl - + EA (3p 5 3p 6 )
The Ionic Lattice Ionic compounds typically: have very high melting points conduct electricity in the molten state and when dissolved in water are brittle (fracture) solids have low volatility The strength of the bonding interaction is the result of the sum of all bonding interactions, every Na + affects every Cl -. While the individual bonds between 2 atoms are quite weak the overall effect of the lattice makes the interactions very strong.
Lattice Energy When an ionic lattice forms it releases a large amount of energy which we call the lattice energy. (Need to put this in to break it apart) The smaller the ions, the tighter the fit an higher the lattice energy. LiF H latt = -1036 kJ/mol NaCl H latt = -787.3 kJ/mol Lattice energy also becomes MUCH stronger as the charge on the ions becomes larger. Na + O 2- (Na 2 O) = -2481 kJ/mol Al 3+ O 2- (Al 2 O 3 ) = - 15916 kJ/mol
A look a physical properties Like dissolves like. A polar substance, such as water will easily dissolve an ionic compound. The electronegative oxygen atoms pickup Na + while the electropositive hydrogens pickup the Cl -. As the lattice energy terms gets larger in magnitude the solids are much harder to break up and dissolve in water.
Ions and Atomic Size Atomic radius largely depends on the effective nuclear charge. Lets consider what happens to the size of an atom when it loses electrons (effective nuclear charge goes up atom gets smaller) When an atom gains electrons the effective nuclear charge decreases, the electrons get farther away it gets bigger Li + Be 2+ B 3+ C 4+ N 3- O 2- F-F- A cation is smaller than the original atom while an anion is larger.
Ionic Compounds 1.Check the ionization energy to see if each molecule will have an ionic bond. 2.Draw the Lewis structures for each element 3.Give the 3 step solutions for the formation of the following compounds: a)Aluminum nitride b)Rubidium sulfide c)Magnesium oxide d)Potassium bromide e)Sodium oxide f)Lithium chloride g)Potassium sulfide h)Francium fluoride i)Strontium iodide j) Cesium phosphide