The element titanium displays an exceptional affinity for oxygen and when heated in air immediately forms TiO2. This substance is the most widely used white pigment in paints, cosmetic skin care products, and sunscreens. The +4 oxidation state for Ti should, because of the high charge and small ionic radius (82pm), be difficult to stabilize in an ionic lattice. Indeed, for years I have taught that ionic charges > +3 are not stable in ionic structures. Perform appropriate calculations, using a Born-Haber type cycle, and an appropriate expression for ΔHL (the lattice energy) to determine whether covalent bonding in TiO2 may be an important component of the bonding and responsible for the stability of this compound.
Ti(s) + O2(g) à TiO2(s)
ΔHt (TiO2) = -519.7 kJ/mol
Data Below
Borne-Lande’ Equation: ΔHL = 138,000 Z+Z- A (1-1/n)
d
TiO2(rutile) 2.3850 (Madelung Constant, A)
O^2- =126 radii- (pm)
Ti^4+ =82 radii+(pm)
r+r=d
z=charges
n=Born Exponent Li+=5, O-=7 average =6
Heats of Atomization (kJ/mol): Ti(s) 425 , O2 498
Ionization Energies (kJ/mol):
1st I.E. 2nd I.E. 3rd I.E. 4th I.E.
Ti 658 1,310 2,653 4,175
ΔHeg(Enthalpy of Electron Gain) ; ΔHeg = -Ea (Electron Affinity)
ΔHeg (kJ/mol): ΔHeg = (-) Ea
O -141 +744
O+1e- à O^1- O^1-+1e- à O^2-
Titanium dioxide has a stable structure due to covalent bonding present in it. If TiO2 were formed solely by ionic bonding, the enathalpy of formation (Hf) for the crystal would be very high and physically not attainable.
The rutile type structure helps the Ti4+ stay at the
body center of the unit cell while the O2- ions
surrounds it.
Calculating the exact heat of formation for TiO2 by
BORN- HABER CYCLE proves the existance of covalent
bonding in the compound which is given in the attachment.
The actual heat of formation for titanium oxide is very low, which
proves that the rutile type staructure exists in covalent bonding
to maintain the stability.
Get Answers For Free
Most questions answered within 1 hours.