December 14, 2019, 08:59:02 AM
Forum Rules: Read This Before Posting


Topic: dsp hybridized bonding in Titanium?  (Read 15375 times)

0 Members and 1 Guest are viewing this topic.

Offline jonogt

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
dsp hybridized bonding in Titanium?
« on: January 28, 2008, 12:48:48 PM »
I'm supposed to explain why Ti has so much higher of a melting point than K. My structural materials textbook talks about how the farther right of the alkali metals you get, the less metallic and the more covalent the bond gets because of more available valence e's to bond.  Makes sense.  It also says Ti, as well as the other transition metals, are subject to dsp hybridized bonding

The only info I can seem to find in my book and online about this is that it's a metallic-covalent mixed bond, which again makes sense, but I'm really not seeing what it is.  I know that d,s, and p are the types of orbitals, but intermolecular bonds aren't occuring on all 3 of these orbitals, are they?   Can someone explain it a little more logically?

thanks
-Jon

Offline Alpha-Omega

  • Full Member
  • ****
  • Posts: 693
  • Mole Snacks: +360/-231
  • Gender: Female
  • Physical Inorganic Chemist
Re: dsp hybridized bonding in Titanium?
« Reply #1 on: February 03, 2008, 05:09:19 AM »
K and Ti both have d shells available for bonding.  K is an alkali metal with an oxidation state of +1.  K, otassum will form ionic bonds with the elements in Group 7 (halogens-salt makers).

Ti is a transion metal.  Transition form coordination complexes.  If you place transition metals in solution you can determine the corresponding oxidation states of the resulting ions. Transition metals have several oxidation states, but nonbonded transition metals only exhibit relatively low oxidation states. Many transition metals share the same oxidation states, and each row follows similar patterns.

Transition metals form coordination complexes....in solution this is often evidenced by highly colored solutions.

See this link:  http://en.wikipedia.org/wiki/Titanium

Yes, Ti has what is called dsp hybridization/bonding.  And yes all three orbitals participate in the bonding.

K:  1s2, 2s2, 2p6 3s2 3p6 4s2   or  [Ar] 4s2

Ti:  1s2 2s2 2p6 3s2 3p6 4s2 3d2  or   [Ar] 4s2 3d2

Ti+2:[Ar]3d2

Ti+3:[Ar]3d1

For Ti we can write Ti:  1s2 2s2 2p6 3s2 3p6 3d2 4s2 

So for Ti we have 3s2 3p6 3d2

Ti forms d2sp3 Hybrids

These particular d2sp3 hybrids are combinations of two 3d, the 3s, and three 3p functions.

The 3s :  3 is the energy level, S is the subshell, 1 orbital

The 3p:  3 is the energy, p is the subshell, 3 orbitals px, py, and pz

The 3d:  3 is the energy, d is the subshell, 5 orbitals dyz, dxz, dxy, dx2-y2, and dz2.

In general:

S subshell has only 1 orbital.
The p subshell has 3 orbitals.
The d subshell has 5 orbitals.
The f subshell has 7 orbitals.

In Ti there are 2 electrons in the 3d shell.

Hence, the designation  d2sp3 Hybrids (TiCl-an octahedral complex)

For octahedral complexes:  There are six d2sp3 hybrid orbitals. They are arranged in an octahedral layout so that each has four other orbitals at 90° to it and one at 180°.

The following links should help you too:

http://faculty.colostate-pueblo.edu/linda.wilkes/111/3c.htm

http://winter.group.shef.ac.uk/orbitron/AO-hybrids/d2sp3/index.html

http://www.chem1.com/acad/webtext/chembond/cb09.html

http://en.wikipedia.org/wiki/Titanium(III)_chloride
« Last Edit: February 03, 2008, 07:45:38 PM by Alpha-Omega »

Sponsored Links