Wikipedia:Reference desk/Archives/Science/2011 May 23

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May 23

sp3d and sp3d2

Do sp3d and sp3d2 hybrid orbitals look different in shape with spx orbitals?. all spx orcitals look the same.--Irrational number (talk) 09:15, 23 May 2011 (UTC)

All spx orbitals do not look the same, sp looks different than sp2 which looks different than sp3. While all sp3 orbitals look identical to each other, sp3 orbitals have a smaller "back lobe" than sp2 orbitals do, and sp2 orbitals have a smaller back lobe than sp orbitals do. This is a function of the amount of "s character" of the orbital; an sp orbital is 50% s character, so it should be more symmetrical around the nucleus. sp3 orbitals are only 25% s, so are more lopsided. If you want to find the actual shapes of sp3d and sp3d2 orbitals, using google images, I turned up this picture: [1] which has pictures of all of the hybrid orbitals from sp up through sp3d2. However, when you get down to the "nitty gritty", expanded octet molecules are better explained by molecular orbital theory, in that their bonding angles and bonding distances are better modeled by that theory than hybridization theory; hybridizing d-orbitals with s and p from the same n-level don't produce realistic results given the large energy gaps involved. --Jayron32 13:00, 23 May 2011 (UTC)

Where does the 1/2 come from?

I was wondering about the classical mechanics formula for kinetic energy (1/2mv^2), and I just can´t seem to find where does that 1/2 come from. I thought of the possibility of it being or having something to do with the integral of momentum (p=mv, making it's integral: 1/2mv^2 + c), but that doesn't seem to make any sense... Thanks in advance.190.25.196.172 (talk) 22:02, 23 May 2011 (UTC)

See Kinetic energy#Derivation. Red Act (talk) 22:19, 23 May 2011 (UTC)

That's a little dense - what they don't say very explicitly is that force is the derivative of momentum.

F = m a , p = m v , F = dp/dt, a = dv/dt . The only difference is the scalar mass. Wnt (talk) 23:24, 23 May 2011 (UTC)

A non-standard derivation from special relativity is given here. What is non-standard here is that I work in natural units, so unlike most textbooks derivations, I don't cheat with the way c appears in my equations. Count Iblis (talk) 23:58, 23 May 2011 (UTC)

The link supplied by Red Act gives a sound explanation based on differential equations. An alternative explanation is as follows:

The purpose of defining kinetic energy is to facilitate the work-energy theorem for classical mechanics which states that the work done on an object is equal to the change in the kinetic energy of the object.

$W=KE_{2}-KE_{1}$

$W=F.s$ where F.s is the dot product of force F and displacement s

Applying Newton’s second law of motion:

$W=m.a.s$ where m is the mass of the object and a is its acceleration

From kinematics $v^{2}=u^{2}+2as$ where u is the initial speed, and v is the final speed after the object has been displaced a distance s.

$as={\frac {v^{2}-u^{2}}{2}}$

$W=m\left({\frac {v^{2}}{2}}-{\frac {u^{2}}{2}}\right)$

$W=KE_{2}-KE_{1}={\frac {mv^{2}}{2}}-{\frac {mu^{2}}{2}}$

Dolphin (t) 00:02, 24 May 2011 (UTC)

That's completely back-to-front, the result is being used to prove the question. The differential equations are required to to obtain the $v^{2}=u^{2}+2as$ result. SpinningSpark 18:46, 24 May 2011 (UTC)

The nature of the question suggests the OP may be just coming to terms with kinetic energy, and therefore is probably completely unfamiliar with calculus and differential equations. Are you able to explain kinetic energy to such a person? Dolphin (t) 01:50, 25 May 2011 (UTC)

The OP talks about integration in their question and is clearly capable of carrying out the simple integration required;

$E=\int F.dx=\int ma.dx=m\int {\frac {dv}{dt}}.dx=m\int v.dv={1 \over 2}mv^{2}$

Everything else is just adding confusion. It really does not give any insight into explaining an equation that has been plucked out of thin air (1/2mv²) by plucking another unexplained equation out of thin air. If I had to explain this to someone who had no calculus abilities at all I would use the special cass of constant acceleration (just as you have done with the v^2 = u^2.... formulation without actually making that clear) and substitute average velocity for x over t:

$E=Fx=max=m{\frac {v-u}{t}}x=m(v-u){\bar {v}}=m(v-u){\frac {v+u}{2}}$

which for zero initial velocity (or alternatively the increase in KE) gives the requried result,

$E={1 \over 2}mv^{2}$ SpinningSpark 06:19, 25 May 2011 (UTC)

Something wrong there. Please check the following:

${\bar {v}}={\frac {v-u}{2}}$ Dolphin (t) 07:53, 25 May 2011 (UTC)

Oops, should be + of course. Now corrected along with another error. SpinningSpark 09:00, 25 May 2011 (UTC)

Thanks for those corrections.

You have made good use of the expressions $(v-u)$ and ${\frac {v+u}{2}}$.

${\frac {(v-u)}{t}}=a$

and therefore $(v-u)=at$

${\frac {v+u}{2}}={\frac {s}{t}}$

Multiplying these two expressions we get:

$(v+u)(v-u)=2as$

and therefore $v^{2}-u^{2}=2as$

Does this look familiar? Dolphin (t) 12:47, 25 May 2011 (UTC)

Don't be condescending. Of course I knew the relationship here. That is completely beside the point. The OP is asking for an explanation, not another equation that would need even more explanation. Now stop cluttering this thread with this nonsense. If you want to criticize me further go do it on my talk page. SpinningSpark 16:46, 25 May 2011 (UTC)