text:
Conservation of angular momentum in electronic transition
My book says the "angular momentum is always conserved in a transition". But if suppose we have a s electron then it's angular momentum is $\hbar\sqrt{l(l+1)}+\hbar\sqrt{s(s+1)}$ $\implies \hbar\frac{\sqrt{3}}{2}$ and suppose it transits to a p electron (which is in accordance with the selection rule as $\Delta l=1$) which has an angular momentum of           $\hbar\sqrt2 + \hbar\frac{\sqrt{3}}{2}$. As the angular momentum of a photon is $\hbar$ then it means (mathematically) that we would have to add $\sqrt2$photons to the atom but that's not possible. Then how is the angular momentum conserved??
Also I can't see why the spin angular momentum of a photon is $\hbar$ as spin angular momentum is $\hbar\sqrt{s(s+1)}$ and as $s=1$ for a photon so it should be $\hbar\sqrt2$.
P.S. the book I am using is Elements of Physical Chemistry by Peter Atkins and Julia de Paula.

I think you have misunderstood some concept. The angular momentum of spin is just $\sqrt{s(s+1)}\hbar$. The spin which we always say is spin projection quantum number in fact not the angular momentum. And I'm not familiar with chemistry, but I know that we can not add up the orbital and spin angular momentum directly because they are not in the same direction. Sometimes we can add up the quantum number. If the spin and orbital angular momentum in opposite direction, we should subtract.

I have some idea about your example. The electron transition is "electron jumps from one energy level to another". So the energy must change. Where does the energy come from? If it is photon, this example has no paradox. But can the energy come from other way? I have no idea.

I had that in the back of my mind because orbital and spin angular momentum are both vectors but I thought it would be better to clarify here.

@ShrayanshJyoti I doesn't understand what you say. In fact, angular momentum in quantum is not a classical angular momentum.

"The classical definition of angular momentum as $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ can be carried over to quantum mechanics, by reinterpreting r as the quantum position operator and p as the quantum momentum operator. L is then an operator, specifically called the orbital angular momentum operator."--Says Wikipedia.

Yes, that's right.


text:
Why don't molecules of ionic compounds exist?
My book says:

As single ions of a metal are not associated in the solid with single ions of a non metal, separate units of ionic compounds do not exist. It is, therefore, wrong to talk of a molecule of an ionic compound.

I know ionic solids exist in form of crystal lattice but why can't we isolate a single molecule of ionic compound?

Actually compounds considered ionic in solid state are molecular in gas phase.

Related: Will gaseous ionic compounds be free moving ions?

Ions are electrostatically bound. One need only separate them with integrated inverse squared distance work. If you mix multiple ionic compounds in solvent, a melt, gas phase, the original paired associations are spontaneously scrambled. Benzene plus xylene does not give you toluene.  "Separating" benzene is not electrostatic attraction.
Quaternized glycine, $\ce{^{+}N(CH3)3-CH2-C(=O)O^{-}}$, is betaine. It is ionic, but the ions are covalently bound into a molecule. Where are the discrete formula units in alumina? You can draw lines to connect closest atoms, but those are not bonds. (Note fractional atom counts for atoms embedded in planes, edges, and corners.)

How is this unit cell $\ce{Al2O3}$?  It seems to have six aluminum and six oxygen atoms.

This oppose the Mithoron comment above. In principle molecular ionic entities exist. The might require exotic conditions but it is not the point.

I agree with Oscar Lanzi in an old message dated November $21, 2017$. The drawn unit cell contains 6  $\ce{Al}$ atoms and $2+1+3$ = $6$ $\ce{O}$ atoms. This cell is $\ce{AlO}$ and not $\ce{Al2O3}$.

Molecules that have strongly ionic bonds actually do exist. In the solid phase where we see most ionic compounds electrostatic bonding with a high density of ions causes the molecules to lose their identity, forming the familiar ionic lattice structures. The vapor phase is a different story. Ionic vapors typically consist of small neutral clusters. Such clusters are, of course, strongly polar, the polarity appearing as dipoles or quadrupoles (two oppositely oriented dipoles attract and join together to form the quadrupole).
Sodium chloride offers an example. Ron's answer to a pervious question identifies sodium chloride vapor as neutral $\ce{NaCl}$ dipoles combined with dimeric $\ce{Na2Cl2}$ quadrupoles.

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text:
Does covalent character increase or decrease bond strength?
On one hand, covalent character increases the bond dissociation enthalpy (i.e., more energy is needed to be put in) and on the other it lowers the melting point of $\ce{Al2O3}$ to a value lower then would be expected with perfect ionic bonding.
So, does it decrease or increase bond strength?

You do realize that you are asking about 2 different energies; the molecular bonds (bond strength) and intermolecular forces (freezing point).

In complex terms we can't ever really describe something as been fully ionic or fully covalent but more of a mix of the two. For example, a HCl molecule will have a small amount of ionic bonding character even though we define it as covalent bonding. This is besides the point though... in this case, we generally assume ionic bonds are stronger than covalent.
If you increased the covalent bonding character in a species you are likely to decrease the ionic bonding character resulting in a lower melting point as forces of attraction are now weaker. However, we do see additional bonding occur in species such as vulcanised rubber where you get cross-linking between carbon atoms on adjacent chains. This then obviously increases the melting point as you have not reduced any ionic bonding character

Vulcanization of rubber actually increases the size of the molecules, so the melting point is expected to be higher.

There was a previous post concerning strength differences between ionic bonds here.
To summarize:

bond strength decrease in the following order covalent > ionic > metallic

Substances with covalent bonds tend to appear weaker than ionically bonded substances (i.e. exhibiting lower boiling points) because ionically bonded substances often form lattice structures from their constituent ions, creating a much more resilient substance, if you will.
You can see the difference in bond strengths, however, when covalent substances form lattice-like structures as well. These network covalent structures are far "stronger" than normal covalent structures because, like ionic substances, they have formed molecular networks that make them stronger. Some examples of these network covalent structures are carbon in the form of diamond or graphene and silicon dioxide (quartz). You will notice that quartz and diamond both exhibit a crystalline structure similar to an ionic substance like NaCl, which is due to their network covalent structures.

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text:
Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium
Is there a complete list of all the half equations for $\ce{H2O2}$ - both oxidation and reduction, in acidic and alkaline conditions? I've looked on the internet but can't seem to find a list with all of them.
These are my first attempts:

$\ce{H2O2 + 2e^- -> 2OH^-}$ (in alkaline conditions)
$\ce{H2O2 + 2H^+ + 2e^- -> 2H2O}$ (in acidic conditions)
$\ce{H2O2 -> 2H^+ + O2 + 2e^-}$ (in acidic conditions)

Are these correct and are there any more?

From Proceedings of the Symposium on Oxygen Electrochemistry and Water Research 2002, 36 (1), 85-94:
$$
\begin{array}{lcc}
\text{Formation}&:&~\ce{2H+ + O2 + 2e- &->& H2O2} &\quad E^0=\pu{0.440V} \text{ vs. SCE}\\
\text{Reduction}&:& ~\ce{H2O2 + 2H+ + 2e- &->& 2H2O} &\quad E^0=\pu{1.534V} \text{ vs. SCE}\\
\text{Direct water}&:& ~\ce{2H+ + 1/2O2 + 2e- &->& H2O}\\
\text{Decomposition}&:&  ~\ce{H2O2 &->& H2O + 1/2 O2}
\end{array}$$
($\text{SCE}$ stands for Saturated Calomel Electrode)
Another strange suggested reaction involves the anion $\ce{HO2-}$ (Abel E.: Über die Selbstzersetzung von Wasserstoffsuperoxyd. Monatshefte für Chemie 1952 83(2) 422–39.):
$$\text{Decomposition 2}:\ce{H2O2 + H2O- -> H2O + O2 + OH-}$$

At the cathode occur at the same time the hydrogen gas evolution:
$$\ce{2H+ + 2e- -> H2~} \qquad\qquad\phantom{00} E^0=\pu{0.242 V} \text{ vs. SCE}$$
Meanwhile, at the anode, oxygen gas is evolved:
$$~\ce{2H2O -> 4H+ + O2 + 4e-} \quad E^0=\pu{0.987V} \text{ vs. SCE}$$

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text:
What exactly happened when I added KOH to hydrogen peroxide?
I recently added a potassium hydroxide, $\ce{KOH}$, pellet to a small droplet of hydrogen peroxide (30 %), $\ce{H2O2}$. You could clearly see the potassium hydroxide dissolving in the droplet when instantaneously the solutions began to fizz strongly (boil and/or release oxygen gas, I'm not sure) and release a lot of heat.
I'm not sure what happened, but I've got a few ideas of what might have happened:

There is some kind of reaction between the $\ce{KOH}$ and $\ce{H2O2}$.
The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of  150.2 °C in which it also triggered the decomposition of $\ce{H2O2}$ into water and oxygen gas.
On Wikipedia I read:

The rate of decomposition increases with rising temperature, concentration and pH.

So when $\ce{KOH}$ is dissolved, the solution has a high pH and this might trigger the decomposition of hydrogen peroxide

I did notice after a pH test that the plastic lid on which I had condensed some of the evaporated water had a very high pH, so this would most likely contain some dissolved potassium hydroxide.
In case a reaction happened, what could that specific reaction be?

I have no time to answer, but I am pretty sure that $\ce{KOH}$ is capable of abstracting a proton from hydrogen peroxide, leaving protonated dioxygen ($\ce{HO2-}$), which should be a very strong acid.

@Martin While $\ce{KOH}$ can certainly deprotonate $\ce{H2O2}$, the resulting hydroperoxide anion $\ce{HO_2^{-}}$ isn't a strong acid. $\ce{H2O2}$ has a $pK_a$ of 11.75 in water according to Wikipedia, and the second ionization is probably several orders of magnitude harder.

@Nicolau You are very right, I confused it with $\ce{HO2+}$ - as I said I was in a hurry.

your cap is basic because of small aerosolized water droplets, not some gas phase product.

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:
$\ce{H2O2 + OH- <=> HO2- + H2O}$
$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.

There is some kind of reaction between the $\ce{KOH}$ and $\ce{H_2O_2}$

There was definitely a reaction.

The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of 150.2 °C in which it also triggered the decomposition of $\ce{H_2O_2}$ into water and oxygen gas.

Water cannot reach a temperature above 100 °C unless it is pressurized.
Yes, $\ce{H_2O_2}$ can disassociate rather quickly under the right conditions. An enzyme called catalase, can break up $\ce{H_2O_2}$ at an astounding rate. If you add an amount you can barely see, like the size of a period at the end of a sentence, to 1 mL of 30% $\ce{H_2O_2}$ in a test tube, the oxygen production will push the liquid out of the tube.

Hydrogen peroxide is stable in acidic conditions. This fact is compounded by the fact that stabilizers (chelates) are commonly acids which do not chelate ions in their basic forms. Furthermore, potassium is a soft largeish ion which catalyzes the autodecomposition.
Bad idea, don't do it again. Hydrogen peroxide is nasty.


text:
How to calculate the rotational constants for different states from a stick spectrum?
Given a stick spectrum for an unknown diatomic molecule, how do you determine the rotational constants $B_\nu$ for both the excited and ground states given that
$$E_{j}=B_{\nu}j(j+1)$$
\begin{array}{cc}
\text{Line position}\, \mathrm{(cm^{-1})} & \text{relative intensity} \\
\hline
27128.20 & 1.47 \\
27171.74 & 2.35 \\
27211.40 & 3.13 \\
27247.18 & 3.34 \\
27279.08 & 2.62 \\
27331.24 & 1.00 \\
27351.50 & 2.62 \\
27367.88 & 3.34 \\
27380.38 & 3.13 \\
27389.00 & 2.35 \\
\hline
\end{array}

I noticed you added a line list, but I don't see how it contributes to the question. Was there some particular reason for it?

In your question, it's not clear whether the quantum number assignments for the transitions are known in advance, so for generality I'll assume that you don't know what they are, and help you find them.
Combination-differences is the most common paper-based method to determine line assignments (for more detailed analysis, a computer-aided fitting program such as PGopher is best). From the line assignments, it's easy to find the $B$-values.
Rotational energy levels are given by
$$F_\nu(J)\approx B_\nu J(J+1)+D_\nu [J(J+1)]^2$$
where $B_\nu$ and $D_\nu$ are the rotational constant and centrifugal distortion constant associated with the manifold $\nu$.
Denoting the initial and final states by $1$ and $2$ respectively, the $P,R$ transition energies are given by
$$P(J)=F_2(J-1)-F_1(J)
\\R(J)=F_2(J+1)-F_1(J).$$
Combination-differences is can be done in two ways:

If your rovibronic transition originates from an initial manifold with well-documented spectral characteristics (including $B_1$), note that
$$R(J-1)-P(J+1)=2 (2 J+1) \left(B_1-2 D_1 \left(J^2+J+1\right)\right)\approx2B_1(2J+1)$$
when $D_1$ is small, and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. This allows you to number the peaks as to their identity.
If instead the upper state has well-known $B_2$, then note that
$$R(J)-P(J)=2 (2 J+1) \left(B_2-2 D_2 \left(J^2+J+1\right)\right)\approx2 B_2 (2 J+1)$$
and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. Again, this allows you to number the peaks as to their identity.

Once the quantum numbers are assigned, finding both $B_1$ and $B_2$ is easy. For example, $R(J)-R(J-1)=2 \left(B_2 (J+1)-B_1 J\right)$, so if you measure it for two values of $J$, then you have two unknowns and two datapoints, so just solve for $B_1$ and $B_2$.
Fun Pictures
Just to give a simple idea of how spectra are simulated, the following Mathematica code does a basic rotational spectrum assuming Boltzmann statistics at a reasonable temperature (hit Cell>Convert To>StandardForm to get it to render the subscripts correctly):
Subscript[F, k_][J_] :=
  Subscript[B, k] J (J + 1) - Subscript[D, k] J^2 (J + 1)^2;
R[J_] := Subscript[F, 2][J + 1] - Subscript[F, 1][J];
P[J_] := Subscript[F, 2][J - 1] - Subscript[F, 1][J];
ListPlot[Transpose[
  Table[{{R[J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}, {P[
       J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}} /. {Subscript[
      D, k_] :> 0, Subscript[B, 1] -> 1, Subscript[B, 2] -> 0.9}, {J,
    0, 15}], {2, 1, 3}], Filling -> Axis]

I think a simple way is 2B is the interval between two spectral lines so subtract line position 2 from line position 1 and then divide by 2 you will get B i.e. rotational constant as $21.77\rm~cm^{-1}$.


text:
How to convert an organic carbonate to its formate ester?
I'm currently working on a chemistry project in which I need to convert ethyl methyl carbonate to ethyl formate.

I only have a simple background in organic chemistry (and I haven't seen anything about reaction mechanisms yet), but I was guessing water could be used in this reaction, to produce $\ce{CH3OH}$ and replace the $\ce{-OCH3}$ group by an $\ce{-OH}$ group. Then the next step would be to remove the oxygen from the  $\ce{-OH}$ group. I've seen that compounds such as $\ce{LiAlH4}$ can do this, but I'm looking for a catalyst to make the process more economically viable, in case the proposed synthetic path is correct.
Does anyone know how to perform the conversion, or even by following another synthetic path? I've got no idea whether this process is correct. I'm also trying to keep the process as low energy-consuming as possible.
I agree it would be better to start from ethanol and formic acid, but I have to start with this molecule.

This is a carbonate derivative, using water will fragment your molecule:$$\ce{EtO-CO-OMe + 2H2O -> EtOH + MeOH + "H2CO3"\ <=> EtOH + MeOH + CO2 ^ + H2O}$$ You need to reduce the central carbon with "$\ce{H-}$" and you also need a soft way to reduce. LAH is the best choice here.

I'm not sure LAH is the right choice here; it's an aggressive reagent and would probably reduce the carbonyl moiety all the way to an alcohol. A more selective reagent (e.g., sterically hindered, electrophilic hydride donors such as DIBAH) is probably called for. There's still the problem of selecting which alkoxy group will leave, however.

There's also the brute force approach: reduce the carbonate ester all the way down to methanol with excess LAH, oxidize the methanol to formic acid, then esterify with ethanol. The nice thing is that the molecule you have already contains within its skeleton two molecules of methanol and one molecule of ethanol (if reduced with LAH and then quenched), which could be recovered, separated, and reused.

Economically, the conversion of ethylmethylcarbonate (98%, 100mL: 212.50 Euro) to ethylformate (97%, 500mL: 44.30 Euro) - prices from Sigma Aldrich, 2014-05-17 - is pointless.

Most likely any reduction of the molecule will result in a mixture of different reduction products, unless total reduction to alcohols is done, so, it is extremely impractical. Moreover, formate will reduce more readily than the original molecule. This said, direct approach is the last thing to consider.

First use $\ce{EtO-/EtOH}$, then lithium-tri-tert-butoxy-aluminium hydride. This will convert it to diethyl carbonate and as the reducing agent is hindered you can manipulate reaction conditions to get the desired product.
Or another way is to hydrolyse it in basic medium to get $\ce{H2CO3}$ and then evaporate $\ce{H2O}$ and $\ce{CO2}$ then after separating it will open a large number if ways as you have got ethyl alcohol and methyl alcohol.


text:
Calculating the pH of a dilluted aqueous KOH solution

Calculate $\mathrm{p}\ce{H}$ of an aqueous solution of $0.01\:\mathrm{nM}~\ce{KOH}$ in water.

I think that water also has $\ce{OH-}$ ions in concentration of $10^{-7}\:\mathrm{M}$ (since $\mathrm{p}\ce{H}=7$)
So the total concentration of $\ce{OH-}$ after mixing the solution with water would be:
$0.01\:\mathrm{nM} + 10^{-7}\:\mathrm{M} = 10^{-11}\:\mathrm{M} + 10^{-7}\:\mathrm{M} \implies 1.0001 \times 10^{-7}\:\mathrm{M}$
and given that $\mathrm{pH} = 14 - \mathrm{p}\ce{OH}$
$\implies 14 - \log(1.0001\times10^{-7})$
$\implies \mathrm{pH} \approx 7$
Does this make sense?

Thanks, I am glad to hear! Has been some time since I had chemistry in high school and couldn't find any example on the internet. Glad that I didn't provide any false information :)

Since this is my first post here on stack chemistry, I am wondering how you guys handle questions like this. Shall I delete it, or can someone just post a "yes" answer that I accept - so that it isn't kept marked as "unanswered". (maybe someone might find this question useful as future reference, maybe it is so trivial that it isn't necessary?)

ok, I added actual answer

Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.
$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.

Your result is correct but I don't agree with your method. Your method would fail under certain circumstances. Which circumstances you might ask? When the pH of the solution is actually between 6.8 and 7.2. In these cases, your assumption that water auto-ionizes to create exactly $\ce{[HO^-]} = 1.0\times10^{-7} M $ will have significant figure implications).
About your assumption:
$\ce{[HO^-]} = 1.0\times10^{-7}$
This is not true when you add any amount of acid or base to water because the addition of either (or both) will disrupt the equilibrium that exists in water (referred to as $\ce{K_{w}}$, the auto-protolysis of water, or the auto-ionization of water):
$\ce{2H_2O <=> HO^- + H_3^+O}$
$\ce{K_{w}=[HO^-][H_3^+O]}=1.0\times10^{-14}$
All you know about your system is that $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$. This is in fact true for any water solution; as we can see from the auto-ionization reaction above, there exists a 1:1 ratio between $\ce{[HO^-]}$ and $\ce{[H_3^+O]}$.
You also know that because $\ce{KOH}$ dissociates completely in water solution, $\ce{[HO^-]_{KOH}} = 1.0 \times 10^{-11} M$
Therefore, the correct way to solve this system would be using $\ce{K_{w}}$. Work below:
$\ce{K_{w}=\sum{[HO^-]\sum[H_3^+O]}}=1.0\times10^{-14}$
$\ce{=([HO^-]_{water}}+\ce{[HO^-]_{KOH})([H_3^+O]_{water})}$
Because $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$, we can make this substitution:
Let $x=\ce{[HO^-]_{water} = [H_3^+O]_{water}}$
Therefore,
$\ce{=(x}+\ce{[HO^-]_{KOH})(x)} = 1.0\times10^{-14}$
And you can solve for x from here. Keep in mind what x equals!

Sorry, I am unfamiliar with the Latex markup here and had to keep refreshing to see the effects of my syntax.

You have a preview under your edit window, there you can see how your post will look like.

You should also use \times ($\times$) when you want to multiply terms.

I never noticed that it changed! You're right; I'll just scroll down next time to see my changes.

@LDC3 or even better \cdot($\cdot$) as $\times$ refers to a vector product

So you want me to edit it again ;)?

@Martin When there are 2 numbers, I'm sure nobody will mistake them for vectors (i.e. $2.4 \times 10^{-5}$).

@LDC3 I agree, that there will be no confusion, bit this should be no excuse to use the correct notation. The main principle behind the notation $\times$ is that a vector product does not necessarily commute, while a product of scalars does.

Glad it helped!


text:
What are similarities and differences among shells, orbitals, subshells, and energy levels?
I think I know what these terms mean:

each row on the periodic table has a new shell of electrons,
a subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.,
and energy levels are the same as shells,
whilst orbitals are the same as subshells.

Is this correct or totally wrong?

I am using the chemistry textbook by Masterton/Hurley (Cengage) as an outline for this answer. It has a side note that reads "Our model for electronic structures is a pragmatic blend of theory and experiment". In the spirit of that quote, there is some experimental data mixed in with the concepts.

Each row on the periodic table has a new shell of electrons

Yes, and you see that in the plot of atomic properties vs. atomic number. For example, atomic radii (image source)increase significantly as you go from noble gas (last element in a period) to group 1 metals (first element with the next shell occupied). The quantum number that defines the shell is the principal quantum number n.

A subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.

Yes. The periodic table is organized into s-block and p-block (main group elements), d-block (transition metals) and f-block (lanthanides and actinides) according to which subshell is partially filled (or just filled). The quantum number that defines the subshell is the second quantum number, $\ell$.

Are energy levels the same as shells?

This is true only for the hydrogen atom and other one-electron systems ($\ce{He+}, \ce{Li^{2+}}$) in the absence of a magnetic field:
$$ E \propto -\frac{Z^2}{n^2} $$
with $Z$ the charge of the nucleus and $n$ the principal quantum number.
For systems with more than one electron, distinct subshells in the same shell correspond to slightly different energy levels. One method to measure these energy levels is photoelectron spectroscopy, with some data on aluminium 2s and 2p shown below (image source).

In general, the energy levels depend on the atomic number (number of protons in the nucleus) and the electron configuration (how many electrons are present, and whether they are at ground state or in an excited state). Going from a hydrogen atom to a larger multi-electron atom, the energy levels of distinct shells can be roughly described as an electron bound to a nucleus surrounded by inner electrons with a combined effective charge $Z_\mathrm{eff}$ (smaller than the actual charge of the nucleus because the attraction by the nucleus is offset to a degree by the repulsion by the other electrons):
$$ E \propto -\frac{Z_\mathrm{eff}^{2}}{n^2} $$
For the innermost shells (n = 1), $Z_\mathrm{eff}$ is almost the same as the nuclear charge, whereas for the outermost shells, $Z_\mathrm{eff}$ can be much lower.
For example, comparing 1s electrons across the periodic table, the binding is tighter and tighter as the atomic number increases. On the other hand, the outermost shell (valence shell) in elements of the same group (e.g. Be, Mg, Ca, Sr, Ba, Ra) corresponds to similar energy levels, even though they differ in the principal quantum number. This explains why elements in a group have similar chemical behavior.

Are orbitals the same as subshells?

No, there are seven different orbitals in the f-subshell, five different orbitals in the d-subshell and three different orbitals in the p-subshell (compare size of blocks in the periodic table). The three orbitals in the p-subshell, for example, are either designated (+1), (0), and (-1), or alternatively $p_x$, $p_y$, and $p_z$. For the s-subshell, there is a single orbital, so only in this special case is the subshell the same as an orbital. The quantum number defining the orbital in a subshell is the third quantum number $m_\ell$.
While the different orbitals of the p, d, and f subshells have distinct, sometimes intricate shapes, a filled subshell corresponds to a spherical distribution of electrons.

How do these concepts help to understand chemical bonds?

We study electronic structure because it explains properties and reactivity of atoms and molecules. It might seem that atomic orbitals of hydrogen are of little use in understanding the electronic structure of larger atoms and of molecules. It turns out, however, that they can be used as building blocks for the more complicated systems, both qualitatively (talking about sigma and pi bonds and lone pairs) and quantitatively (computational methods use atomic orbitals as basis functions).
Surprisingly, the changes in electronic structure going from atoms to molecules are moderate, with spherical electron distribution around the atoms explaining most of the electron density observed, for example, in crystallographic studies. The small deviations from this are called electron deformation densities, and they give an experimental view of the subtle shifts in electron density upon bond formation. The picture below shows a contour diagram of the total electron density on the left (near the atoms, there would be many more contour lines, but they are not shown), and the deformation density on the right (lines: higher density than atomic density, dashed: lower density than atomic density, dot-dash: same density as atomic density).
(Image source)

If you are interested in deformation densities and the electrostatic interpretation of the chemical bond, this is a good starting point: https://www.chemistry.mcmaster.ca/esam/Chapter_7/section_3.html

Historically, the first of these concepts to emerge was the shell of the Bohr model. This very nicely rationalized the observed spectral lines of the hydrogen atom, by restricting electrons to be in orbits with fixed energy levels. In modern parlance, the shell corresponds to the principal quantum number $n$.
Later experiments (also, it's enough just to look a the spectral lines for helium) suggested that this was not enough and that each shell had multiple energy levels within it. These levels within the shell are called subshells. In modern parlance, this corresponds to a fixed value of $l$, the azimuthal quantum number. We refer to the different subshells by letters ($s$, $p$, $d$, $f$, $g$, ...), and these correspond to respective values of $l$ (0, 1, 2, 3, 4, ...).
At this point, it is helpful to draw a division because the concept of the orbital is only loosely associated with the prior concepts. The previous concepts come from models that are considered out-of-date by modern standards. This is because quantum mechanics provided a new model whose principal features are incompatible with the fundamental assumptions of these prior models (for example, orbits are not compatible with the uncertainty principle) or simply unaddressed at all (e.g., spin).
Quantum mechanics arrives at a solution that is the following (or at least, this is roughly what one of the major formulations says). The behavior of matter is governed by the Schrödinger equation. For a hydrogen-like atom, we can parametrize the solutions of the Schrödinger equation for that lone electron with several quantum numbers ($n, l, m$). The $n,l,m$ fully describe the probabilities of finding the electron in any point in space, and therefore, it roughly defines where the electron is likely to be. This region is not an orbit or shell, since the electron is not really restricted in its distance from the nucleus. It is therefore an orbital.
The orbital has some features that are similar to the older models (it has to since we still need to rationalize older experimental outcomes). It has a fixed energy level corresponding to $n$ (corresponding to a shell). Different orbitals may be associated with the same shell but still have different energy levels dictated by $l$ (akin to the subshell).

There is an arcane lore of quantum mechanics, that says that movement of electron in a hydrogen-like atom (with exactly one electron) is fully specified by four numbers:

Principal quantum number (n) (1, 2, 3 ...) - specifies a shell
Azimuthal quantum number(ℓ) (1 .. n) - specifies a subshell within the shell
Magnetic quantum number (m) (-ℓ .. 0 .. ℓ) - specifies an orbital within the subshell
Spin quantum number (s) (+/- 0.5) specifies one of two possible positions of electron on the orbital

While movement of electrons in non-hydrogen-like atoms is somewhat different, the form of movement, i.e. orbitals, are mostly same, so it is generally said, that electrons in a multielectron atom may be assigned the same numbers (with a side-note that electrons cannot be distinguished) and no two electrons can have all four numbers the same.
'Energy level' is a completely different concept: it is a group of states possible for some quantum chemical system and having the same energy. This term may (randomly) pop-up in discussions where the electron is viewed as a singular particle with other parts of the system being 'framework'. While somewhat working, such considerations are incorrect and should better be avoided.

There is no movement of electrons in orbitals in a strict sense. It is just a probability density of electron around a positively charged  nucleus.

@M.Farooq Orbitals in particular and wave function in general is quantum equivalent of trajectory. In particular, each orbital corresponds to some specific kinetic energy.

Interestingly, just two days ago, I was discussing similar questions with physical chemists. Both explicitly said that orbital is nothing more than a probability density around a nucleus. The kinetic energy is no longer classical and hence the concept of trajectory, moving path or anything else related to orbitals is rather a misconception.

@M.Farooq You shouldn't discuss it with phys. chemists, but with physicists. Quantum mechanics is derived from classical mechanics in Hamilton formulation by direct replacement of trajectory with wave function and fitting replacement of impulse and coordinate variable with respective operators. Since it is directly derived, terminology is inherited and thus, "movement of electrons" is still named this way even if the wavefunction is stationary.

Of course, you understand it. I was writing for the students (perhaps the original poster). They have tons of misconceptions regarding orbitals. You might enjoy the article "The nature of the chemical bond—1990: There are no such things as orbitals!" https://pubs.acs.org/doi/abs/10.1021/ed067p280

But they are a valid conceptual tool in many cases: Jenkins

Shells are the major grouping of the orbitals (1, 2, 3, 4, etc). Subshells are the groupings inside each shell (s, p, d, f, etc). Energy levels correspond to the orbitals and vary with the element and the number of electrons in the atom ($\ce{Fe}$, $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$).


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Determining central atom in NCl3
Less electronegative atom is the central atom. $\ce{N}$ is more electronegative than $\ce{Cl}$ as $\ce{Cl}$ has larger atomic radius. So why in $\ce{NCl3}$, $\ce{N}$ is the central atom?

Cl is slightly more electronegative according to the tables I've seen: http://en.wikipedia.org/wiki/Electronegativity
But I wouldn't say it's an absolute rule that the less electronegative atom is central.  For example, $\ce{NH3}$ has the N central and $\ce{Cl_2O}$ has the O central.

The relative positions of chlorine and nitrogen on the electronegativity scale can switch depending on the definition of electronegativity being used. Allen scale electronegativities are quite recent but also popular, and there nitrogen is higher than chlorine.  But yeah, the asker is just taking a rule of thumb too strictly.

Acc. to Pauling's scale, both have electronegativity 3.0 but as $\ce N $ has lesser atomic radius,it is more electronegative.But here it seems contradiction as a lesser electronegative atom ought to be the central atom.But it is also accurate that in nitrogen tri chloride , nitrogen is the central atom .Is the 'less electronegative atom is the central atom ' a wrong concept?

How can I determine accurately  the central atom ?

@user36790 I would say that the example of Cl and O basically proves there is no simple set of rules, since Cl can be in the center for chlorate, perchlorate and other compounds, while $\ce{Cl_2O}$ has the O central. Also, the symmetrical choice or lone atom in the center, isn't always correct as there is NNO. The rule "H is never in the center" you can usually count on, although a hydride can bridge two metal or boron atoms. I would say that carbon and nitrogen tend to be central slightly more than their electronegativities suggest and halogens non-central. SeCSe, CAt4, NI3.

Being "central" has nothing to do with electronegativity.  It is a matter of geometric symmetry. The atom at the "center" of a molecule is the atom with the highest geometric symmetry. But if you look at benzene, there is no atom at the "center" of the molecule.


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What is defined by part by weight in the definition of equivalent weight?
According to my book, equivalent weight is defined as the ‘number of parts by weight’ of an element that interacts with 1.008 parts by weight of hydrogen. It is dimensionless while gram equivalent weight has gram as units. But in https://en.wikipedia.org/wiki/Equivalent_weight, equivalent weight and gram equivalent weight are same. I’ve two questions: Which one is correct, my book or Wikipedia? And what is ‘number of parts by weight’?

The term "parts by weight" is how you would specify in a recipe the relative proportions of the ingredients. So if the recipe called for one part by weight of flour and two parts by weight of sugar, you could use 1 ounce of flour and two ounces of sugar (to make a small batch) or 1/2 pound of flour and 1 pound of sugar to make a large batch.
So the "recipe" for water calls for exactly 1.008 parts by weight of hydrogen and 8.000 parts by weight of oxygen. If you use too much of either ingredient, there will be an excess of it after the reaction is run.
"Parts by volume" would apply if you were combining liquids and measuring them by volume rather than by weight.


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How does a lone pair of a central atom affect the dipole moment?
Dipole moment is the degree of polarity, i.e. the seperation of positive and negative charges. But I am not getting the intuition why and how lone pairs affect the polarity and dipole moment. I cannot connect the  separation of positive and negative charges  definition with lone pair.

How does lone pair of a central atom affect the dipole moment?

There is no single answer to your question, let me explain.  Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around the atom that has the lone pair of electrons.  This increased electron density could lead to a more significant contribution from the lone pair electrons to the molecular dipole moment than from electrons spread out more diffusely in a covalent bond.  Next we must understand the directionality of the lone pair of electrons.  Consider the two molecules pictured below, ammonia and phosphine.  The molecules appear to be very similar, they are in the same column in the Periodic Table.

However in ammonia the $\ce{H-N-H}$ angle is around 107 degrees and the molecule is roughly $\ce{sp^3}$ hybridized, the lone pair and the 3 $\ce{N-H}$ bonds roughly pointing towards the corners of a tetrahedron.  You can see that in this case (as shown by the arrows, the "arrowhead" end representing the negative end of a dipole), the lone pair on nitrogen will make a contribution to the molecular dipole moment.  Next, let's examine phosphine.  The $\ce{H-P-H}$ angle is around 90 degrees and the molecule can be viewed as being unhybridized, the lone pair is an $\ce{s}$ orbital and the 3 $\ce{P-H}$ bonds are constructed from phosphorous $\ce{p}$ orbitals.  You can see that in this case, the lone pair on phosphorous, due to its spherical symmetry will not make a contribution to the overall molecular dipole moment.
So in summary, a lone pair of electrons can make a significant contribution to the magnitude of a molecular dipole moment due to the fact that they are more localized than bonding electrons and consequently there is a high electron density.  But, the directionality (or lack thereof) of the lone pair must also be assessed, since a lack of directionality may preclude it from making a significant contribution to the overall molecular dipole moment.

Could you elaborate a bit more on the directionality part? I don't understand why the dipole moment of the lone pair in ammonia points away from the nitrogen atom. Shouldn't more electron density be closer to the nitrogen atom (where nuclear attraction is greater)? This would imply that the dipole moment's direction (due to the lone pair) would be opposite to the one actually observed.

@Gerard  The nitrogen nucleus is positively charged.  The lone pair electrons are contained in an orbital directed away from the nucleus.  The vector describing the nitrogen nucleus - lone pair dipole would be a straight line having it's positive end around the nucleus and its negative end around the center of the lone pair orbital - as I've drawn in the figure above.

@ron , JD lee says, “ the lone pair has no contribution to he dipole moment of the molecule , if the lone pair is present in the pure s or p orbital as s is spherically symmetrical and p is objected equally in opposite directions”
Does this statement agree with your answer?

@AaryanDewan Yes, read what I say about lone pair "directionality".

@ron You say the nitrogen nucleus is positively charged. How? As far as I can see, nitrogen has no formal charge here and it is receiving electron density from the nitrogen-hydrogen bonds.

@YUSUFHASAN  The key word is nucleus.  The nucleus contains protons and neutrons and is always positively charged.  Add the electrons and you then have a neutral atom.

@ron As a rule of Thumb, Can I say that when the lone pair rests in an hybridized orbital, it will contribute to molecules dipole moment but when it is in the pure orbital (unhybridized) it will not contribute to the molecules dipole moment ?

@Tushar I think that might be a bit too binary.  I would prefer to say that the lone pair will contribute less when in an unhybridized orbital.

Dipole moment can be defined as the product of magnitude of charges and the distance of separation between the charges. Dipole moment may refer to:

Electric dipole moment, the measure of the electrical polarity of a system of charges
Transition dipole moment, the electrical dipole moment in quantum mechanics
Molecular dipole moment, the electric dipole moment of a molecule.
Bond dipole moment, the measure of polarity of a chemical bond
Electron electric dipole moment, the measure of the charge distribution within an electron
Magnetic dipole moment, the measure of the magnetic polarity of a system of charges
Electron magnetic dipole moment
Nuclear magnetic moment, the magnetic moment of an atomic nucleus
Topological dipole moment, the measure of the topological defect charge distribution
The first order term (or the second term) of the multipole expansion of a function

(from wikipedia)

The dipole moment is actually affected by the presence of a lone pair of electrons because the electrons on the central atom can cause shielding effect as the inner orbital electrons does for the outer-orbitals.
It can cause hindrances by charge imbalances which could result in imbalance of charges and which could destabilise the atom due to decrease in dipole moment.
Consider a molecule of carbon dioxide where two oxygens are present at each side of carbon:

Here the formal charge on carbon is zero and the same formal charge is there on both the oxygens. Coming to the topic of dipole moments you should consider the lone pair of electrons on oxygen. The lone pair is generally resided on the other side of the bond [as shown in figure]. These electrons are a part of oxygen and these makes oxygen more electronegative because of which the dipole moment increases ans hence the bond becomes stronger.
In most of the cases the central atom in any compound of oxygen is more electropositive atom and electropositive atoms do not contain lone pairs. You can just imagine there are lone pairs on the central atom carbon which could generate some partial negative character in them and hence the bond between them must not be as strong as the bond between an electropositive and electronegative atom.
This all is governed by dipole moment only.

the pure shielding effect is not seen in the MOLECULES they are just seen in ATOMS in order to make you understand just used the term but it does'nt mean that it is as same as in atoms here it just hinders the central atom [make's a sort of covering ] from the attached atoms so that the complete atom is'nt exposed to the central atom and there will surely be a difference in the lenght of bond but this is not what you asked keep this topic aside

i will try to make the same same thing with ammonia

This was hard to read, edit, and I believe most of it is wrong. For example the central atom in $\ce{NH3}$ is the most electronegative element. The picture of $\ce{CO2}$ is just utter garbage and has no source given. The lone pairs of oxygen do not make oxygen more electronegative.


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Do isotones share any similarities?
A trivial research in atoms and their basic theories led me to this term: "isotones"
Nuclides sharing the same number of neutrons but different atomic numbers.
A simple request or shall I say, quest:
please indicate me to any feasible similarities existing between isotones and please, DO NOT use jargon.

Chemistry is determined entirely by electron interactions. Since the number of electrons in an atom is determined by the number of protons in the nucleus, the atomic number (number of protons) is what defines the chemical behavior of a given element. This means that in terms of chemical properties, isotones would have no similarities unless they happened to be in the same group (column) of the periodic table, but that would be a coincidence and would have nothing to do with the number of neutrons.
As DavePhD and ron said, the nuclear stability could be similar, since nuclear stability has a lot to do with the total number of nucleons (neutrons + protons) as well as the relative proportion of protons to neutrons. The wikipedia article on isotones discusses stability in a little more detail.

I suspect that having the same number of neutrons may mean something at the nuclear chemistry and nuclear physics level, but in terms of general chemistry, including inorganic, organic and biochemistry, it's all about the number of electrons and protons.  So I don't think that isotones would show any differences in terms of (non-nuclear) reaction chemistry.

In chemistry, if an atom has a closed valence electron shell, as do the noble gases, it is particularly stable.
Similarly, there is a shell model of the nucleus.
Certain numbers of neutrons (and also protons) are referred to as magic numbers: 2,8,20,28,50,82,126, all even numbers, because nuclei with these numbers of neutrons (or protons) are particularly stable.
The similarity between isotones is that stability of a nucleus is somewhat correlated with the number of neutrons.


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Does the distance increase as the polarity increases?
According to the dipole moment's definition, as polarity increases, the distance between the centres of positive and negative charges increases. In the ionic bond, as the polarity is maximum, the distance between the two charged centres is maximum. Is this concept correct? If not, what is the relationship between distance and polarity?

The dipole moment $\vec\mu$ is
$$\vec\mu = q\cdot\vec d,$$
where $q$ is the magnitude of charge on any particle and $\vec d$ is the distance between them.
This must imply that if charge increases or distance between the charges increases, dipole moment increases. The larger the dipole moment, the greater the polarity.

Actually when the more electronegative elements pulls electrons, the partial charge, $q\le e$, increases and in case of $100\,\%$ polar compound, $q=e,$ whereas in $100\,\%$ covalent compounds $q = 0.$
So, you can see dipole moment and polarity is maximum for ionic compounds. But if two atoms have greater bond length, surely dipole moment and polarity increases.

If suppose a compound can exist in covalent and go to ionic form[totally hypothetical]in which one is more electronegative,then as dipole moment is max. in ionic compound,is this because there is whole charge than partial charge or as it goes from covalent to ionic,the distance increases or both? Thanks again.

can't say about the distance, as in real there is no change of covalent to ionic, so changing nature will change distance, can't say.

I'm not sure if the guideline actually works out in empirical observations. There are at least two factors affecting bond length - atomic radii - and polarity of the bond. A more polar bond will have a less even distribution of electrons. The electron density will generally be closer to the more electronegative atom. This isolation of electrons may have the effect of "stretching" the bond.
Consider the following two molecules, both of which are linear and polar:

From polarity considerations alone, we might expect the $\ce{H-F}$ bond to be longer than the $\ce{H-Cl}$ bond because fluorine is more electronegative than chlorine. However, chlorine is a bigger atom than fluorine by nature of having additional sublevels of electrons.
Also, consider the below data: bond length actually increases with decreasing polarity (quantified through dipole moments).
Fluoromethane Bond length 136pm, Bond strength 108 kcal/mol, Dipole moment (D) 1.85
Bromomethane Bond length 178pm, Bond strength 84 kcal/mol, Dipole moment (D) 1.87
Chloromethane Bond length 193pm, Bond strength 70 kcal/mol, Dipole moment (D) 1.81
Iodomethane Bond length 214pm, Bond strength 56 kcal/mol, Dipole moment (D) 1.62
The unfortunate thing about your guideline is that I thought that it would work at least across periods - in which atomic size stays relatively constant. However, the facts do not fit the case. Consider $\ce{H-N}$ and $\ce{H-O}$. Oxygen is more electronegative than nitrogen; we'd expect the $\ce{H-O}$ bond to be more polar, and thus longer than the $\ce{H-N}$, at least according to your guideline. The opposite is actually true; the $\ce{H-O}$ bond is actually shorter than the $\ce{H-N}$ bond.

So what is the role of distance in measuring the degree of polarity.According to u , polarity is inversely proportional to the distance?Can u plz explain?

I didn't say that. I think that you should consider bond length on a case by case basis as there are several factors.

Can u give me the actual answer:what is the role of $d$ in measuuring the polarity ?

How does $d$ affect the dipole moment?

d is a measure of the dipole moment.

What is the difference between dipole moment and polarity?

The bond length decreases as polarity increases because in a polar bond, one side is partially negative and one side is partially positive, so these two sides attract even more, get closer together, and the bond length decreases. So, more polarity means more attraction (or attractive pull), which brings the atoms closer together, hence decreasing the bond length.
That is why when there is more difference in electronegativity between atoms, their bond length is shorter; when there is more difference in electronegativity, there is more polarity, and shorter bond length.


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Does hybridization take place in other atoms apart from central atom in a molecule?
I have seen my textbooks always give formula and description of hybridization of central atom of a molecule. But I wonder will other atoms present (not Hydrogen) go in hybridization or only the central atom only uses hybrid orbital and other use pure orbital?

Yes, all atoms with p or d orbitals can hybridize in a molecule, not just the central atom.  Take acetone for example, the carbonyl carbon is $\ce{sp^2}$ hybridized and the carbonyl oxygen is also $\ce{sp^2}$ hybridized.  Another example is methyl chloride, the methyl carbon is $\ce{sp^3}$ hybridized and the chlorine is also $\ce{sp^3}$ hybridized.  As a final example consider dimethyl ether, each of the carbon atoms is $\ce{sp^3}$ hybridized and the central oxygen is also $\ce{sp^3}$ hybridized.

Ok,then can I say all atoms in a molecule use hybridised orbitals instead of pure atomic orbitals?

How to find the state of hybridization of other atoms apart from central atom?

For your first question, the answer is "yes" when we are talking about orbitals that 1) contain electrons and 2) are used to form bonds.  For your second question, look at the various bond angles (including those involving hydrogen) around the atom in question.  From this information you can infer hybridization.  For example, if the angles around an atom are all 120 degrees, then the atom is $\ce{sp^2}$ hybridized.  See my answer here for a more detailed example  http://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test/10666#10666

I do partially disagree with that: Terminal atoms are almost always (lack of external field) better described as having at the most sp hybrid orbitals. In acetone the oxygen would therefore (roughly) have 1 sp hybrid orbital forming the sigma bond, 1 p orbital for the pi bond, 1 perpendicular p orbital for the lone pair (stabilised by hyperconjugation), another sp lone pair orbital.

Hybridization is the rearrangement of the orbitals predicted by the Schrodinger Equations to coincide with the observed orbital arrangements. So any atom with electrons in the S and P orbitals (lithium and heavier elements) can have hybridization rearrangements.


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Why does the lone pair of phosphine not contribute to its dipole moment?
Unlike ammonia, the lone pair of phosphorous does not add to the dipole moment of phosphine. My textbook says this is due to the presence of lone pair of phosphorous in $\ce{PH3}$ in an s-orbital. I couldn't understand my book's explanation. How can the orbital of lone pair determine its contribution to the dipole moment?

I elaborate on that in the second half of my answer here. Also, user ron already seems to have answered that in another question you made, so you might want to reread it.

Ron has beautifully answered that ; i also understood the 1st part but couldn't decipher how the orbital of lone pair is important in determining the dipole moment. I only want to know why it happens so?

@Nikolau Saker :How does the asymmetry of an orbital give rise to dipole moment?

@user36790  Do you understand why the lone pair in ammonia contributes to the dipole moment?

@ron : Sir,yes but no! Sir, I could not understand how an asymmetric hybridised orbital can cause polarity (in case of ammonia) and why does a symmetric one cannot add to polarity (in case of phosphine).

Let's see if we can use the following picture to help clarify things.

Let's look at the molecule $\ce{H3X-Z}$.  Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$.  Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment.  Are you with me so far?
If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased.  If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$.  These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture.  The orbital and the electrons in it still "point" towards where $\ce{Z}$ used to be.  The orbital and the electrons in it still have directionality.  They still contribute to a dipole moment.  Can you see this?
Let's continue assuming you can.  Let's repeat all that we've done above with one change.  Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized.  Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital.  An $\ce{s}$ orbital is spherically symmetric.  It has no directionality, it doesn't "point" in any specific direction because it is spherically symmetric.  So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.

@ ron:sir,if instead of taking $s$ ,if I take other pure atomic orbitals,then also will I get the same result?

@user36790  No, all other orbitals, p, d and f have directionality; the distribution of electrons in them is not spherically symmetric.  Only the electrons in an s orbital are distributed in a spherically symmetric manner.

@ ron:Thanks for such an excellent answer. It is clear the orbital which has direction can be responsible for polarity.

Great, I think you've got it now!


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Removing prenylated phenolics from clothing, shoes, and upholstery
I live in Southern California and sometimes hike in areas that can be overgrown with a plant called purple poodle-dog bush. As I found out the first time I went crashing through a stand of the stuff, it's a horrific skin irritant, much worse than poison oak. It has microscopic hairs that get stuck in your skin and act like syringes. The hairs have chemicals on them that are skin irritants. Wikipedia says:

This contact dermatitis is due to prenylated phenolics exuded by hairs (glandular trichomes) of the plant. The principal irritants are derivatives of farnesyl hydroquinone and 3-farnesyl-P-hydroxybenzoic acid.

I'm not a chemist, and all the nomenclature is Greek to me. My question is this. When I hike in an area where it will be hard to avoid the bush, I try to wear long sleeves and change my clothes immediately afterwards. Will throwing my clothes in the washing machine remove these chemicals and/or break them down into harmless ones? What if my shoes aren't machine washable? Will the chemicals break down naturally within a certain period of time? Would other methods such as exposure to sunlight help? Are these chemicals soluble in water, so that simple rinsing would help? How about making sure they aren't lurking in the upholstery of my car?

I don't have time to answer fully, but I imagine the compounds have a non-zero but rather low solubility in water. The actual amount of irritant substances present may be low enough that it could dissolve even in a regular wash, however. Another thing is that relativity caustic media will cause some degree of deprotonation, forming a salt which is likely far more soluble. The hydroxybenzoic acid derivative is particularly susceptible to this, though even the hydroquinone is slightly acidic.

The ultimate problem is likely the hairs which contain the irritants; they might not be easily removed mechanically or chemically, and may impede the access of solvent to the irritants if they are stored in cavities on the hairs.

If the solubility is low but the amount of the substance is small, would it make sense to leave the clothes soaking in water for a long time so that the substance would gradually leach out and go into solution? What would be an example of a caustic medium that I could use to soak my clothes?

Soaking could possibly slowly dissolve everything, but if you were to do that then I would recommend adding a basic compound to the water to possibly increase the speed and efficiency of the process. A safe, cheap and easily washed off base that you could use would be sodium carbonate, $\ce{Na2CO3}$. A solution containing 2 g/100 mL of water will reach somewhere around pH 11 at ambient conditions. Sodium bicarbonate $\ce{NaHCO3}$ can be used too, though it's less basic.

I cannot completely guarantee that your clothes will resist the alkaline solution entirely unharmed, though it should be quite safe. At least one commercial non-chlorinated bleaching agent for clothes actually contains sodium carbonate as part of its formulation (though the bleaching compound is actually the hydrogen peroxide). Furthermore, as I mentioned earlier, the hairs themselves may act as a physical barrier and hamper dissolution. In that case, it may be best to strip the hairs off directly using some adhesive (sticky tape?)

@NicolauSakerNeto: How about making that an answer?

As a starting point, I shall assume that the irritant compounds are freely exposed to the environment (which is not necessarily the case, as I mention further on). As such, one of the conceivably simplest ways to remove them is to wash them off, i.e. disperse them in a solvent and discard the liquid mixture. For several reasons (abundance, extensive solvation ability, non-toxicity, etc.) water is a good choice.
First of all, how well would pure water work at dispersing the prenylated phenolics? Probably not too great. The representative parent compounds of the irritants, hydroquinone and p-hydroxybenzoic acid have modest solubilities in water (59 g and 5 g per 1 L, respectively), and the addition of farnesyl chains in the compounds would quite likely cause a severe reduction in water solubility, unless the farnesyl chains are highly substituted with hydrophilic groups.
Of course, there is no substance that is entirely insoluble in any solvent, though it is very common for certain solutes to require impractical amounts of solvent to disperse completely. It could well be that a washing machine's load of water (of order 100 L) is actually capable of dissolving the farnesylated toxins, especially because natural products are generally present in rather small quantities. However, there are some techniques which can possibly help the process. I list a few in the following:

Use hot water - The majority of substances become more soluble in hot solvents because dissolution very often increases the entropy of a system ($\Delta S>0$), and hence from the Gibbs' free energy relation $\Delta G=\Delta H - T \Delta S$, an increase in temperature leads to a larger decrease in free energy, making the process more favourable. Of course, this only really works well if the water is hot throughout the soaking/washing process; adding hot water and letting it cool off will just cause part of the solute to precipitate out of solution again.
Use a surfactant (detergent) - Since the offending compounds potentially have large hydrophobic chains, they can conceivably removed by the same process we use to wash away oils and fats. An amphipatic molecule contains a hydrophobic part (such as the hydrocarbon chain of sodium dodecylbenzenesulphonate) responsible for interacting with the oils and a hydrophilic part (in SDBS, the sulphonate group), which is well solvated by water. Thus, the surfactant bridges the interaction between oil and water, allowing effective dispersion.
Use a basic solution - In this specific case, the undesirable compounds have some acidic character. In particular, the hydrogen atom in the -COOH (carboxyl) group of the p-hydroxybenzoic acid derivatives are about as acidic as acetic acid. The -OH (hydroxyl) groups in both p-hydroxybenzoic acid and hydroquinone are also slightly acidic, though weaker by some 5-6 orders of magnitude. Thus, mixing in a basic solution may allow an acid-base reaction to take place, forming salts of the organic compounds with significantly increased solubility in water. Many washing powders/liquids are already slightly alkaline, but it is also easy to prepare basic solutions separately. A cheap and very safe base would be sodium carbonate, $\ce{Na2CO3}$, though a few tens of grams may be required to prepare a decent volume of basic medium for soaking. Sodium bicarbonate, $\ce{NaHCO3}$, also works though it is a weaker base and would require a higher amount to reach a given pH value. Caustic soda, $\ce{NaOH}$, is yet another cheap option, but care must be taken to not add too much, as it is a strong base (in contrast to the previous two). I don't know how resistant different fabrics are but, very approximately, any more than a couple of grams of lye per litre of water risks partially hydrolysing the fabric and weakening its structural integrity. Use at your own risk!
Use a bleaching agent (chlorinated or non-chlorinated) - Bleaches work by partially oxidizing organic matter, breaking organic molecules into smaller chunks and making each piece more hydrophilic (due to added oxygenated functionalities). Even if the compounds aren't removed entirely, the reaction products may be much less harmful.
Miscellaneous - You suggest exposure to sunlight. Light is certainly capable of decomposing organic substances, especially UV light, but there's no guarantee that the exposure required to eliminate the irritants won't also break down fabric and bleach dyes. Waiting is also probably not very efficient, as there is no guarantee the irritant compounds are unstable enough to undergo spontaneous decomposition in ambient conditions at a significant rate.

Unfortunately, all this discussion is of limited use if the initial assumption (freely accessible irritants) is incorrect. It could be that the compounds are stored in cavities in the plant hair's surface or interior, and the hairs themselves are insoluble and impermeable to water, effectively locking the toxins away. In this case, the compounds will likely be removed only slowly, as the hairs are progressively worn out by abrasion and other processes. Thus a physical removal method may be preferable. I have heard that adhesives (sticky tape, waxing strips, etc.) can be successfully used to peel out painful hairs. This is also a useful option for surfaces which cannot be washed easily (such as car upholstery).
In practice, I am not sure how this will all work out. It's been a few days since I wrote my comments, so perhaps you've had time to test something out? If so, I would be curious to hear some feedback!

Sorry about not answering before. I felt an answer needed more polish, but I was having trouble making time for it. Even now I don't think what I wrote is as good as it should be. Also, I suspect a fair amount of the above is not particularly useful to you, but I like my answers to be as thorough as I can make them.

Thanks for the very helpful and complete answer. I soaked the clothes overnight in a sodium carbonate solution, then washed them again in the washing machine. Unfortunately it's difficult to tell whether it worked or not. My immune system is currently lit up like a Christmas tree, so I've put the clothes aside and will not try wearing them until my skin is back to normal. If they cause the reaction to flare back up at that point, then I guess we'll know the method didn't work!

Yikes, I didn't know it was that bad. Do be extra careful then! I wouldn't want to unwittingly inconvenience you a second time.

I realize it's been a year, but I thought I'd make a small addition: it may be that dry cleaning could be effective at removing the offending compounds in this case. Dry cleaning often uses non-polar chlorinated solvents like tetrachloroethylene as a washing liquid, which would excel at dissolving the fatty prenylated phenolics, again under the assumption that the solvent can reach the compounds.


text:
Why is the bond angle H-P-H smaller than H-N-H?
$\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.
But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference?

An answer to a closely related question is given by my answer  here http://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h%E2%82%82o-h%E2%82%82s-h%E2%82%82se/57973#57973

related http://chemistry.stackexchange.com/questions/38599/why-is-the-inversion-barrier-larger-in-ph3-than-it-is-in-nh3

Starting point: 2s orbitals are lower in energy than 2p orbitals.
The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees.  Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.
In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees.  This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.
So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.
There are two choices for the central atom:
remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another.  As a result,

substituents will be arranged closer together and destabilizing steric interactions will be increased
due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy
the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital

or
hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another.  As a result,

steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart
due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy
the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character

An example:
Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$.  Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine.  Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density.  Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital.  The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!).  Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds.  As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$.  Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!
Back to our problem:
Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2).  In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$.  Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed.  The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.
In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the  hybridized case the lowest energy.  Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.

It is not true that hybridisation costs energy, as it is only a mathematical concept, which has no real physical meaning, it is impossible to assign an energy value to hybridisation. It is also not possible, that an element chooses to hybridise. Hybridisation is a result of a certain - the thermodynamically most stable - geometric arrangement of atoms in a molecule.

@Martin  Thanks for the comment. Martin wrote 1) "It is not true that hybridisation costs energy" - I agree, see my edited answer. 2) "Hybridisation is a result of a certain - the thermodynamically most stable - geometric arrangement of atoms in a molecule" - I agree, in fact I said that in my initial version, "Whichever of these two pathways is most exothermic..."

Old post, I know, so there's a good chance I don't get an answer here, but does the decrease in H-P-H angle due to lack of hybridization necessarily mean more steric hindrance here? Yes, the cone angle is smaller, but an $sp^3$-orbital is shorter than a p-orbital, and with small atoms like hydrogen, I would think a small change in bond length would have a pretty significant effect on sterics around them phosphorus atom.

@DGS asked, "does the decrease in H-P-H angle due to lack of hybridization necessarily mean more steric hindrance"? If the bond angle was the only thing changing, then increased steric interaction would result.  But, as you note the bond length changes too (1.42 angstroms in phosphine) making it hard to answer your question without calculations.

@Martin-マーチン "It is not true that hybridisation costs energy, as it is only a mathematical concept..." I agree. Yet this might lead to some confusion. Hybridisation is conceptually taught (at secondary school level) via two mechanisms. The first option distinguishes initial promotion ($\neq$ hybridisation) from the mixing itself ($=$ hybridisation). The second way is to combine promotion and mixing by calling both together as hybridisation.

In the first line of reasoning, hybridisation does not involve an energy cost (as you said); whereas in the second it's essentially the promotion even though not called as such. This arguement assumes that both options are correct in their own right, which is what I've been lead to believe. (Please correct me if I'm wrong.)

@LinearChristmas Why would hybridisation not require an energetic cost? If an electron was initially in a 2s orbital while later on, it is assigned to be in a hybrid of 2s and 2p orbitals, then it is necessarily raised in energy since the hybrid is certainly of higher energy. I understand that the concept of "electron promotion" followed by hybridisation may not be correct as I have learnt from here: https://chemistry.stackexchange.com/questions/98910/electron-promotion-in-hybridisation-a-misconception

@Martin-マーチン Yes, the mixing of orbitals is a mathematical concept. But the re-organisation of electrons does require energy inputs, especially so when electrons of lower energy states now have to occupy higher energy states. Certainly, you would have to agree that there is a re-organisation of electrons when atoms form chemical bonds. Hybridisation describes this re-organisation. Thus, I think it is fair to say that hybridisation incurs energetic costs.

@TanYongBoon No, absolutely not. I do not agree with anything you have written in your comment. The formation of bonds (and the associated costs and gains) is completely independent of the mathematical description. A lone carbon atom would never, ever choose to have hybrid orbitals in the absence of an external field. Once the hydrogens form such a field in methane (for example), it is a completely different state, in which a description in terms hybrid orbitals may make sense, but is not necessary at all.

"It will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital. " Can you explain a bit whats rehybridization( i know whats hybridization) and why electrons do so? Its new to me (term) also why less electron density region will have more p character bonding chances? @ron

@Orion_Pax  "Rehybridization" is just a shorthand way of saying that the geometry and hybridization will be different when we make some change.  For example, if we compare ammonia and trifluoroamine the geometries and hybridizations will be different.  We can say that, as we replace the hydrogens in ammonia with fluorines, the molecule will adopt a different geometry, the bonds will involve different hybridizations in order for the molecule to adopt the lowest energy configuration possible.  We can say all of this quickly and simply by saying that the molecule will "rehybridize".

@Orion_Pax you asked, "why less electron density region will have more p character bonding chances?" Each carbon atom uses 1 2s and 3 2p orbitals to make all of its bonds.   Therefore, if a bond uses less s character, then the other bonds around this carbon will have to use more s character.

I see but Sir in hybridization all four hybrid are degenerate orbitals isnt ? So if some bond uses less s character as for formation will not it be having a different energy compared to what a hybrid should be ?

@Orion_Pax  In some cases like methane, hybridization produces 4 equivalent (degenerate) orbitals that are all at the same energy.  In other cases like methyl fluoride, the 4 hybridized orbitals are not equivalent. Bent's Rule tells us that the $\ce{C-F}$ bond will have less s character and be at higher energy relative to the 3 degenerate $\ce{C-H}$ bonds which will each have a bit more s character and consequently lie at lower energy.

First note, that hybridisation is a mathematical concept which can be applied to interpret a bonding situation. It has no physical meaning whatsoever. Instead it helps us to understand the direction of bonds better.
Second note, that the second period usually behaves quite differently from the remaining elements in a group. So in a way, ammonia behaves unnatural or anomalous.

If you compare nitrogen with phosphorus, you will note, that the former is much smaller than the latter, i.e. van der Waals radii $r(\ce{N})=155~\mathrm{pm};\ r(\ce{P})=180~\mathrm{pm}$ (ref. wikipedia), covalent radii $r(\ce{N})=71~\mathrm{pm};\ r(\ce{P})=107~\mathrm{pm}$ (ref. wikipedia). Therefore also the orbitals in nitrogen are smaller, and $\ce{s}$ and $\ce{p}$ orbitals will occupy more of the same space than in phosphorus. As a result the $\ce{N-H}$ bond distance will naturally also be shorter.
A lone pair is usually most stable in an orbital that has high $\ce{s}$ character. Bonds will most likely be formed with the higher lying $\ce{p}$ orbitals. The orientation of these towards each other is exactly $90^\circ$.
In ammonia this would lead to very close $\ce{H\cdots{}H}$ contacts, which are repulsive and therefore the hydrogen atoms are pushed away from each other. This is possible since in the second period the $\ce{s-p}$ splitting is still very small and the nitrogen $\ce{s}$ orbital is accessible for the hydrogen atoms. This will ultimately result in mixing $\ce{s}$ and $\ce{p}$ orbitals for nitrogen in the respective molecular orbitals. This phenomenon can be referred to as hybridisation - the linear combination of orbitals from the same atom. This term is therefore somewhat independent from its most common usage.
It is also very important to know, that the molecular wavefunction of a molecule has to reflect its overall symmetry. In this case it is $C_{3v}$, which means there is a threefold rotational axis and three vertical mirror planes (the axis is element of these planes). This gives also rise to degenerate orbitals. A canonical orbital picture has to reflect this property (BP86/cc-pVDZ; valence orbitals are ordered with increasing energy from left to right).

Note that the lowest lying valence molecular orbital is formed only from $\ce{s}$ orbitals (There is one additional $\ce{1s^2-N}$ core orbital.)
Now Natural Bond Orbital (NBO) Theory can be used to transform these delocalised molecular orbitals to a more common and familiar bonding picture, making use of atomic hybrid orbitals. This method is called localising orbitals, but it has the expense of losing the energy eigenvalue that may be assigned to canonical orbitals (NBO@BP86/cc-pVDZ; valence NBO cannot be ordered by energy levels).

In this theory you will find three equivalent $\ce{N-H}$ bonds, that are composed of $32\%~\ce{1s-H}$ and $68\%~\ce{s^{$0.87$}p^3-N}\approx\ce{sp^3-N}$ orbitals. Note that the lone pair orbital at nitrogen has a slightly higher $\ce{s}$ orbital contribution, i.e. $\ce{s^{1.42}p^3-N}\approx\ce{sp^3-N}$.
So the thermodynamically most favoured angle is found to be $107^\circ$ due to a compromise between optimal orbital overlap and least internuclear repulsion.

The canonical bonding picture in phosphine is very similar to ammonia, only the orbitals are larger.  Even in this case it would be wrong to assume, that there is no hybridisation present at all. However, the biggest contribution to the molecular orbitals stems from the $\ce{p}$ orbitals at phosphorus.

Applying the localisation scheme, one end up with a different bonding picture. Here are three equal $\ce{P-H}$ bonds that are composed of $48\%~\ce{1s-H}$ and $52\%~\ce{s^{$0.5$}p^3-P}$ orbitals. The lone pair at phosphorus is composed of $57\%\ce{s} + 43\%\ce{p}$ orbitals.

One can see the difference of the molecules also in their inversion barrier, while for ammonia the inversion is readily available at room temperature, $\Delta E \approx 6~\mathrm{kcal/mol}$, it is very slow for phosphine, $\Delta E \approx 40~\mathrm{kcal/mol}$.
This is mostly due to the fact, that the nitrogen hydrogen bonds have already a significant $\ce{s}$ orbital contribution, which can be easily increase, to form the planar molecule with formally $\ce{sp^2}$ hybrids.

@ martin : so hybridization is totally hypothetical and not practical,right?

@user36790 In the most common usage, this is true, but it can give you somewhat more insight into the shape of molecules. In a more genaral meaning, hybridisation just refers to the mixing of orbitals at the same atom in a molecule, and that might be observable.

Since Phosphorus has vacant 3d orbitals, the s-orbitals of hydrogen can directly overlap with the orbitals of Phosphorus. The procedure of bonding does not see much benefit in hybridisation as it is energetically expensive. Thus the hybridised character of the bonds is less dominant than in ammonia and the structure resembles the Lewis structure that we would expect instead of the tetreahedral (VSEPR) structure. This effect is called Drago's Rule and can also be observed in hydrogen Sulfide.

d-orbitals are too far removed to do anything in phosphorus and sulphur. Check out Martin's answer.


text:
Henry's Law with varying temperature
I am trying to determine what the derived partial pressure would be for CO2 in the global atmosphere from historical inconstant temperatures and CO2 concentrations of the global atmosphere:

Is it possible to do this?  If so, how?

An expression for partial pressure, or some sort of useful single value? Did you mean partial pressure of CO2 or something else?

@rch Thank you for looking rch!  I don't think I was concentrating when I wrote that.  I'll edit...

The short answer is no.
The long answer is:
Henry's law works for small concentrations of ideal mixtures at equilibrium. Henry's Law constant varies with temperature according to the Van't Hoff equation:
$$
k(T)=k(T^{\circ})exp\left[-C\left(\frac{1}{T}-\frac{1}{T\circ}\right)\right]
$$
Here $C$ is a constant related to the enthalpy of solvation for each gas, and $^{\circ}$ represents the standard state, T = 298 K and 1 atm.
Using standard state data, you could get an estimate of the partial pressure at a given temperature. However, two key assumptions that are implicit to Henry's law and the Van't Hoff equation do not hold for global atmospheric gas composition:

The atmosphere and the oceans are not at equilibrium (not even close when you consider atmospheric layers and convection)
$\ce{CO2}$ in seawater is not an ideal mixture (very non-ideal since $\ce{CO2}$ reacts with water).

This is one of the many reasons that climate modeling is difficult - the oceans act as huge carbon sinks with a very complex and not-very-stable relationship between dissolved $\ce{CO2}$ content, temperature, composition of rock on the sea floor, convection currents in the atmosphere and the ocean, amount of shellfish (no kidding), and probably lots of other factors (I'm not a climate expert).
The net result is that while Henry's law can get you a quick idea of how much $\ce{CO2}$ could be dissolved at a given temperature under pseudo-equilibrium conditions (or what the corresponding partial pressure would be for a given concentration in water), it isn't enough to get you anywhere close to a good prediction of what the global atmospheric composition will be, particularly for $\ce{CO2}$ (due to its reactivity with water and incorporation into biomass).
Now, if you really just want to convert $\ce{CO2}$ concentration in ppm to a partial pressure, you can do that.
$\ce{CO2}$ in air is close enough to ideal that we can use the ideal gas law. For an ideal gas, the partial pressure is the same as whatever the pressure would be if it were in a container by itself:
$$
P_1 = \frac{n_1RT}{V}
$$
You can find V for a given mass of air, find the mass of $\ce{CO2}$ from the concentration, then convert to moles and plug it into the above.

You currently have numbers in ppm (of ${CO_2}$); ppm values for gas tend to be equivalent to ${\mu mol/mol}$ units. Can't you just use that, the the definition of partial pressure, and the atmospheric pressure and be set (though it varies by altitude and a little over time too)? Some forum has a similar homework-level question. Not sure if I'm missing something with that.
I take it you wanted to use Henry's Law to relate concentration to partial pressure. typically that's used in the realm of vapor-liquid systems; not sure how it fares in a mixture of many gases that I'd qualify the atmosphere as. See things to note: (Chemwiki)
Looking at a (pretty crude IMHO) pressure vs altitude graph of the atmosphere (Wikipedia: Atmospheric pressure): you can see the pressure even within the troposphere ranges from ~12kPa all the way to ~100kPa (sea level: ~101 kPa). The troposphere has 80% of the atmosphere's mass and you want the whole atmosphere anyways, so just take 101kPa as atmospheric pressure? I also assume atmosphere's mass is constant over the span of 400 million years - no clue on whether I can neglect losses and gains from weather, the biosphere, and space over this time span. Then you will have a bunch of carbon dioxide partial pressures over time.

On short-term constancy of the atmosphere's mass, see SE.Physics. The mean even changes by season. It's comparatively small comparing 1.5E+15 kg to the total of 5.15E+18 kg.


text:
Wavelength of X-rays
What will be the wavelength of an X-ray, if we apply a potential difference of, say $\pu{20 kV}$ across the ends of the X-ray tube?
Alright, I started with the modified de-Broglie equation-
$$\Lambda=\frac{h}{\sqrt{meV}}$$
I pinned the values of Planck constant, electronic charge and potential difference but can't guess what would be the mass $m$ here.
The answer comes $\pu{0.62*10^-10m}$ if solved perfectly. Can anybody confirm?

check  Planck–Einstein equation.

But actually, the question provides insufficient information. X-ray from X-ray tubes is part of hi-energy electronic spectra of the anode material, so in reality the wavelength of most photons would be defined by anode material, and the only requirement for the potential difference is to be enough to cause excitation of the anode material. (It is somewhat oversimplified description, but reasonably adequate)

$$\lambda=\frac h{\sqrt{meV}}$$
$$E=mc^2$$
Also $$E=\frac{hc}\lambda$$
Therefore,
$$mc^2=\frac{hc}\lambda$$
$$mc=\frac h\lambda$$
Hence $$m=\frac h{\lambda c}$$
Plugging the value of $m$ from here we get an equation with all known constants. Solving for $\lambda$, we can get the result as $0.62\times10^{-10}\ \mathrm m$.

This gives the shortest (highest energy) x-ray. The overall spectra from the x-ray tube would be Bremsstrahlung radiation (sort of like blackbody radiation) with line spectra from the element of the anode material of the x-ray tube. The electron beam of an x-ray tube is in a vacuum. The x-rays are generally emitted through a thin beryllium window which absorbs lower energy (long wavelength) x-rays.

Firstly as @MaxW pointed out, using the given information, it is possible to find the shortest wavelength (or maximum frequency) X-ray only.

In an X-ray tube, electrons are accelerated in a vacuum by an electric field and shot into a piece of heavy metal (e.g., $\ce{W, Rh, Mo, Cu, Ag}$) plate. X-rays are emitted as the electrons decelerate in the metal. The output spectrum consists of a continuous spectrum of X-rays, with sharp peaks at certain energies as in the graph. The continuous spectrum is due to bremsstrahlung (German for "deceleration radiation"), while the sharp peaks are characteristic X-rays associated with the atoms in the target.
The spectrum has a sharp cutoff at low wavelength (high frequency), which is due to the limited energy of the incoming electrons (which is equal to the voltage on the tube times the electron charge). This cutoff applies to both the continuous (bremsstrahlung) spectrum and the characteristic sharp peaks, i.e. there is no X-ray of any kind beyond the cutoff. However, the cutoff is most obvious for the continuous spectrum.
Hence, the Duane-Hunt law comes handy in here, which gives the maximum frequency of X-rays that can be emitted by Bremsstrahlung in an X-ray tube by accelerating electrons through an excitation voltage $V$ into a metal target.
The maximum frequency $\nu_\text{max}$ is given by-
${\displaystyle \nu _{\text {max}}={\frac {eV}{h}},}$
which corresponds to a minimum wavelength
${\displaystyle \lambda _{\text {min}}={\frac {hc}{eV}},}$
where $h$ is Planck's constant, $e$ is the charge of the electron, and $c$ is the speed of light.
Thus putting the values given-
$$\lambda _{\text {min}}={\frac {hc}{eV}}={\frac {\pu{1240 eV nm}}{\pu{20*1000 eV}}}= \pu{0.62*10^-10 m}$$
For further references check the Wikipedia page.

I speculate what reads like a (of course incomplete) list of (solid) targets required the correction Mb -> Mo, and benefits the addition of Cu and Ag so often used for the characterization of organic molecules (proteins), regardless if «classical» tube, rotating anode (especially in biochem), or microfocus X-ray tube.

@Buttonwood Ah...sure! I was in a hurry, and I simply omitted that part while doing the corrections. Thanks! I wasn't aware of the usage of $\ce{Ag}$ though.

Some of them are listed in IUCr's teaching, and of course by the manufactures of X-ray tubes. Beside micro focus tubes (e.g. on diffractometers with both a Cu and a Mo source), be aware of the brighter liquid targets, e.g. background for small crystals.


text:
Why is HSO4- shown as an example of a weak acid instead of HSO4?
I'm watching some (really awesome) chemistry courses, and I think I have a fairly decent handle on acids and bases (or maybe not, and I'm fooling myself). But the teacher just showed a table of weak/strong acids. $\ce{HSO4-}$  was one of them.
I'm wondering, why hydrogen sulfate ion instead of just hydrogen sulfate? Why is $\ce{HSO4-}$ an acid and $\ce{HSO4}$ apparently isn't? This was the only ion on the table. Or does $\ce{HSO4}$ simply not exist?

The bisulfate molecule doesn't exist in solution. I can only find references to the hydrogen sulfate molecule as part of salts, which implies that hydrogen sulfate is usually found as an ion.
The bisulfate (hydrogen sulfate) ion is indeed an acid. It is a relatively strong weak acid too, with a $\ce{K_a}$ value of $1.2 * 10^{-2}$.
Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. This makes the hydrogen more partially positive and thus more reactive (more likely to be taken by a base).

I have to add, that while $\ce{HSO4}$ does not exist, the doubled molecule $\ce{H2S2O8}$ (peroxodisulfuric acid) does exist and is apparently a quite strong acid (its double ammonium salt is stable solid)

I have to add, that $\ce{HSO4}$ might well exist, just not in aqueous solution - in the gas phase, why not ?

I agree Martin!

@Martin - it would have an odd number of valence electrons, and so it probably wouldn't exist for long!

"Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. " is that actually true? How would that work? Isn't it just acidic because the negative charge can be fairly good delocalized into the $\pi$-bonds? If we'd were to replace one of the two  S-O double bond with a SOH group, I think it would be that acidic anymore...

It is true. Do you think a hydroxide ion with its significant negative charge density would be more likely to attack a hydrogen atom with no partial charge or a hydrogen atom with a partial positive charge? Your reason is also true. The sulfate anion has several resonance structures.

@Jori there is no significant pi bonding in the sulfate anion. I think that was discussed with the phosphate ion here on the site. I am on mobile now, so it is hard to find the reference. However, the electron of hydrogen is already very much localised at the oxygen. The hydrogen oxygen bond is quite ionic.

@thomij Agreed, but it would certainly exist longer than an anion in the gasphase.

" I think it wouldn't be that acidic anymore..." is what I meant to say. Do these structures exist? Replacing one double S-O bond with a SOH group. I think you understood me wrong. I meant that the reason sulfuric acid is so acidic, is not the amount of oxygen atoms close to the SOH group, but the delocalisation of the negative charge when hydrogen takes off. Resulting in a more stable structure. Is the force of the other surrounding oxygen atoms (that are negatively charged) of suck influence over that distance, and wouldn't the strongly positively charged sulfur get in the way?

I'm just thinking here a bit. You guys are probably way better educated than me in chemistry haha. We coudl confirm the truth (or at least get an indication of the truth) if we knew the acidity of the hypothetical structures H3SO4 and H4SO4 (do they even exist?)

I notice that there is no mention of polyprotic acids in the other answers!
$\ce{H2SO4}$ (sulfuric acid) is diprotic - meaning it has two "detachable" protons which can come off in aqueous solution. The first one is strongly acidic - the $Ka = 2.4 × 10^6$ - this means the odds of finding a non-dissociated $\ce{H2SO4}$ molecule in solution are something like 1 out of 2.4 million.
On the other hand, the second proton is weakly acidic - the $Ka = 1.0 × 10^{-2}$. This means that there will mostly be $\ce{HSO4-}$ instead of $\ce{SO4^2-}$ in solution (roughly 100 times as much).
The other answers are correct in saying that $\ce{HSO4}$ does not exist in solution - but it is important to realize that we list the bisulfate ion ($\ce{HSO4-}$) as a weak acid because it is the conjugate base of sulfuric acid, and is in turn the conjugate acid of the sulfate ion. This becomes very important when you start to study acid/base equilibria.

$\ce{HSO4}$ does not exist.  There is $\ce{H2SO4}$ and $\ce{HSO4-}$  Try drawing the Lewis structure for sulfate ion.  You'll see that there are two oxygens with formal charges.  They can be left as is or they can have $\ce{H}$ bonded to them.

I think you mean H- bonded to them (H forms single bonds).

The adequate Lewis structure for the sulfate ion has one negative charge at each oxygen and a double positive charge at sulfur. Everything else is just too far from the truth.