Chemistry Question + Answer

[srindogg@hotmail.com]

Yes, that is correct. The identities switch when the cell is recharging.

[gamma]

So ALL 3rd row elements regardless of their place in the table (ie even row 3 ones in the p section) have d orbitals? why is S neutral atom then in the p section of the 3rd row?

[goose]

So ALL 3rd row elements regardless of their place in the table (ie even row 3 ones in the p section) have d orbitals? why is S neutral atom then in the p section of the 3rd row?

 

All elements technically have all orbitals - even hydrogen has a d orbital, but it's electrons are in an orbital far lower in energy that it is nearly impossible for them to be excited to the d orbital. As a general rule, all elements in the third row and higher have electrons that are in, or are capable of being in a d orbital. This means that they can have expanded octets by filling their empty unhybridized d orbitals because their bonding electrons are "close enough" to d. The elements in the d block have valence electrons in the d orbital. Elements like Bromine have electrons in the d orbital, but are in the p block because that's where it's valence electrons are.

 

Hopefully my explanation wasn't too convoluted.

[gamma]

^oh/k, so when we have the ion Cl^3-, that means that the 1st electron would fill the p orbital and the other 2 would enter the d orbital because Cl would not have a 4s orbital?

i don't feel like any chem prof or ta has ever mentioned this before now i don't know what else i might've missed

 

ok I know that the d orbital is higher in energy than the s (that's why the order goes 3p4s3d) but my question is if it's higher in energy, why is it that ions are formed by removing electrons from the s shell vs d?

[dan0105]

Errr, you are asking us to teach you a whole course, where were you in chem 101!

 

Cl^3- still does NOT have anything close to what you just said.

Cl is normally filled up 1s2s2p3s3p with ONE electron short in the 3p (remember the s's hold 2 and the p's hold 6). If it were charged with an extra three electrons, the first one would go to filling the octet rule, well basically just filling the orbital that isn't full yet. So 3p is full now. The next two would go to the next orbital in line, which is lowest in energy (but still higher than the others that are already full) this would be, 4s.

 

There is ONE special case of this that you are eluding to. That is, if for some odd reason removing electrons, or rearranging electrons between s and d orbitals can create half-filled or fully filled d orbitals than it will, because it just so happens to be that they are very close in energy and half/fully filled d orbitals are unusually stable. So, say something like Fe comes along which normally has 2 electrons in its s (full) and 6 in its d (not so full), and then it loses three electrons. It COULD lose 3 from the d, and that would make sense, but what ends up happening is it loses the 2 from the s, and then 1 from the d, to leave an empty s orbital and a d orbital with 5 electrons. BUT 5 electrons in a d is half-full which is sort of stable, so this ends up being preferred.

[eng_dude786]

can someone confirm this:

 

K<1, delta g (std) >0

 

reaction is non-spontaneous

 

K>1, delta g (std) <0

 

reaction is spontaneous

 

K=1, delta g (std) =0

 

what is special about this condition?

[Avinash]

can someone confirm this:

 

K<1, delta g (std) >0

 

reaction is non-spontaneous

 

K>1, delta g (std) <0

 

reaction is spontaneous

 

K=1, delta g (std) =0

 

what is special about this condition?

 

Those are correct.

 

 

G(std) refers to the change in free energy as the reaction proceeds from standard state, to equlibrium state.

 

Standard state is define here, as the state in which all reactants and products are at a concentration of 1M. This means, that at standard state, the reaction quotient (Q) is 1.

 

If the K for a reaction is less than one, it means that, at equilibrium, there are more reagents than products. So as the reaction proceeds from standard state (equal reactants and products) towards equlibrium (more reactants than products), it is spontaneous in the direction that converts products to reactants. In other words, the back reaction is spontaneous, and delta G is positive.

 

Try to apply that to the other scenarios.

[oohpsjin]

For the simple reaction:

 

2NaClO3 + SO2 + H2SO4 -> 2NaHSO4 + 2ClO2

 

how do you know which is the reducing agent? Couldn't SO2 AND H2SO4 both qualify as reducing agents as they are both oxidized?

I.e. SO2 oxidation number increases in NaHSO4

H2SO4 loses a proton in NaHSO4...

 

The answer is that SO2 is the only reducing agent, but I don't understand why H2SO4 would not be considered a reducing agent as well.

 

Thanks guys.

[dan0105]

2NaClO3 + SO2 + H2SO4 -> 2NaHSO4 + 2ClO2

 

how do you know which is the reducing agent? Couldn't SO2 AND H2SO4 both qualify as reducing agents as they are both oxidized?

I.e. SO2 oxidation number increases in NaHSO4

H2SO4 loses a proton in NaHSO4...

 

Oxidation states of all those parts:

NaClO3 is 0, therefore the Na is +1 and the rest is -1. ClO3 is chlorate and NOT a peroxide so all is good and each O gets -2 and thus the Cl is +5 (weird but ok).

 

SO2 is 0 (we can go further and say each O is -2 and S must be +4)

 

H2SO4 is 0, each H is +1 so the SO4 is -2 (we can go further and say each O is -2 and thus the S is +6).

 

NaHSO4 is 0, the Na is +1, so HSO4 is -1. We can take this further and say H is +1 and the SO4 is then -2 (because +1 + -2 = -1). (Again, we can go further and say each O is -2 and thus the S is +6)

 

ClO2 is 0, thus Cl is +4 and each O is -2.

 

Anyways, you are right about SO2 increasing when it goes to NaHSO4 thus it IS the reducing agent (it undergoes oxidation, loss of electrons, thus gets more negative in number) but thing you said about H2SO4 losing a proton in NaHSO4, it gained an Na instead! so the number for the rest of the molecule really didn't change at all!!! The SO4 part went from -2 to -2....

 

cheers, more examples,

http://www.mymcat.com/wiki/Oxidation_Numbers

[gamma]

when Cu (s) reacts with H20 (l), no reaction occurs. When Cu(s) is placed with HNO3, a gas evolves. What's the likely identity of this gas?

 

Ans: NO (g)

 

but ..how? don't redox reactions give off H2(g)?

[The Law]

Redox reactions don`t have to give off hydrogen. What you`re looking for is a change in the oxidation numbers. In this case, copper is oxidized with oxygen and the nitrate is reduced because it loses oxygens (decreasing its oxidation number).

 

 

A metal cation has the following electron configuration:

 

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1

 

What will be the cation it forms...

 

A) +1

+2

C) +3

D) +4

 

(answer: C)

 

 

I`m confused, I said +1. I knew for sure the p1 had to get removed, but why 4s2... that`s a full orbital... if someone could explain, would really appreciate it!

[avenir001]

Redox reactions don`t have to give off hydrogen. What you`re looking for is a change in the oxidation numbers. In this case, copper is oxidized with oxygen and the nitrate is reduced because it loses oxygens (decreasing its oxidation number).

 

 

A metal cation has the following electron configuration:

 

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1

 

What will be the cation it forms...

 

A) +1

+2

C) +3

D) +4

 

(answer: C)

 

 

I`m confused, I said +1. I knew for sure the p1 had to get removed, but why 4s2... that`s a full orbital... if someone could explain, would really appreciate it!

 

uhhh i think the question is kinda wrong cuz it should say "a metal has bla bla configuration" (not a metal cation)..but anyway, from the PT, the metal with that configuration is Ga which is in the Aluminum family which form +3 cations.

[The Law]

My bad, I wrote cation because I`m dumb.

 

Anyway, yes, I was too lazy to look at the group and thought I would be able to solve without that. Dumb Law, I should have just done that lol. What if it was the same thing, but with no p electrons... (again too lazy to look at periodic table, would it lose the s electrons then)

[avenir001]

i think generally u lose 3d electrons before 4s ones cuz 3d is higher in energy.

[The Law]

Yeah, but is disrupting 5 stable electrons better than disrupting one lower energy orbital... hmm. lol

[avenir001]

actually i think u're right..i do remember smth weird happening with transition metals in that to form ions, u lose electrons from subshells with higher n first (ie 4s loses electrons before 3d).

[The Law]

actually i think u're right..i do remember smth weird happening with transition metals in that to form ions, u lose electrons from subshells with higher n first (ie 4s loses electrons before 3d).

 

Did everybody catch that!

[avenir001]

Did everybody catch that!

 

aaahahahaha...poor law..he's so used to being wrong that he can't believe his eyes. will u all now plz join me in a round of applause for this poor guy

 

*earth-shattering applause from audience*

[eng_dude786]

According to VSEPR theory, what is the shape of SO2?

 

A. Bent

B. linear

C. trigonal planar

D. tigonal bipyramidal

 

 

The right answer is A, but why is it not C. I'm confused. The answer says something along the lines that electron configuration arrangement is different than the actual shape of molecule.

 

I know the different electron configuration can be: linear, trigonal planar, tetrahedral.

 

What about the "shapes of molecules" I should know for the MCAT? What else are there beside "bent"?

[always_panicked]

I would think because SO2 and H2O are the same (2 lone pairs), therefore SO2 is bent....you should know:

 

linear, bent, trigonal planer, tetrahedral, trigonal pyramidal, trigonal bipyramidal, see-saw, t-shape, octahedral, square pyramidal, and square planer

[eng_dude786]

i can't seem to find this in my notes:

 

if you are given Enthalpies of 2 reactions and the reaction you are interested in is essentially the summation of the 2 reactions.

 

Do you multiply the 2 enthalpies or add them?

 

i know its too basic (pun intended) of a question but i can't find the notes 1 day before the exam..

[kz87]

i believe you add them

[The Law]

i can't seem to find this in my notes:

 

if you are given Enthalpies of 2 reactions and the reaction you are interested in is essentially the summation of the 2 reactions.

 

Do you multiply the 2 enthalpies or add them?

 

i know its too basic (pun intended) of a question but i can't find the notes 1 day before the exam..

 

Yes, you add them! But don't forget, you have to make sure you have the right moles of everything and that would results in you multiplying the standard enthalpies. Also, you have to make sure the reaction is in the right order.

[eng_dude786]

last one hopefully.

 

I attached the question as a picture. The back of the book gives the following explanation:

 

B is correct. Beaker B will have a lower vapor pressure than Beaker A. All the

solvent will go to Beaker B.

 

My confusion is that I thought that the solvent will only go into the second beaker just enough to equilibrate the pressure in the two beakers. Someone help!

[avenir001]

all the solvent will eventually go into Beaker B because the vapour pressure above beaker B is always less than the vapour pressure above Beaker A...remember some stupid law (sorry law ) (raoult's law i think) says that vapour pressure of a solution (eg volatile solvent + nonvolatile solute) is proportional to the mole fraction of the solvent and the vapour pressure of the pure solvent...so if 90% of the solution in beaker B is solvent, then the vapour pressure above Beaker B will be 90% of the vapour pressure of pure solvent (which is the case above Beaker A)...as more volatile solvent goes into Beaker B, the mole fraction of the volatile solvent in B increases but remember it never gets to be pure like Beaker A...so all the solvent in beaker A will go into Beaker B (altho with a decreasing rate of diffusion) until eventually nothing is left in A.