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hammer [34]
3 years ago
13

How would you describe the behavior of particles in a solid?

Chemistry
1 answer:
ehidna [41]3 years ago
4 0
A.

Particles within a solid only vibrate in place.
You might be interested in
1s^2 2s^2 2p^6 3s^2 3p^6 how many unpaired electrons are in the atom represented by the electron configuration above?
Sedbober [7]
It's a combination of factors:
Less electrons paired in the same orbital
More electrons with parallel spins in separate orbitals
Pertinent valence orbitals NOT close enough in energy for electron pairing to be stabilized enough by large orbital size
DISCLAIMER: Long answer, but it's a complicated issue, so... :)
A lot of people want to say that it's because a "half-filled subshell" increases stability, which is a reason, but not necessarily the only reason. However, for chromium, it's the significant reason.
It's also worth mentioning that these reasons are after-the-fact; chromium doesn't know the reasons we come up with; the reasons just have to be, well, reasonable.
The reasons I can think of are:
Minimization of coulombic repulsion energy
Maximization of exchange energy
Lack of significant reduction of pairing energy overall in comparison to an atom with larger occupied orbitals
COULOMBIC REPULSION ENERGY
Coulombic repulsion energy is the increased energy due to opposite-spin electron pairing, in a context where there are only two electrons of nearly-degenerate energies.
So, for example...
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−− is higher in energy than
↑
↓
−−−−−

↓
↑
−−−−−

↑
↓
−−−−−
To make it easier on us, we can crudely "measure" the repulsion energy with the symbol
Π
c
. We'd just say that for every electron pair in the same orbital, it adds one
Π
c
unit of destabilization.
When you have something like this with parallel electron spins...
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−
It becomes important to incorporate the exchange energy.
EXCHANGE ENERGY
Exchange energy is the reduction in energy due to the number of parallel-spin electron pairs in different orbitals.
It's a quantum mechanical argument where the parallel-spin electrons can exchange with each other due to their indistinguishability (you can't tell for sure if it's electron 1 that's in orbital 1, or electron 2 that's in orbital 1, etc), reducing the energy of the configuration.
For example...
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−− is lower in energy than
↑
↓
−−−−−

↓
↑
−−−−−

↑
↓
−−−−−
To make it easier for us, a crude way to "measure" exchange energy is to say that it's equal to
Π
e
for each pair that can exchange.
So for the first configuration above, it would be stabilized by
Π
e
(
1
↔
2
), but the second configuration would have a
0
Π
e
stabilization (opposite spins; can't exchange).
PAIRING ENERGY
Pairing energy is just the combination of both the repulsion and exchange energy. We call it
Π
, so:
Π
=
Π
c
+
Π
e

Inorganic Chemistry, Miessler et al.
Inorganic Chemistry, Miessler et al.
Basically, the pairing energy is:
higher when repulsion energy is high (i.e. many electrons paired), meaning pairing is unfavorable
lower when exchange energy is high (i.e. many electrons parallel and unpaired), meaning pairing is favorable
So, when it comes to putting it together for chromium... (
4
s
and
3
d
orbitals)
↑
↓
−−−−−
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−
compared to
↑
↓
−−−−−
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−
is more stable.
For simplicity, if we assume the
4
s
and
3
d
electrons aren't close enough in energy to be considered "nearly-degenerate":
The first configuration has
Π
=
10
Π
e
.
(Exchanges:
1
↔
2
,
1
↔
3
,
1
↔
4
,
1
↔
5
,
2
↔
3
,

2
↔
4
,
2
↔
5
,
3
↔
4
,
3
↔
5
,
4
↔
5
)
The second configuration has
Π
=
Π
c
+
6
Π
e
.
(Exchanges:
1
↔
2
,
1
↔
3
,
1
↔
4
,
2
↔
3
,
2
↔
4
,
3
↔
4
)
Technically, they are about
3.29 eV
apart (Appendix B.9), which means it takes about
3.29 V
to transfer a single electron from the
3
d
up to the
4
s
.
We could also say that since the
3
d
orbitals are lower in energy, transferring one electron to a lower-energy orbital is helpful anyways from a less quantitative perspective.
COMPLICATIONS DUE TO ORBITAL SIZE
Note that for example,
W
has a configuration of
[
X
e
]
5
d
4
6
s
2
, which seems to contradict the reasoning we had for
Cr
, since the pairing occurred in the higher-energy orbital.
But, we should also recognize that
5
d
orbitals are larger than
3
d
orbitals, which means the electron density can be more spread out for
W
than for
Cr
, thus reducing the pairing energy
Π
.
That is,
Π
W
5 0
3 years ago
Read 2 more answers
Calculate how many moles of Ca(OH)2 are needed to produce 4.8 x 1024 NH3 molecules, according to the following equation
bonufazy [111]

Answer:

66.33

Explanation:

5 0
3 years ago
A pH scale has how many sections on it?
leva [86]

Answer:

0-14

Explanation:

3 0
3 years ago
Radon (Rn) is the heaviest, and only radioactive, member of Group 8A(18) (noble gases). It is a product of the disintegration of
Serggg [28]

Given data                Atomic mass of Ra= 226g/mol

no. of moles =1.0/226g/mol           =0.04424moles

no. of atoms in 0.044moles

no. of atoms =no. of moles x avogadro's number

= 0.044x 6.022 x10^23                  = 0.264968 x 10^22

 If 10^15  atoms of Ra produce 1,373*10^4  atoms of<u> Rn per second</u> then 2,66 *10^21  forms 3,658*10^10 atoms of Rn per second.

Day has 246060=86400 s

That means that 2,66x10^21  atoms of Ra produces 3,16 x10^15  atoms of Rn in a day.

N(Rn)=3.16* 10 ^15                           n(Rn)=N/NA

n(Rn)=5,25*10−9                              pV=nR*T

T=273.15K                                        R=8,314

p=101325Pa                                      V=n∗R∗T/p

V=5.25∗10^−9 ∗ 8.314 ∗ 273.15  /  101325

V=1.1810^−10 m^3 =   118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra

To know more about Ra here

brainly.com/question/9112754

#SPJ4

6 0
2 years ago
Hydrochloric acid reacts with sodium hydroxide to form water and sodium chloride. Hydrochloric acid is an extremely acidic, clea
Brut [27]

The answer is C.

You're welcome!

7 0
3 years ago
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