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26. If each orbital contain 3 electrons the
number of elements present in 5th period wi
(1) 18
(2) 27
(3) 32
(4) 36
Answers
Well, this is kind of open to interpretation... but if I interpreted it correctly, I get
27
elements, compared to the original
18
elements, in the 4th period of the periodic table.
(If you already knew that the number of electrons allowed in a given orbital is derived from the properties of electrons, and NOT of the orbitals themselves, the answer quickly follows.)
PRELIMINARY THINGS
I think there's a typo in the question... I looked this up elsewhere, and it's probably...
If each orbital can hold a maximum of 3 electrons, the number of elements in the 4th period of the periodic table (long form) is...?
Also, note that this is entirely theoretical, as all electrons only have two possible spins (
m
s
=
±
1
2
) in a given orbital, and no two electrons can share the same quantum state (Pauli Exclusion Principle); that restricts each orbital to contain only two electrons in real life.
That aside, when we suppose three electrons are "allowed" in a single orbital (assuming the other three quantum numbers are as normal), we suddenly "allow"
50
%
more elements in a given quantum level.
(It's not really why the periodic table was arranged historically, but... I suppose that's what the intent of this question was...)
EXPANDING THE PERIOD...
An electron configuration for the fourth period in a generalized manner is written as:
4
s
x
3
d
x
4
p
x
where:
x
=
(
2
l
+
1
)
⋅
N
m
s
is the total number of electrons in all the orbitals in a given subshell.
l
is the angular momentum quantum number.
l
=
0
,
1
,
2
,
3
,
.
.
.
corresponds to
s
,
p
,
d
,
f
,
.
.
.
orbitals.
2
l
+
1
is known as the degeneracy of the subshell; it is how many orbitals are in that subshell.
N
m
s
is an arbitrary number of spins the electron could have, as it also then gives the maximum number of electrons per orbital. In this case we SUPPOSE that
N
m
s
=
3
, but in real life it is just
2
.
Now, the number of allowed
m
s
values derives from the properties of the electron, not of the orbitals themselves, so having more spins allowed does NOT change the orbital shapes or relative energies.
For the
4
s
orbital:
l
=
0
,
⇒
2
l
+
1
=
1
For the
3
d
orbitals:
l
=
2
⇒
2
l
+
1
=
5
For the
4
p
orbitals:
l
=
1
,
⇒
2
l
+
1
=
3
Thus, the hypothetical electron configuration we would then write is...
4
s
1
⋅
3
3
d
5
⋅
3
4
p
3
⋅
3
=
4
s
3
3
d
15
4
p
9
And that would apparently expand the fourth period of the periodic table from
18
elements to
27
elements.
write the numbers or equation togather because of my phone malfunfunction it happened