How can we determine a particular element using the quantum no.s
Perfect answers wd be marked as brainliest
Useless answers shall be reported
Answers
m
l
is the magnetic quantum number, corresponding to the projection of the angular momentum of an orbital, i.e. its orientation in space.
As the symbol suggests, it has to do with
l
, the angular momentum quantum number.
l
describes the shape of the orbital. Let's look at various values of
l
and their corresponding
m
l
.
l
=
0
→
m
l
=
0
, orbital =
s
l
=
1
→
m
l
=
−
1
,
0
,
+
1
, orbital =
p
l
=
2
→
m
l
=
−
2
,
−
1
,
0
,
+
1
,
+
2
, orbital =
d
l
=
3
→
m
l
=
−
3
,
−
2
,
−
1
,
0
,
+
1
,
+
2
,
+
3
, orbital =
f
and so on.
The general pattern is that we have:
m
l
=
−
l
,
−
l
+
1
,
−
l
+
2
,
.
.
.
,
0
,
+
1
,
+
2
,
.
.
.
,
+
l
−
2
,
+
l
−
1
,
+
l
or
m
l
=
0
,
±
1
,
±
2
,
.
.
.
,
±
l
In short, we have
2
l
+
1
values of
m
l
for a particular
l
for a particular orbital.
If, let's say, we chose boron (
Z
=
5
), it has access to the valence orbitals
2
s
and
2
p
, but it also has the
1
s
technically as a core orbital.
1
s
:
(
n
,
l
,
m
l
)
=
(
1
,
0
,
0
)
Hence, there is only one
1
s
orbital.
2
s
:
(
n
,
l
,
m
l
)
=
(
2
,
0
,
0
)
So, there is only one
2
s
orbital.
2
p
:
(
n
,
l
,
m
l
)
=
(
2
,
1
,
[
−
1
,
0
,
+
1
]
)
Therefore, there are only three
2
p
orbitals (
2
p
x
,
2
p
y
, and
2
p
z
).
For its valence orbitals, since it has one
2
s
and three
2
p
orbitals, it can have up to
2
×
1
+
3
×
2
=
8
valence electrons. Thus, it is not expected to exceed
8
valence electrons in its molecular structures.m
l
is the magnetic quantum number, corresponding to the projection of the angular momentum of an orbital, i.e. its orientation in space.
As the symbol suggests, it has to do with
l
, the angular momentum quantum number.
l
describes the shape of the orbital. Let's look at various values of
l
and their corresponding
m
l
.
l
=
0
→
m
l
=
0
, orbital =
s
l
=
1
→
m
l
=
−
1
,
0
,
+
1
, orbital =
p
l
=
2
→
m
l
=
−
2
,
−
1
,
0
,
+
1
,
+
2
, orbital =
d
l
=
3
→
m
l
=
−
3
,
−
2
,
−
1
,
0
,
+
1
,
+
2
,
+
3
, orbital =
f
and so on.
The general pattern is that we have:
m
l
=
−
l
,
−
l
+
1
,
−
l
+
2
,
.
.
.
,
0
,
+
1
,
+
2
,
.
.
.
,
+
l
−
2
,
+
l
−
1
,
+
l
or
m
l
=
0
,
±
1
,
±
2
,
.
.
.
,
±
l
In short, we have
2
l
+
1
values of
m
l
for a particular
l
for a particular orbital.
If, let's say, we chose boron (
Z
=
5
), it has access to the valence orbitals
2
s
and
2
p
, but it also has the
1
s
technically as a core orbital.
1
s
:
(
n
,
l
,
m
l
)
=
(
1
,
0
,
0
)
Hence, there is only one
1
s
orbital.
2
s
:
(
n
,
l
,
m
l
)
=
(
2
,
0
,
0
)
So, there is only one
2
s
orbital.
2
p
:
(
n
,
l
,
m
l
)
=
(
2
,
1
,
[
−
1
,
0
,
+
1
]
)
Therefore, there are only three
2
p
orbitals (
2
p
x
,
2
p
y
, and
2
p
z
).
For its valence orbitals, since it has one
2
s
and three
2
p
orbitals, it can have up to
2
×
1
+
3
×
2
=
8
valence electrons. Thus, it is not expected to exceed
8
valence electrons in its molecular structures.