Write the electronic configuration of the following. Also explain whether they are paramagnetic / diamagnetic :-
Cr+4 ,Sc+ , Cu+2 , Cu+ , Ag+
Answers
Answer:
Explanation:
we can start from the electron configuration of the neutral atoms. I'm assuming you meant
Cr
3
+
,
Ca
2
+
, and
Fe
3
+
...
Cr
:
[
A
r
]
3
d
5
4
s
1
Ca
:
[
A
r
]
4
s
2
Na
:
[
N
e
]
3
s
1
Fe
:
[
A
r
]
3
d
6
4
s
2
The rightmost orbitals listed here are highest in energy, so we ionize these atoms by booting off the highest-energy electrons. Thus:
Cr
→
Cr
3
+
+
3
e
−
[
A
r
]
3
d
5
4
s
1
→
[
A
r
]
3
d
3
Since there are
5
3
d
orbitals, they are not all at least singly filled yet, and thus, all three electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.
The original atom is also paramagnetic. The
4
s
subshell contains
1
electron (in one
4
s
orbital) and the
3
d
subshell contains
5
electrons, one in each
3
d
orbital. No valence electrons are paired here.
That's in agreement with our expectations from Hund's Rule (generally, for the lowest-energy configuration, maximize parallel spins where possible by singly-filling all orbitals of very similar energies first and then doubling up afterwards).
Ca
→
Ca
2
+
+
2
e
−
[
A
r
]
4
s
2
→
[
A
r
]
⇒
1
s
2
2
s
2
2
p
6
3
s
2
3
p
6
This is a noble gas configuration, so no electrons are unpaired Thus, this is diamagnetic.
Na
→
Na
+
+
e
−
[
N
e
]
3
s
1
→
[
N
e
]
⇒
1
s
2
2
s
2
2
p
6
This is a noble gas configuration, so no electrons are unpaired. Thus, this is diamagnetic.
Fe
→
Fe
3
+
+
3
e
−
[
A
r
]
3
d
6
4
s
2
→
[
A
r
]
3
d
5
Since there are
5
3
d
orbitals, in accordance with Hund's Rule, all five electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.