bond
Q. In which of the following
order has increased
and the magnetic pour
has changed and How?
1.) No Not
2) O2 - O₂ t
Na Na
4)
C₂ ezt
†
3)
Answers
Explanation:
A) During the ionisation C2
→C
2
+
the bond order decreases from 2 to 1.5 and the magnetic behaviour changes from diamagnetic to paramagnetic.
For C2
, electronic configuration is
1s
2
1s
∗2
2s
2
2s
∗2
2px
2
2py
2
.
N
2
has maximum bond order of 3 and hence, has shortest bond length.
Bond Order =
2
4−0
=2
For C
2
+
1s
2
1s
∗2
2s
2
2s
∗2
2px
2
2py
1
Bond Order =
2
3−0
=1.5
(B) During the ionisation NO→NO
+
, the bond order
increases from 2.5 to 3 and the magnetic character changes from
paramagnetic to diamagnetic.
For NO,
1s
2
∗1s
2
2s
2
∗2s
2
2pz
2
2px
2
2py
2
∗2px
1
∗2py
0
Bond Order =
2
6−1
=2.5
For NO
+
,
1s
2
∗1s
2
2s
2
∗2s
2
2pz
2
2px
2
2py
2
Bond Order =
2
10−4
= 3
(C) During the ionisation O
2
→O
2
+
, the bond order increases from 2 to 2.5 and the magnetic character remains same as both are paramagnetic.
The electronic configuration of the O
2
+
containing 15 electrons can be written as:
1s
2
∗1s
2
2s
2
∗2s
2
2pz
2
2px
2
2py
2
∗2px
1
Bond Order =
2
8−3
=2.5
For O
2
1s
2
∗1s
2
2s
2
∗2s
2
2pz
2
2px
2
2py
2
∗2px
1
∗2py
1
Bond Order =
2
6−2
=2
(D) During the ionisation N
2
→N
2
+
, the bond order decreases from 3 to 2.5 and the magnetic character changes from diamagnetic to paramagnetic.
For N
2
, electronic configuration is
1s
2
∗1s
2
2s
2
∗2s
2
2px
2
2py
2
2pz
2
.
N
2
has maximum bond order of 3 and hence, has shortest bond length.
Bond Order =
2
6−0
= 3
For N
2
+
1s
2
∗1s
2
2s
2
∗2s
2
2px
2
2py
2
2pz
1
Bond Order =
2
5−0
= 2.5