Use molecular orbital theory to explain why the Be2 molecule does not exist. Write the molecular orbital electronic configuration of Be2 molecule. Also calculate the bond order in Be2 molecule. Is Be2 a paramagnetic molecule? Explain your answer
Answers
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration of beryllium will be,
As there are 4 electrons present in beryllium.
The number of electrons present in molecule = 2(4) = 8
The molecular orbital configuration of molecule will be,
The number of unpaired electron in molecule is, 0. So, this is diamagnetic.
The formula of bonding order =
The bond order of =
The bond order of is, zero. So,
will not be exist.
Answer:
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration of beryllium will be,
(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)(σ
1s
),(σ
1s
∗
),(σ
2s
),(σ
2s
∗
),[(π
2p
x
)=(π
2p
y
)],(σ
2p
z
),[(π
2p
x
∗
)=(π
2p
y
∗
)],(σ
2p
z
∗
)
As there are 4 electrons present in beryllium.
The number of electrons present in Be_2Be
2
molecule = 2(4) = 8
The molecular orbital configuration of Be_2Be
2
molecule will be,
(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^0=(\pi_{2p_y})^0],(\sigma_{2p_z})^0,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0(σ
1s
)
2
,(σ
1s
∗
)
2
,(σ
2s
)
2
,(σ
2s
∗
)
2
,[(π
2p
x
)
0
=(π
2p
y
)
0
],(σ
2p
z
)
0
,[(π
2p
x
∗
)
0
=(π
2p
y
∗
)
0
],(σ
2p
z
∗
)
0
The number of unpaired electron in Be_2Be
2
molecule is, 0. So, this is diamagnetic.
The formula of bonding order = \frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})
2
1
×(Number of bonding electrons−Number of anti-bonding electrons)
The bond order of Be_2Be
2
= \frac{1}{2}\times (4-4)=0
2
1
×(4−4)=0
The bond order of Be_2Be
2
is, zero. So, Be_2Be
2
will not be exist.