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Chemistry Question on Chemical bonding and molecular structure

Which one of the following species does not exist under normal conditions?

A

Be2+Be_2^+

B

Be2Be_2

C

B2B_2

D

Li2Li_2

Answer

Be2Be_2

Explanation

Solution

Molecules with zero bond order, do not exist. According to molecular orbital theory,
(a) Be2+(4+4−1=7)Be_2^ + (4+4-1=7)
             =σ1s2,σ∗1s2,σ2s2,σ∗2s1\, \, \, \, \, \, \, \, \, \, \, \, \, = \sigma 1s^2, _{\sigma}^{*} 1s^2, \sigma 2s^2,_{\sigma}^{*} 2s^1
Bond order (BO) = 4−32=0.5 \frac{4-3}{2}=0.5
(b) Be2(4+5=8)=σ1s2,σ∗1s2,σ∗1s2,σ2s2,σ∗2s2Be_2(4+5=8) = \sigma 1s^2 , _{\sigma}^{*} 1s^2, _{\sigma}^{*} 1s^2, \sigma 2s^2, _{\sigma}^{*} 2s^2
        BO=4−42=0\, \, \, \, \, \, \, \, BO = \frac{ 4-4}{2}= 0
(c) B2(5+5=10)B_2 (5+5=10)
     =σ1s2,σ∗1s2,σ2s2,σ∗2s2,π2PX1≈π2PY1\, \, \, \, \, = \sigma 1s^2 , _{\sigma}^{*} 1s^2, \sigma 2s^2 ,_{\sigma}^{*} 2s^2,\pi 2 P_X^1 \approx \pi 2 P_Y^1
Bond order (BO)= 6−42=1\frac{6-4}{2} =1
(d)Li2(3+3=6)=σ1s2,σ∗1s2,σ2s2 Li_2(3+3=6) = \sigma1s^2,_{\sigma}^{*} 1s^2,\sigma2s^2
       BO=4−22=1\, \, \, \, \, \, \, BO = \frac{4-2}{2} =1
Thus, Be2Be_2 does not exist under normal conditions.