Question

molecular orbital of ne2

Answer 1

Molecular orbital diagram of

•Neon atom has 10 electrons and its electronic configuration is considered, it has two neon atoms and thus is composed of

The electronic configuration and bond order ofElectronic configuration:

$Ne_2 = (\pi^*_{2{py}})^2(\sigma^*_{2pz})^2$

Bond order =

$\frac{1}{2}[N_b-N_a]$

=

$\frac{1}{2}[8-8] = 0$

As the bond order value forN_2, it is unstable and cannot exist.

The molecular orbital diagram of hypothetical molecule is given in the attachment.

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Answer 2

Explanation:

Formation of N

2

molecule: Electronic Configuration,

σ1s

2

<σ∗1s

2

<σ2s

2

<σ∗2s

2

<[π2p

x

2

=π2p

x

2

]<<σ2p

z

2

Bond order = (N

b

−N

a

)/2=(10−4)/2=3

Bond order indicates the number of bonds in diatomic molecule is 3, hence TripleBond

Formation of F

2

molecule: Electronic Configuration:

σ1s

2

<σ∗1s

2

<σ2s

2

<σ∗2s

z

2

σ∗2pz

2

<[π2p

x

2

=π2p

x

2

]<[π∗2p

x

2

=π∗2p

x

2

]<σ2p

z

Bond order = (N

b

−N

a

)/2=(10−8)/2=1

Single bond F−F.

Formation of Ne

2

molecule: Electronic Configuration,

σ1s

2

<σ∗1s

2

<σ2s

2

<σ∗2s

z

2

<σ∗2pz

2

<[π2p

x

2

=π2p

x

2

]<[π∗2p

x

2

=π∗2p

x

2

]<σ2p

z

2

Bond order = (N

b

−N

a

)/2=(10−10)/2=0

Hence, no bond between Ne atoms