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According to molecular orbital theory, the number of unpaired electron(s) in O 2 2 −  is :

 

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago

    O 2 − 2 = 8 × 2 + 2 = 1 8 e −

    σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 σ 2 p z 2 π 2 p x 2 = π 2 p y 2 π 2 p x * 2 = π 2 p y * 2      

    Number of unpaired e- = 0

    Ans. = 0

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Bonding in which of the following diatonic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

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Bond strength ∝  Bond order

N 2 → σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 2 ≡ π 2 p y 2 σ 2 p z 2  

O 2 → σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 σ 2 p z 2 π 2 p x 2 ≡ π 2 p y 2 π 2 p x * 1 ≡ π 2 p y * 1

C 2 → σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 2 ≡ π 2 p y 2  

B 2 → σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 1 ≡ π 2 p y 1  

NO ® Number of electron = 7 + 8 = 15

B.O. Similar to  N 2 −  

N 2 − → σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 2 ≡ π 2 p y 2 σ 2 p z 2 π 2 p x * 1 ≡ π 2 p y *  

B.O. of N2 = 3     B.O of C2 = 8 − 4 2 = 2  

Removal of e- form antibonding molecular orbital increases bond order.

In NO & O2 has valance e- in p orbital.

Which amongst the following is incorrect statement?
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O? (15) will have configuration σ1s²σ1s²σ2s²σ2s²σ2p? ² (π2p? ²=π2p? ²) (π*2p? ¹). This ion is paramagnetic.

The difference between bond orders of CO and NO⁺ is x/2 where x is _______. (Round off to the Nearest integer)
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B.O. of CO = 3
B.O. of NO? = 3
Both are isoelectronic
So difference = 0
∴ x = 0

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