Main axis of a diatomic molecule is , molecular orbital p_x and p_y overlap to form which of the following orbitals.

Option 1 - π molecular orbital

Option 2 - 0 molecular orbital

Option 3 -  6 molecular orbitals

Option 4 - No bond will form.

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Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

For overlap, the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

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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?

(A) NO

(B) N2

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(E) B2

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Bond strength $\propto$ Bond order

$N_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{2}$

$O_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{* 1}$

$C_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2}$

$B_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{1} \equiv π_{2 p y}^{1}$

NO ® Number of electron = 7 + 8 = 15

B.O. Similar to $N_{2}^{-}$

$N_{2}^{-} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{*}$