B 2 molecule will be diamagnetic when : (I) s − p mixing is NOT operative (II) s − p mixing is operative and Hunds rule is violated (III) s − p mixing is operative and Hunds rule is NOT violated

Question

B 2 ​ molecule will be diamagnetic when : (I) s − p mixing is NOT operative (II) s − p mixing is operative and Hunds rule is violated (III) s − p mixing is operative and Hunds rule is NOT violated

TestHub · Accepted Answer

The B₂ molecule has 10 electrons.

(I) If s-p mixing is NOT operative, the MO configuration is $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p}^{4}$ . All electrons are paired, making B₂ diamagnetic.

(III) If s-p mixing IS operative, the MO configuration is $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p}^{2} σ_{2 p}^{0}$ . According to Hund's rule, the two electrons in the $π_{2 p}$ orbitals will be unpaired, making B₂ paramagnetic.

(II) If Hund's rule is violated , the electrons could be paired making it diamagnetic.

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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