Which molecule has the highest bond order?

N_{2}

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L i_{2}

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H e_{2}

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O_{2}

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Solution

The correct option is A $N_{2}$

Bond order=3 is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. The bond order of $N_{2}$ is 3. which is Nitrogen molecule.
The molecule $L i_{2}$ is a stable molecule in the gas phase, with a bond order of one.
Bond Order=2(bonding electrons)−0(anti−bonding e−)2=1.
Bond order is the number of bonding electrons minus the number of antibonding electrons, "all over 2", hence, (2-0)/2 = 1. The bond order for $H e_{2}$ is 0.
$O_{2}$ has two unpaired electrons in its π* orbitals, and a bond order of 2.
therefore , the correct answer here is Option A. $N_{2}$

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Similar questions

C_{2}, L i_{2}, O_{2}^{+}, H e_{2}^{+}

(i) O_{2}^{+} (i i) C O (i i i) B_{2} (i v) O_{2}^{-}

(v) N O^{+} (v i) H e_{2}^{+} (v i i) C_{2}^{2 +} (v i i i) C N^{-}

(i x) N_{2}^{-}

H e_{2}

H e_{2}

N_{2}^{+}

N_{2}

O_{2}^{+}

O_{2}

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