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Question

(B) To determine which oxide contains an odd electron,we calculate the total number of valence electrons for each molecule:$1$. $N_{2}O$: $(2 	imes 5

Vedclass · Accepted Answer

$NO_{2}$

Vedclass · Answer

(B) To determine which oxide contains an odd electron,we calculate the total number of valence electrons for each molecule:$1$. $N_{2}O$: $(2 	imes 5) + 6 = 16$ valence electrons (even).$2$. $NO_{2}$: $5 + (2 	imes 6) = 17$ valence electrons (odd).$3$. $N_{2}O_{3}$: $(2 	imes 5) + (3 	imes 6) = 28$ valence electrons (even).$4$. $N_{2}O_{5}$: $(2 	imes 5) + (5 	imes 6) = 40$ valence electrons (even).Since $NO_{2}$ has an odd number of valence electrons $(17)$,it is a paramagnetic molecule with an odd electron on the nitrogen atom.

The oxide which contains an odd electron at the nitrogen atom is ..... .

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