A) \[NO_{2}^{-}<NO_{2}^{+}<N{{O}_{2}}\]
B) \[NO_{2}^{-}<N{{O}_{2}}<NO_{2}^{+}\]
C) \[NO_{2}^{+}<N{{O}_{2}}<NO_{2}^{-}\]
D) \[NO_{2}^{+}<NO_{2}^{-}<N{{O}_{2}}\]
Correct Answer: B
Solution :
Key Idea: As the number of lone pair of electrons increases, bond angle decreases. \[NO_{2}^{+}\] ion is isoelectronic with \[C{{O}_{2}}\] molecule. It is a linear ion and its central atom \[({{N}^{+}})\]undergoes sp-hybridisation, hence bond angle is 180°. In \[NO_{2}^{-}\]ion, N-atom undergoes sp2- hybridisation. The angle between hybrid orbital should be \[{{120}^{o}}\]but one lone pair of electrons is lying on N-atom, hence bond angle decreases to 115°. In \[N{{O}_{2}}\]molecule, N-atom has one unpaired electron in \[\text{s}{{\text{p}}^{\text{2}}}\text{-}\]hybrid orbital. The bond angle, should be 120° but actually it is 132°. It may be due to one unpaired electron in \[\text{s}{{\text{p}}^{\text{2}}}\text{-}\]hybrid orbital. Therefore, the increasing order of bond angles is:\[\underset{({{115}^{o}})}{\mathop{NO_{2}^{-}}}\,<\underset{({{132}^{o}})}{\mathop{NO_{2}^{{}}}}\,<\underset{({{180}^{o}})}{\mathop{NO_{2}^{+}}}\,\]
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