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. |
| x\[NO_{2}^{+}\] 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 \[180{}^\circ \]. |
| In \[NO_{2}^{-}\] ion, N-atom undergoes \[s{{p}^{2}}\]-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\[115{}^\circ \]. |
| In \[N{{O}_{2}}\] molecule, N-atom has one unpaired electron in \[s{{p}^{2}}-\]hybrid orbital. The bond angle, should be \[120{}^\circ \] but actually it is \[132{}^\circ \]. It may be due to one unpaired electron in \[s{{p}^{2}}-\]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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