Nitrogen forms a number of oxides. One of the important oxides of nitrogen is NO2, a red-brown color — Physical Chemistry Chemistry Question
Compounds of nitrogen
Nitrogen forms a number of oxides. One of the important oxides of nitrogen is NO2, a red-brown colored reactive gas.
Draw the Lewis structure of NO2 and predict its shape using valence shell electron pair repulsion theory.
Model Answer
15.1 NO2: The number of electrons in the valence shell around the nitrogen atom = 5 + 0 + 2 = 7
The Lewis structure for NO2 is as shown below.
[VISUAL]
According to VSEPR, the molecule ideally should have linear geometry. However, this molecule has one single unpaired electron present on nitrogen. Due to the repulsion between the unpaired electron and the other two bonded pairs of electrons, the observed bond angle is less than 180° (132°). Thus, the shape of the molecule is angular as shown below.
[VISUAL]
Using VSEPR, predict the shapes of the NO2- and NO2+ ions. Compare the shapes of these two ions with that of NO2.
Model Answer
15.2 NO2+: The number of electrons in the valence shell around nitrogen atom = (5 + 2 + 2 – 1) = 8
The Lewis structure is as shown below
[VISUAL]
Thus, there are no non-bonded electrons present on nitrogen. The two σ-bonds will prefer to stay at 180° to minimize repulsion between bonded electron pairs giving a linear geometry (180°). The π-bonds do not influence the shape.
[VISUAL]
NO2-: The number of electrons in the valence shell around the nitrogen atom = 5 + 2 + 1 = 8
The Lewis structure is as shown below.
[VISUAL]
In case of anion NO2-, there is a lone pair of electrons present on the nitrogen atom. Due to strong repulsion between the lone pair of electrons and the bonded pairs of electrons the angle between the two bond pairs shrinks from the ideal 120° to 115°.
[VISUAL]
Consider two other compounds of nitrogen, trimethylamine (Me3N) and trisilylamine (H3Si)3N. The observed bond angles at nitrogen in these compounds are 108° and 120° respectively. Explain the difference in the bond angles.
Model Answer
15.3 In case of trimethylamine, the shape of the molecule is pyramidal with a lone pair present on nitrogen. Due to the lone pair Me-N-Me angle is reduced from 109°4’ to 108°.
[VISUAL]
However, in case of trisilylamine, d orbital of silicon and p orbital of nitrogen overlaps giving double bond character to the N-Si bond. Thus, delocalisation of the lone electron pair of the nitrogen atom takes place and the resultant molecule is planar with 120° bond angle.
[VISUAL]
Both nitrogen and boron form trifluorides. The bond energy in BF3 is 646 kJ mol-1 and that in NF3 is only 280 kJ mol-1. Account for the difference in bond energies.
Model Answer
15.4 Both N and B are tricovalent. However, NF3 is pyramidal in shape. In case of BF3, the B-F bond gets double bond character due to the overlapping of p orbitals present on boron and fluorine. The observed bond energy is, therefore, much greater in BF3.
[VISUAL]
The boiling point of NF3 is –129 °C while that of NH3 is –33 °C. Ammonia acts as a Lewis base whereas NF3 does not. The observed dipole moment of NF3 (0.24 D) is much less than that of NH3 (1.46 D), even though fluorine is much more electronegative than hydrogen. Explain the differences between boiling points and basicities of NF3 and NH3. Account for the low dipole moment of NF3.
Model Answer
15.5 The difference in boiling points of NF3 and NH3 is due to hydrogen bonding which is present in ammonia.
High electronegativity of fluorine decreases the basicity of nitrogen in NF3. Thus, NF3 does not act as a Lewis base.
In NF3, the unshared pair of electrons contributes to a dipole moment in the direction opposite to that of the net dipole moment of the N-F bonds. See figure (a). [VISUAL]
In NH3, the net dipole moment of the N-H bonds and the dipole moment due to the unshared pair of electrons are in the same direction. See figure (b). [VISUAL]
The reaction of aqueous sodium nitrate with sodium amalgam as well as that of ethyl nitrite with hydroxylamine in presence of sodium ethoxide give the same product. This product is the salt of a weak unstable acid of nitrogen. Identify the acid and write down its structure. This acid isomerises into a product, which finds use in propellant formulations. Write the structure of the isomer.
Model Answer
15.6 NaNO3 + 8 Na(Hg) + 4 H2O → Na2N2O2 + 8 NaOH + 8 Hg
NH2OH + EtNO2 + 2 NaOEt → Na2N2O2 + 3 EtOH
Na2N2O2 is the salt of H2N2O2 (Hyponitrous acid).
Structure: [VISUAL]
Isomer is: H2N—NO2 (Nitramide)
[VISUAL]