Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7. (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. \((\mathbf{b})\) The \(\mathrm{B}-\mathrm{N}\) bond is obviously more polar than the C-C bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4 ). (c) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(\mathrm{B}-\mathrm{N}\) bond in the trichloride reaction product would be more or less polar than the \(\mathrm{B}-\mathrm{N}\) bond in the trifluoride product, and justify your reasoning.

Short Answer

Expert verified
The reaction product of ammonia (NH3) and boron trifluoride (BF3) forms a polar B-N bond with a partial negative charge on nitrogen (Nδ-) and a partial positive charge on boron (Bδ+), represented as N(H3)δ- -- Bδ+(F3). Comparing to the trichloride reaction product, the B-N bond is less polar due to the lower electronegativity of chlorine attached to boron compared to fluorine.

Step by step solution

01

Draw the Lewis structure of ammonia-boron trifluoride reaction product

First, let's recall the Lewis structures for ammonia (NH3) and boron trifluoride (BF3). Ammonia has one lone pair and three bonded electrons, while boron trifluoride has three bonded electrons and an empty orbital. When they react, the lone pair of electrons from ammonia donates to the empty orbital of boron in BF3. The resulting Lewis structure looks like this: N(H3)--BF3 Here, the B-N bond is formed by the nitrogen atom in ammonia donating its lone pair of electrons to the boron atom in boron trifluoride.
02

Determine the polarity of the B-N bond

The B-N bond is more polar than the C-C bond because nitrogen and boron have different electronegativities, which means they have different tendencies to attract electrons. Nitrogen (with an electronegativity of 3.04) is more electronegative than boron (with an electronegativity of 2.04). Therefore, the electrons within the B-N bond are more attracted to the nitrogen atom than the boron atom, resulting in a polar bond.
03

Draw the charge distribution

Since the B-N bond is polar, and nitrogen is more electronegative than boron, the charge distribution within this bond can be represented with delta plus (δ+) and delta minus (δ-) symbols as follows: N(H3)δ- -- Bδ+ (F3) This means that there is a partial negative charge on the nitrogen atom and a partial positive charge on the boron atom in the reaction product.
04

Predict the polarity of the B-N bond in the trichloride reaction product

Boron trichloride (BCl3) is similar to boron trifluoride (BF3) in structure, having three bonded atoms and an empty orbital. It reacts with ammonia in a similar fashion, with the lone pair of nitrogen in ammonia donating to the empty orbital of boron in BCl3, forming a B-N bond. In order to determine whether the B-N bond in the trichloride product is more or less polar than in the trifluoride product, we need to consider the electronegativity of the atoms attached to the boron. Fluorine (F) has an electronegativity of about 3.98, while chlorine (Cl) has an electronegativity of about 3.16. Since fluorine is more electronegative, it will attract the electrons in the boron-fluorine bond more strongly than the chlorine in BCl3 will. As a result, the boron atom in BF3 will likely carry a stronger partial positive charge than in BCl3. Therefore, the B-N bond in the trichloride reaction product is expected to be less polar than the B-N bond in the trifluoride product. This is because the partial positive charge on boron in BCl3 is weaker due to the lower electronegativity of chlorine compared to fluorine.

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Most popular questions from this chapter

(a) Which of these compounds is an exception to the octet rule: carbon dioxide, water, ammonia, phosphorus trifluoride, or arsenic pentafluoride? (b) Which of these compounds or ions is an exception to the octet rule: borohydride \(\left(\mathrm{BH}_{4}^{-}\right),\) borazine $\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\right.$ which is analogous to benzene with alternating \(\mathrm{B}\) and \(\mathrm{N}\) in the ring \(),\) or boron trichloride?

(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Arrange the following substances not listed in Table 8.1 according to their expected lattice energies, listing them from lowest lattice energy to the highest: MgS, KI, GaN, LiBr.

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{SiF}_{4}\) and \(\mathrm{LaF}_{3}\), (b) \(\mathrm{FeCl}_{2}\) and \(\mathrm{ReCl}_{6}\), (c) \(\mathrm{PbCl}_{4}\) and RbCl.

Which ionic compound is expected to form from combining the following pairs of elements? (a) calcium and nitrogen, (b) cesium and bromine, (c) strontium and sulfur, (d) aluminum and selenium.

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround them: $(\mathbf{a}) \mathrm{HCl},(\mathbf{b}) \mathrm{ICl}_{5},\( (c) \)\mathrm{NO}\( (d) \)\mathrm{CF}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{I}_{3}^{-}$

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