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Chapter 9: Problem 51

Which charge(s) for the \(\mathrm{N}_{2}\) molecule would give a bond order of 2.5\(?\)

Short Answer

Expert verified
None of the charge configurations for the N₂ molecule (N₂⁺, N₂²⁺, N₂⁻, and N₂²⁻) give a bond order of 2.5. The calculated bond orders are 3.5, 2, 4.5, and 4, respectively.

Step by step solution

01

Determine nitrogen's atomic orbital configuration

Nitrogen has 7 electrons, so its atomic orbital configuration is \[1s^{2}2s^{2}2p^{3}\]. Each nitrogen atom contributes 7 electrons, and in the N₂ molecule, there will be a total of 14 electrons. Step 2: Determine the molecular orbital diagram for nitrogen
02

Construct nitrogen's molecular orbital diagram

In molecular orbital theory, atomic orbitals of the same energy level combine to form molecular orbitals. For nitrogen, we have the following molecular orbitals: - One σ1s and one σ1s* from the combination of two 1s atomic orbitals - One σ2s and one σ2s* from the combination of two 2s atomic orbitals - One σ2p, two π2p, one σ2p*, and two π2p* from the combination of six 2p atomic orbitals Step 3: Determine N₂ electronic configuration
03

Fill molecular orbitals for N₂

Filling the molecular orbitals of N₂ with 14 electrons, the electronic configuration for the neutral N₂ molecule is: \[σ_{1s}^{2}σ_{1s*}^{2}σ_{2s}^{2}σ_{2s*}^{2}σ_{2p}^{2}π_{2p}^{4}\] Step 4: Determine different charged N₂ electronic configurations
04

List electronic configurations for different charged N₂ molecules

- For N₂⁺ (13 electrons): \[σ_{1s}^{2}σ_{1s*}^{2}σ_{2s}^{2}σ_{2s*}^{2}σ_{2p}^{2}π_{2p}^{3}\] - For N₂²⁺ (12 electrons): \[σ_{1s}^{2}σ_{1s*}^{2}σ_{2s}^{2}σ_{2s*}^{2}σ_{2p}^{2}π_{2p}^{2}\] - For N₂⁻ (15 electrons): \[σ_{1s}^{2}σ_{1s*}^{2}σ_{2s}^{2}σ_{2s*}^{2}σ_{2p}^{2}π_{2p}^{5}\] - For N₂²⁻ (16 electrons): \[σ_{1s}^{2}σ_{1s*}^{2}σ_{2s}^{2}σ_{2s*}^{2}σ_{2p}^{2}π_{2p}^{6}\] Step 5: Calculate the bond order for each of these configurations
05

Calculate bond order with different charged N₂ molecules

Remember that bond order is given by the following formula: Bond Order = \(\frac{(number\:of\:bonding\:electrons - number\:of\:antibonding\:electrons)}{2}\) Let's calculate the bond order for each of the charged N₂ molecules: - For N₂⁺: \(\frac{(10-3)}{2}\) = 3.5 - For N₂²⁺: \(\frac{(8-4)}{2}\) = 2 - For N₂⁻: \(\frac{(12-3)}{2}\) = 4.5 - For N₂²⁻: \(\frac{(12-4)}{2}\) = 4 Step 6: Identify which charge(s) gives a bond order of 2.5
06

Find the correct charged N₂ molecules

As we can see from our bond order calculations, none of the charged N₂ molecules has a bond order of 2.5. Therefore, there is no charge configuration for the N₂ molecule that would give a bond order of 2.5.

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Most popular questions from this chapter

Use the localized electron model to describe the bonding in \(\mathrm{CCl}_{4}\) .

Which of the following statements concerning \(\mathrm{SO}_{2}\) is (are) true? a. The central sulfur atom is \(s p^{2}\) hybridized. b. One of the sulfur-oxygen bonds is longer than the other(s). c. The bond angles about the central sulfur atom are about \(120^{\circ} .\) d. There are two \(\sigma\) bonds in \(\mathrm{SO}_{2}\) . e. There are no resonance structures for SO.

Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some Holecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why He_ molecules can come to exist and why they dissociate.

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity $$ \text {a} C F_{4} \quad \text { e. BeH }_{2} \quad \text { i. } \operatorname{KrF}_{4} $$ $$ \text {b} \mathrm{NF}_{3} \quad \text { f. } \mathrm{TeF}_{4} \quad \text { j. SeF }_{6} $$ $$ \text {c} \mathrm{OF}_{2} \quad \text { g. AsF_ } \quad \text { k. } \mathrm{IF}_{5} $$ $$ \text {d} \mathrm{BF}_{3} \quad \text { h. } \mathrm{KrF}_{2} \quad \text { L. } \mathrm{IF}_{3} $$

Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triodide \(\left(\mathrm{N} \mathrm{I}_{3}\right)\) is known to be a highly explosive material. \(\mathrm{NI}_{3}\) can be synthesized according to the equation $$ \mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g) $$ a. What is the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) given the enthalpy of reaction \((-307 \mathrm{kJ})\) and the enthalpies of formation for $\mathrm{BN}(s)(-254 \mathrm{kJ} / \mathrm{mol}), \operatorname{IF}(g)(-96 \mathrm{kJ} / \mathrm{mol})\( and \)\mathrm{BF}_{3}(g)(-1136 \mathrm{kJ} / \mathrm{mol}) ?$ b. It is reported that when the synthesis of \(\mathrm{NI}_{3}\) is conducted using 4 moles of IF for every 1 \(\mathrm{mole}\) of \(\mathrm{BN}\) , one of the by-products isolated is \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .\) What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?
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