Problem 30 - Chemistry The Molecular Science

Question

For each pair of bonds, predict which is the shorter. (a) \(\mathrm{Si}-\mathrm{N}\) or \(\mathrm{P}-\mathrm{O}\) (b) \(\mathrm{Si}-\mathrm{O}\) or \(\mathrm{C}-\mathrm{O}\) (c) \(\mathrm{C}-\mathrm{F}\) or \(\mathrm{C}-\mathrm{Br}\) (d) The \(\mathrm{C}=\mathrm{C}\) or the \(\mathrm{C} \equiv \mathrm{N}\) bond in acrylonitrile, \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{N}\)

Step-by-Step Solution

Answer

(a) P-O; (b) C-O; (c) C-F; (d) C≡N.

Step 1: Analyze Periodicity and Bond Length

In general, bond length decreases across a period from left to right on the periodic table, and increases down a group. This is because atomic size decreases across a period and increases down a group.

Step 2: Apply Periodicity to Si-N and P-O Bonds

For pair (a), silicon (Si) and nitrogen (N) are in Period 3 and Period 2, respectively, in rows 3 and 2 of the periodic table. Phosphorus (P) is in Period 3 and oxygen (O) is in Period 2. Since both bonds have elements from the same rows, compare the average bond distances. P-O typically forms shorter bonds compared to Si-N due to smaller atomic size of oxygen relative to nitrogen. Hence, P-O is shorter.

Step 3: Apply Periodicity to Si-O and C-O Bonds

For pair (b), silicon (Si) is larger than carbon (C) since it is below carbon in the periodic table (Period 3 vs Period 2). When bonded to oxygen (O), which is in the same period for both, C-O bond will be shorter as carbon is smaller and forms shorter, stronger bonds than silicon.

Step 4: Compare C-F and C-Br Bonds

For pair (c), compare fluorine (F) and bromine (Br). Fluorine is smaller and in Period 2, whereas bromine is larger and in Period 4. Hence, the C-F bond is shorter than C-Br because fluorine's smaller atomic radius results in a shorter bond.

Step 5: Compare Bond Orders for C=C and C≡N Bonds

For pair (d), higher bond order usually results in a shorter bond. A C=C double bond has a bond order of 2, while a C≡N triple bond has a bond order of 3. Therefore, the C≡N bond is expected to be shorter than the C=C bond due to the higher bond order.

Key Concepts

Periodicity in Chemistry

Chemical elements are arranged in the periodic table in such a way that their properties exhibit trends or periodic variations. This is known as periodicity. One of these trends is atomic size, which is key in predicting bond lengths. In general, as you move from left to right across a period, atoms become progressively smaller. This is because, while the number of protons (and thus nuclear charge) increases, the electrons are being added to the same energy level without an increase in screening. This stronger pull shrinks the atomic radius.

• **Across a Period:** Atomic size decreases.
• **Down a Group:** Atomic size increases.

When you understand how size changes within the periodic table, you can predict how this affects bond lengths, such as looking at silicon–nitrogen compared to phosphorus–oxygen or carbon–fluorine compared to carbon–bromine. Knowing which atoms are smaller helps in predicting shorter bond lengths.

Bond Order

Bond order is a concept from molecular orbital theory that describes the number of chemical bonds between a pair of atoms. Essentially, it's a count of how many electron pairs are shared between the atoms. Higher bond orders mean stronger and usually shorter bonds. This is key when evaluating pairs of bonds, as a bond with more shared electrons will be tighter and closer together.

• **Single Bonds:** Bond order of 1 (e.g., C-C).
• **Double Bonds:** Bond order of 2 (e.g., C=C).
• **Triple Bonds:** Bond order of 3 (e.g., C≡C or C≡N).

In exercises, you'll often determine that a higher bond order, like the triple bond in carbon–nitrogen (C≡N), results in a significantly shorter bond than a double bond, evidenced in compounds like acrylonitrile.

Atomic Size and Bond Length

The size of an atom greatly influences the length of the bond it forms. Atomic size determines how close the bonded atoms will be in space. Larger atoms, like bromine compared to fluorine, will inevitably form longer bonds because their larger atomic radii push the nuclei of bonded atoms further apart.

• **Larger Atoms:** Longer bond lengths.
• **Smaller Atoms:** Shorter bond lengths.

This principle is essential in determining bond lengths in comparisons between elements like silicon–oxygen and carbon–oxygen. Despite both silicon and carbon bonding to oxygen, the smaller atomic size of carbon results in a generally shorter bond, illustrating how crucial atomic size is in predicting bond lengths across different elements in chemistry.