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Question: Find the sum of bond order in that given species which have fractional bond order. NO, O$_2^-$, N$_...

Find the sum of bond order in that given species which have fractional bond order.

NO, O2−_2^-, N2_2, H2+_2^+, CO, C2−2_2^{-2}, O2_2, CO3−2_3^{-2}

Answer

8.5

Explanation

Solution

To solve this problem, we need to calculate the bond order for each given species and then sum the bond orders of only those species which have a fractional bond order.

We will use Molecular Orbital Theory (MOT) for diatomic species and resonance structures for the polyatomic species (CO3−2_3^{-2}).

  1. NO (Nitric Oxide):
    Total electrons = 7 (N) + 8 (O) = 15
    MOT configuration: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗1=π2py∗0)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*1} = \pi_{2p_y}^{*0})
    Bond order (BO) = (Number of bonding electrons - Number of antibonding electrons) / 2
    Bonding electrons = 2 + 2 + 2 + 4 = 10
    Antibonding electrons = 2 + 2 + 1 = 5
    BO(NO) = (10 - 5) / 2 = 2.5. This is a fractional bond order.

  2. O2−_2^- (Superoxide ion):
    Total electrons = 8 (O) + 8 (O) + 1 (charge) = 17
    MOT configuration: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗2=π2py∗1)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*1})
    Bonding electrons = 2 + 2 + 2 + 4 = 10
    Antibonding electrons = 2 + 2 + 3 = 7
    BO(O2−_2^-) = (10 - 7) / 2 = 1.5. This is a fractional bond order.

  3. N2_2 (Nitrogen molecule):
    Total electrons = 7 (N) + 7 (N) = 14
    MOT configuration: σ1s2σ1s∗2σ2s2σ2s∗2(π2px2=π2py2)σ2pz2\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^2 = \pi_{2p_y}^2) \sigma_{2p_z}^2
    Bonding electrons = 2 + 2 + 4 + 2 = 10
    Antibonding electrons = 2 + 2 = 4
    BO(N2_2) = (10 - 4) / 2 = 3. This is an integer bond order.

  4. H2+_2^+ (Hydrogen molecular ion):
    Total electrons = 1 (H) + 1 (H) - 1 (charge) = 1
    MOT configuration: σ1s1\sigma_{1s}^1
    Bonding electrons = 1
    Antibonding electrons = 0
    BO(H2+_2^+) = (1 - 0) / 2 = 0.5. This is a fractional bond order.

  5. CO (Carbon Monoxide):
    Total electrons = 6 (C) + 8 (O) = 14
    CO is isoelectronic with N2_2.
    BO(CO) = 3. This is an integer bond order.

  6. C2−2_2^{-2} (Acetylide ion):
    Total electrons = 6 (C) + 6 (C) + 2 (charge) = 14
    C2−2_2^{-2} is isoelectronic with N2_2.
    BO(C2−2_2^{-2}) = 3. This is an integer bond order.

  7. O2_2 (Oxygen molecule):
    Total electrons = 8 (O) + 8 (O) = 16
    MOT configuration: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗1=π2py∗1)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1})
    Bonding electrons = 2 + 2 + 2 + 4 = 10
    Antibonding electrons = 2 + 2 + 2 = 6
    BO(O2_2) = (10 - 6) / 2 = 2. This is an integer bond order.

  8. CO3−2_3^{-2} (Carbonate ion):
    This is a polyatomic ion with resonance. The Lewis structure has carbon as the central atom bonded to three oxygen atoms. There are three resonance structures, where a double bond is delocalized over the three C-O bonds.
    Total valence electrons = 4 (C) + 3 * 6 (O) + 2 (charge) = 4 + 18 + 2 = 24.
    In the resonance hybrid, each C-O bond is equivalent.
    Bond order of each C-O bond = (Total number of bonds in all resonance structures) / (Number of equivalent bond positions)
    In the three resonance structures, there is one C=O bond (bond order 2) and two C-O single bonds (bond order 1) in each structure.
    Total bond count over the three positions across all three structures = (2 + 1 + 1) * 3 / 3 = 4. Or simply, sum of bond orders in one structure = 2 + 1 + 1 = 4. Number of bonds = 3.
    Average BO(C-O) = 4 / 3. This is a fractional bond order.

The species with fractional bond order are:

  • NO (BO = 2.5)
  • O2−_2^- (BO = 1.5)
  • H2+_2^+ (BO = 0.5)
  • CO3−2_3^{-2} (average C-O BO = 4/3)

The question asks for the sum of bond order in that given species which have fractional bond order. For diatomic species, the "bond order in the species" is the calculated bond order. For polyatomic species with resonance, the "bond order in the species" could refer to the sum of the bond orders of all bonds within the species, or the average bond order. Given the context of summing values, the most reasonable interpretation is to sum the total bond contribution within each identified species.
For diatomic species, this is simply the bond order.
For CO3−2_3^{-2}, there are three C-O bonds, each with an average bond order of 4/3. The total bond order contribution from all bonds in the species is 3 * (4/3) = 4.

Sum of bond orders = BO(NO) + BO(O2−_2^-) + BO(H2+_2^+) + (Sum of BO in CO3−2_3^{-2})
Sum = 2.5 + 1.5 + 0.5 + 4
Sum = 4.0 + 0.5 + 4
Sum = 4.5 + 4
Sum = 8.5

The final answer is 8.5\boxed{8.5}.

Explanation of the solution:

  1. Calculate the bond order for each species using MOT for diatomic species and resonance theory for polyatomic species.
  2. Identify the species with fractional bond order: NO (2.5), O2−_2^- (1.5), H2+_2^+ (0.5), CO3−2_3^{-2} (average C-O BO = 4/3).
  3. For each species with fractional bond order, determine the total bond order within the species. For diatomic species, this is the calculated bond order. For CO3−2_3^{-2}, this is the sum of the bond orders of all C-O bonds (3 * 4/3 = 4).
  4. Sum these total bond orders: 2.5 + 1.5 + 0.5 + 4 = 8.5.

The final answer is 8.5\boxed{8.5}.