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Question: Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT...

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

A

Bond order of Ne₂ is zero.

B

The highest occupied molecular orbital (HOMO) of F₂ is σ-type.

C

Bond energy of O₂⁺ is smaller than the bond energy of O₂.

D

Bond length of Li₂ is larger than the bond length of B₂.

Answer

(B), (C)

Explanation

Solution

Rationale for each statement:

(A) Bond order of Ne₂ is zero. Ne has 10 electrons, so Ne₂ has 20 electrons. The molecular orbital configuration for molecules with > 14 electrons is (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(σ2pz)2(π2px)2(π2py)2(π2px∗)2(π2py∗)2(σ2pz∗)2(\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}^*)^2 (\sigma_{2p_z}^*)^2.

Number of bonding electrons (NbN_b) = 2 + 2 + 2 + 2 + 2 = 10. Number of antibonding electrons (NaN_a) = 2 + 2 + 2 + 2 + 2 = 10. Bond order = (Nb−Na)/2=(10−10)/2=0(N_b - N_a) / 2 = (10 - 10) / 2 = 0.

A bond order of zero means the molecule is unstable and does not exist. Statement (A) is correct.

(B) The highest occupied molecular orbital (HOMO) of F₂ is σ-type. F₂ has 18 electrons. Using the MO configuration for > 14 electrons: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(σ2pz)2(π2px)2(π2py)2(π2px∗)2(π2py∗)2(\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}^*)^2.

The last electrons are filled into the π2px∗\pi_{2p_x}^* and π2py∗\pi_{2p_y}^* molecular orbitals. These are the highest occupied molecular orbitals (HOMO). These orbitals are π\pi-type, specifically antibonding π\pi orbitals.

Statement (B) says the HOMO of F₂ is σ-type. This is incorrect.

(C) Bond energy of O₂⁺ is smaller than the bond energy of O₂. O₂ has 16 electrons. MO configuration: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(σ2pz)2(π2px)2(π2py)2(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. Nb=2+2+2+2+2=10N_b = 2+2+2+2+2 = 10. Na=2+2+1+1=6N_a = 2+2+1+1 = 6. Bond order of O₂ = (10−6)/2=2(10 - 6) / 2 = 2.

O₂⁺ is formed by removing one electron from O₂, which comes from the highest occupied antibonding π∗\pi^* orbital. O₂⁺ has 15 electrons. MO configuration: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(σ2pz)2(π2px)2(π2py)2(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 (or vice versa). Nb=10N_b = 10. Na=5N_a = 5. Bond order of O₂⁺ = (10−5)/2=2.5(10 - 5) / 2 = 2.5.

Bond energy is generally proportional to bond order. Higher bond order means stronger bond and higher bond energy. Since the bond order of O₂⁺ (2.5) is greater than the bond order of O₂ (2), the bond energy of O₂⁺ is larger than the bond energy of O₂.

Statement (C) says the bond energy of O₂⁺ is smaller than the bond energy of O₂. This is incorrect.

(D) Bond length of Li₂ is larger than the bond length of B₂. Bond length is generally inversely proportional to bond order.

Li₂ has 6 electrons. MO configuration (for <= 14 electrons): (σ1s)2(σ1s∗)2(σ2s)2(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2. Nb=2+2=4N_b = 2+2 = 4. Na=2N_a = 2. Bond order of Li₂ = (4−2)/2=1(4 - 2) / 2 = 1.

B₂ has 10 electrons. MO configuration (for <= 14 electrons): (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2px)1(π2py)1(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p_x})^1 (\pi_{2p_y})^1. Nb=2+2+1+1=6N_b = 2+2+1+1 = 6. Na=2+2=4N_a = 2+2 = 4. Bond order of B₂ = (6−4)/2=1(6 - 4) / 2 = 1.

Both Li₂ and B₂ have a bond order of 1. When bond orders are equal, bond length is primarily influenced by atomic size. Li is in Period 2, Group 1, and B is in Period 2, Group 13. Atomic radius decreases across a period. Li atom is larger than B atom. Larger atoms form longer bonds.

Therefore, the bond length of Li₂ is larger than the bond length of B₂. Statement (D) is correct.

The incorrect statements are (B) and (C).