According to molecular orbital theory,
According to molecular orbital theory,
A
`C_(2)^(2+)` is expected to be diamagnetic
B
`O_(2)^(2+)`is expected to have a smaller bond length than `O_(2)`
C
`N_(2)^(2+) and N_(2)^(-)` have the same bond order.
D
`He_(2)^(+)` has the same energy as two isolated He atoms.
Text Solution
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To solve the question regarding molecular orbital theory step by step, we will analyze the options provided in the question, focusing on the molecular configurations and bond orders of the species mentioned.
### Step 1: Analyze C2 2+
1. **Determine the number of electrons**:
- Carbon (C) has 6 electrons. For C2, the total is 12 electrons. Since we have C2 2+, we remove 2 electrons, leaving us with 10 electrons.
2. **Write the molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x¹, π2p_y¹ (where π2p_x and π2p_y are degenerate orbitals).
3. **Identify the highest occupied molecular orbital (HOMO)**:
- The HOMO is π2p_x and π2p_y, each containing 1 electron.
4. **Check for unpaired electrons**:
- There are 2 unpaired electrons (1 in each π orbital), indicating that C2 2+ is **paramagnetic**.
**Conclusion for Step 1**: The statement that C2 2+ is expected to be diamagnetic is incorrect.
### Step 2: Analyze O2 2+
1. **Determine the number of electrons**:
- Oxygen (O) has 8 electrons. For O2, the total is 16 electrons. For O2 2+, we remove 2 electrons, leaving us with 14 electrons.
2. **Write the molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x², π2p_y².
3. **Calculate the bond order**:
- Bond order = 1/2 (bonding electrons - antibonding electrons).
- Bonding electrons = 10 (σ1s², σ2s², σ2p_z², π2p_x², π2p_y²).
- Antibonding electrons = 4 (σ*1s², σ*2s²).
- Bond order = 1/2 (10 - 4) = 3.
4. **Compare bond lengths**:
- O2 has a bond order of 2 (calculated similarly), which is less than O2 2+.
- Higher bond order corresponds to a shorter bond length.
**Conclusion for Step 2**: The statement that O2 2+ is expected to have a smaller bond length than O2 is correct.
### Step 3: Analyze N2 2+ and N2 -
1. **Determine the number of electrons**:
- For N2 2+, we have 10 electrons (N2 has 14 electrons, removing 4).
- For N2 -, we have 16 electrons (adding 2 to N2).
2. **Write the molecular orbital configurations**:
- N2 2+: σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y².
- N2 -: σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x², π2p_y², π*2p_x¹.
3. **Calculate the bond orders**:
- N2 2+: Bond order = 1/2 (8 - 4) = 2.
- N2 -: Bond order = 1/2 (10 - 5) = 2.5.
**Conclusion for Step 3**: The statement that N2 2+ and N2 - have the same bond order is incorrect.
### Step 4: Analyze He2 + and He2
1. **Determine the number of electrons**:
- He2 has 4 electrons, while He2 + has 3 electrons.
2. **Write the molecular orbital configurations**:
- He2: σ1s², σ*1s².
- He2 +: σ1s², σ*1s¹.
3. **Calculate the bond orders**:
- He2: Bond order = 1/2 (2 - 2) = 0 (unstable).
- He2 +: Bond order = 1/2 (2 - 1) = 0.5 (stable).
**Conclusion for Step 4**: The statement that He2 + has the same energy as two isolated He2 atoms is incorrect.
### Final Conclusion:
- The only correct statement is that O2 2+ is expected to have a smaller bond length than O2. All other statements are incorrect.
---
To solve the question regarding molecular orbital theory step by step, we will analyze the options provided in the question, focusing on the molecular configurations and bond orders of the species mentioned.
### Step 1: Analyze C2 2+
1. **Determine the number of electrons**:
- Carbon (C) has 6 electrons. For C2, the total is 12 electrons. Since we have C2 2+, we remove 2 electrons, leaving us with 10 electrons.
2. **Write the molecular orbital configuration**:
- The configuration is:
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Knowledge Check
Which of the following statement is not correct from the view point of molecular orbital theory?
Which of the following statement is not correct from the view point of molecular orbital theory?
A
`Be_(2)` is not a stable molecule .
B
`He_(2)` is not stable of `N_(2)` is maximum amongst the homonuclear diatomic molecules belonging to the second period .
C
Bond strength of `N_(2)` is maximum amongst the homonuclear diatomic molecules belonging to the second period.
D
The order of energies of molecular orbitals in `N_(2)` molecule is ` sigma 2s lt sigma^(***) 2s lt sigma2p_(z) lt (pi 2p_(x) = pi 2p_(y)) lt (pi^(***) 2p_(x) = pi^(***) 2p_(y)) lt sigma^(***) 2p_(z)`
Which of the following statements is not true regarding molecular orbital theory?
Which of the following statements is not true regarding molecular orbital theory?
A
The atomic orbitals of comparable energies combine to form molecular orbitals.
B
An atomic orbital is monocentric while a molecular orbital is polycentric.
C
Bonding molecular orbital has higher energy than antibonding molecular orbital.
D
Molecular orbitals like atomic orbitals obey Aufbau principle for filling of electrons.
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According to the molecular orbital theory which of the following statement is incorrect? [LUMO = lowest unoccupied molecular orbital]
According to the moleular orbital theory, all atomic orbitals combine to form molecular orbital by LCAO (linear combination of atomic orbitals) method When two atomic orbitals have additive (constructive) method When two atomic orbitals have additive (constructive) overlapping they form bonding molecular orbitals (BMO) which have lower energy than atomic orbitals whereas when atomic orbitals overlap subtractive higher energy antibonding molecular orbitals (ABMO) are formed Each MO occupies two electrons with opposite spin Distribution of electrons in MO follows Aufbau principle as well as Hund's rule MO theory can successfully explain the magnetic behaviour of molecules Bond strength increases when
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