Let's break down the solution step-by-step for drawing and labelling the **molecular orbital energy level diagram for \( \text{KrBr}^+ \)**. --- ## Step 1: **Determine the Atomic Configuration** ### **Krypton (Kr)** - Atomic number = 36 - Electron configuration: [Ar] 3d¹⁰ 4s² 4p⁶ ### **Bromine (Br)** - Atomic number = 35 - Electron configuration: [Ar] 3d¹⁰ 4s² 4p⁵ --- ## Step 2: **Identify the Valence Orbitals** Both **Kr** and **Br** are in period 4. The relevant valence orbitals are: - **4s** and **4p** But in MO diagrams for diatomics, only the outermost s and p orbitals are considered (4s, 4p). --- ## Step 3: **Count the Valence Electrons** - **Kr:** 8 valence electrons (4s² 4p⁶) - **Br:** 7 valence electrons (4s² 4p⁵) **Total (neutral KrBr):** 8 + 7 = 15 **But this is KrBr⁺,** so remove one electron: **Total = 15 – 1 = 14 valence electrons** --- ## Step 4: **Order of Orbital Energies** For diatomics with atomic number > 8, the order of energy levels for p-block elements is: - \(\sigma_{4s}\) - \(\sigma_{4s}^*\) - \(\sigma_{4p_z}\) - \(\pi_{4p_x} = \pi_{4p_y}\) - \(\pi_{4p_x}^* = \pi_{4p_y}^*\) - \(\sigma_{4p_z}^*\) But, since Kr and Br are heavy (Z>8), **the σ(4p_z) is lower than the π(4p_x), π(4p_y)**. --- ## Step 5: **Draw the MO Diagram** Let's focus on **the 4s and 4p orbitals**: ``` Kr (4s, 4p) MO Levels Br (4s, 4p) | | |σ4s (bonding) | | | | | |σ4s* (antibonding)| | | | | |σ4p_z (bonding) | | | | |π4p_x (bonding) | | | | |π4p_y (bonding) | | | | |π4p_x* (antibonding)| | | | |π4p_y* (antibonding)| | | | |σ4p_z* (antibonding)| | | ``` ### **Label the Energies (from bottom to top):** 1. **σ(4s)** 2. **σ*(4s)** 3. **σ(4p_z)** 4. **π(4p_x) = π(4p_y)** 5. **π*(4p_x) = π*(4p_y)** 6. **σ*(4p_z)** --- ## Step 6: **Fill the Molecular Orbitals** We have **14 electrons** to fill in the following order: 1. **σ(4s)** : 2 electrons 2. **σ*(4s)** : 2 electrons 3. **σ(4p_z)** : 2 electrons 4. **π(4p_x)** : 2 electrons 5. **π(4p_y)** : 2 electrons 6. **π*(4p_x)** : 2 electrons 7. **π*(4p_y)** : 2 electrons Sum: 2 + 2 + 2 + 2 + 2 + 2 + 2 = **14 electrons** So, up to **π*(4p_y)** is filled. --- ## Step 7: **Final Diagram (with Electron Filling)** Here’s a text version you can use to sketch: ``` Kr (4s, 4p) MO Diagram Br (4s, 4p) 4p (6e) σ*(4p_z) (empty) 4p (5e) 4s (2e) π*(4p_x) (2e) 4s (2e) π*(4p_y) (2e) π(4p_x) (2e) π(4p_y) (2e) σ(4p_z) (2e) σ*(4s) (2e) σ(4s) (2e) ``` ### **Electron Configurations in MOs:** - σ(4s)² σ*(4s)² σ(4p_z)² π(4p_x)² π(4p_y)² π*(4p_x)² π*(4p_y)² All electrons are paired. --- ## Step 8: **Calculate Bond Order** Bond order = ½[(bonding electrons) – (antibonding electrons)] - **Bonding MOs:** σ(4s), σ(4p_z), π(4p_x), π(4p_y) = 2 + 2 + 2 + 2 = **8** - **Antibonding MOs:** σ*(4s), π*(4p_x), π*(4p_y) = 2 + 2 + 2 = **6** **Bond Order = ½(8 – 6) = 1** --- ## **Final Answer:** ### **Molecular Orbital Energy Level Diagram for \( \mathrm{KrBr}^+ \):** 1. Draw the atomic orbitals of Kr and Br (4s and 4p). 2. Draw the central MO energy levels in this order (bottom to top): - σ(4s) (2e) - σ*(4s) (2e) - σ(4p_z) (2e) - π(4p_x), π(4p_y) (2e each) - π*(4p_x), π*(4p_y) (2e each) - σ*(4p_z) (empty) 3. Fill 14 electrons in order as above, 2 per orbital. 4. Label all orbitals fully on your diagram. 5. **Bond Order = 1** 6. All electrons are paired (so the molecule is diamagnetic). --- ### **Summary Table:** | MO Level | Electrons | |-------------------|-----------| | σ(4s) | 2 | | σ*(4s) | 2 | | σ(4p_z) | 2 | | π(4p_x), π(4p_y) | 2 each | | π*(4p_x), π*(4p_y)| 2 each | | σ*(4p_z) | 0 | **Bond order = 1** --- #### **Diagram Sketch (vertical energy axis):** ``` σ*(4p_z) (empty) π*(4p_x),π*(4p_y) (2e each) π(4p_x), π(4p_y) (2e each) σ(4p_z) (2e) σ*(4s) (2e) σ(4s) (2e) ``` --- **In summary:** - Draw the MO diagram as above, with full labels and electron filling. - For KrBr⁺, there are 14 electrons; bond order = 1. - All electrons are paired. If you need a hand-drawn or digital diagram, let me know!