Indicate which orbitals would be able toparticipate in bonding and what theirenergy difference would be:Example: BeH, (H) 1s / (Be) 2s (0.4 eV), (H) 1s/ (Be) 2pz (5.32 eV)1. HСІ2. NaF3. C12.Orbital Energieskey: s orbitalsp orbitalsd orbitalsunits: eVНеH-13.61-24.59LiВеBNe-19.43-10.66-32.38-15.8540.17-18.65-48.47-21.59-5.39-25.5613.18-9.32-14.05-8.30AlSCIAr-29.24-15.82NaMgSiP-14.89-7.78-18.84-9.65As-5.14-7.65-11.32-5.98-22.71-11.62-25.23-13.67Ti-7.12KVMnFeNiCuZn-9.39-17.30Ga-12.61-5.93In-11.89Br-24.37-12.49СаScCrCoGeSeKr-16.05-7.54-18.94-9.17-21.37-10.824.34-6.11-6.60-7.92-7.32-10.11-7.46-7.84-7.91-11.66-8.10-8.22-8.42-27.51-13.47Ag-7.58-9.22-10.75-11.14-12.12-12.93-14.22Rb-4.18Sr-5.70Sb-16.74YZr-7.31NbMo-7.24Tc-7.21Ru-7.13RhPd-7.44Cd-8.99-17.86HgSn-14.56Te-18.71Xe-23.40-6.70-7.22-8.86-7.28-20.89-6.49-8.31-9.14-9.25-9.31-9.45-9.58-12.78-5.60-7.01-8.41-9.79-11.18-12.56HfWOsPtCs-3.90ВаLaTaRe-8.76-9.70IrAu-9.22-11.86TI-13.14-5.47PbBiPo-20.05-9.42At-22.69-10.71Rn-5.21-7.04-5.28-7.52-6.14-8.45-7.58-8.52-8.76-8.82-10.00-8.83-10.22-8.75-10.37-10.44-15.59-15.12-6.81-17.52-25.47-8.15-12.03FrRaAc??

In the formation of bond, outermost orbitals of atoms can participate and form bonds.

Indicate which orbitals would be able to participate in bonding and what their energy difference would be: Example: BeH, (H) 1s / (Be) 2s (0.4 eV), (H) 1s / (Be) 2pz (5.32 eV) 1. НСІ 2. NaF 3. C12. Orbitel Enereles

Indicate which orbitals would be able to participate in bonding and what their energy difference would be: Example: BeH, (H) 1s / (Be) 2s (0.4 eV), (H) 1s / (Be) 2pz (5.32 eV) 1. НСІ 2. NaF 3. C12. Orbitel Enereles

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter6: Covalent Bonding

Section: Chapter Questions

Problem 112QRT

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Similar questions

The molecular orbital diagram of NO shown in Figure 10.47 also applies to OF. Draw the complete molecular orbital diagram for OF. What is the OF bond order? Figure 10.47 Molecular orbital diagram for nitric oxide (NO). The molecular orbital diagram for NO predicts a bond order of 2.5 and predicts that the molecule is paramagnetic with one unpaired electron. These predictions are verified by experimental measurements.
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