Why iron is taken as 3 for second excited state of hydrogen atom?
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I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, E2s=E2pE2s=E2p, E3s=E3p=E3dE3s=E3p=E3d, etc. Thus,
The first excited state of hydrogen atom would be one in which either 2s2sor one of the three 2p2p orbitals is occupied and it will be 4-fold degenerate: 1s02s11s02s1, 1s02p1x1s02px1, 1s02p1y1s02py1, 1s02p1z1s02pz1.
Analogously, the second excited state of hydrogen atom would be one in which either 3s3s or one of the three 3p3p or one of the five 3d3d orbitals is occupied and it will be 9-fold degenerate: 1s02s02p0x2p0y2p0z3s11s02s02px02py02pz03s1, 1s02s02p0x2p0y2p0z3s03p1x1s02s02px02py02pz03s03px1, 1s02s02p0x2p0y2p0z3s03p1y1s02s02px02py02pz03s03py1, 1s02s02p0x2p0y2p0z3s03p1z1s02s02px02py02pz03s03pz1, plus 5 configurations with only 3d3d-orbitals occupied.
For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do not have same energy anymore. For instance, E2s<E2pE2s<E2p, E3s<E3p<E3dE3s<E3p<E3d, etc. Thus,
The first excited state of the hydride would be one in which one electron populates 1s1s orbital and another 2s2sone, i.e. a non-degenerate 1s12s11s12s1state.
The second excited state of the hydride would be one in which one electron populates 1s1s orbital and another one of the three 2p2p ones, i.e. a 3-fold degenerate state: 1s12s02p1x1s12s02px1, 1s12s02p1y1s12s02py1, 1s12s02p1z1s12s02pz1
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The first excited state of hydrogen atom would be one in which either 2s2sor one of the three 2p2p orbitals is occupied and it will be 4-fold degenerate: 1s02s11s02s1, 1s02p1x1s02px1, 1s02p1y1s02py1, 1s02p1z1s02pz1.
Analogously, the second excited state of hydrogen atom would be one in which either 3s3s or one of the three 3p3p or one of the five 3d3d orbitals is occupied and it will be 9-fold degenerate: 1s02s02p0x2p0y2p0z3s11s02s02px02py02pz03s1, 1s02s02p0x2p0y2p0z3s03p1x1s02s02px02py02pz03s03px1, 1s02s02p0x2p0y2p0z3s03p1y1s02s02px02py02pz03s03py1, 1s02s02p0x2p0y2p0z3s03p1z1s02s02px02py02pz03s03pz1, plus 5 configurations with only 3d3d-orbitals occupied.
For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do not have same energy anymore. For instance, E2s<E2pE2s<E2p, E3s<E3p<E3dE3s<E3p<E3d, etc. Thus,
The first excited state of the hydride would be one in which one electron populates 1s1s orbital and another 2s2sone, i.e. a non-degenerate 1s12s11s12s1state.
The second excited state of the hydride would be one in which one electron populates 1s1s orbital and another one of the three 2p2p ones, i.e. a 3-fold degenerate state: 1s12s02p1x1s12s02px1, 1s12s02p1y1s12s02py1, 1s12s02p1z1s12s02pz1
PLEASE MARK MY ANSWER AS A BRAINLIEST.
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