Return to Question

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$ $$S_{ij}>0\Rightarrow\text{bonding}$$ $$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_{ij}}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_{ij}}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coefficient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_{ij}}}>\frac{1}{\sqrt{1+S_{ij}}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$ $$S_{ij}>0\Rightarrow\text{bonding}$$ $$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_{ij}}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_{ij}}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coefficient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_{ij}}}>\frac{1}{\sqrt{1+S_{ij}}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$ $$S_{ij}>0\Rightarrow\text{bonding}$$ $$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_ij}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_ij}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coeffcient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_ij}}>\frac{1}{\sqrt{1+S_ij}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$ $$S_{ij}>0\Rightarrow\text{bonding}$$ $$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_{ij}}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_{ij}}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coefficient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_{ij}}}>\frac{1}{\sqrt{1+S_{ij}}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding.

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_ij}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_ij}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coeffcient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_ij}}>\frac{1}{\sqrt{1+S_ij}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$ $$S_{ij}>0\Rightarrow\text{bonding}$$ $$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_ij}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_ij}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coeffcient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_ij}}>\frac{1}{\sqrt{1+S_ij}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?