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Here are the H-X-H$\ce{H-X-H}$ bond angles and the H-X$\ce{H-X}$ bond lengths: \begin{aligned} &\ce{H2O} & 104.5^\circ && \pu{ 96 pm}\\ &\ce{H2S} & 92.3^\circ && \pu{134 pm}\\ &\ce{H2Se} & 91.0^\circ && \pu{146 pm}\\ \end{aligned}\begin{array}{lcc} \text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\ \hline \ce{H2O} & 104.5 & 96 \\ \ce{H2S} & 92.3 & 134 \\ \ce{H2Se}& 91.0 & 146 \\ \hline \end{array}

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the H-X-H$\ce{H-X-H}$ angle to contract slightly. So instead of the H-O-H$\ce{H-O-H}$ angle being the perfect tetrahedral angle (109.5°$109.5^\circ$) it is slightly reduced to 104.5°$104.5^\circ$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridizedhave no orbital hybridization. That is, The S-H$\ce{S-H}$ and Se-H$\ce{Se-H}$ bonds use pure p$\ce{p}$-orbitals from sulfur and selenium respectively. Two p$\ce{p}$-orbitals are used, one for each of the two X-H$\ce{X-H}$ bonds; this leaves another p$\ce{p}$-orbital and an s$\ce{s}$-orbital to hold the two lone pairs of electrons. If the S-H$\ce{S-H}$ and Se-H$\ce{Se-H}$ bonds used pure p$\ce{p}$-orbitals we would expect an H-X-H$\ce{H-X-H}$ interorbital angle of 90°$90^\circ$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H$\ce{X-H}$ bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H$\ce{H-S-H}$ angle being slightly larger than the corresponding H-Se-H$\ce{H-Se-H}$ angle. Since the H-Se$\ce{H-Se}$ bond is longer then the H-S$\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two O-H$\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} &\ce{H2O} & 104.5^\circ && \pu{ 96 pm}\\ &\ce{H2S} & 92.3^\circ && \pu{134 pm}\\ &\ce{H2Se} & 91.0^\circ && \pu{146 pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.5°) it is slightly reduced to 104.5°. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90°. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths: \begin{array}{lcc} \text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\ \hline \ce{H2O} & 104.5 & 96 \\ \ce{H2S} & 92.3 & 134 \\ \ce{H2Se}& 91.0 & 146 \\ \hline \end{array}

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization. That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively. Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons. If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider. This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

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Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} \ce{&H2O~& 104$.$5^{o} && 96 pm}\\ \ce{&H2S~& 92$.$3^{o} && 134 pm}\\ \ce{&H2Se~& 91$.$0^{o} && 146pm}\\ \end{aligned}\begin{aligned} &\ce{H2O} & 104.5^\circ && \pu{ 96 pm}\\ &\ce{H2S} & 92.3^\circ && \pu{134 pm}\\ &\ce{H2Se} & 91.0^\circ && \pu{146 pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.5$\ce{^{o}}$) it is slightly reduced to 104.5$\ce{^{o}}$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90$\ce{^{o}}$90°. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} \ce{&H2O~& 104$.$5^{o} && 96 pm}\\ \ce{&H2S~& 92$.$3^{o} && 134 pm}\\ \ce{&H2Se~& 91$.$0^{o} && 146pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.5$\ce{^{o}}$) it is slightly reduced to 104.5$\ce{^{o}}$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90$\ce{^{o}}$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} &\ce{H2O} & 104.5^\circ && \pu{ 96 pm}\\ &\ce{H2S} & 92.3^\circ && \pu{134 pm}\\ &\ce{H2Se} & 91.0^\circ && \pu{146 pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.) it is slightly reduced to 104.. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90°. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

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Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} \ce{&H2O~& 104$.$5^{o} && 96 pm}\\ \ce{&H2S~& 92$.$3^{o} && 134 pm}\\ \ce{&H2Se~& 91$.$0^{o} && 146pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.5$\ce{^{o}}$) it is slightly reduced to 104.5$\ce{^{o}}$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90$\ce{^{o}}$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital. See page 18/20 in this excellent tutorial for a more detailed explanation with helpful pictures - the entire article is worth reading.

Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} \ce{&H2O~& 104$.$5^{o} && 96 pm}\\ \ce{&H2S~& 92$.$3^{o} && 134 pm}\\ \ce{&H2Se~& 91$.$0^{o} && 146pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.5$\ce{^{o}}$) it is slightly reduced to 104.5$\ce{^{o}}$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90$\ce{^{o}}$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital. See page 18/20 in this excellent tutorial for a more detailed explanation with helpful pictures - the entire article is worth reading.

Here are the H-X-H bond angles and the H-X bond lengths: \begin{aligned} \ce{&H2O~& 104$.$5^{o} && 96 pm}\\ \ce{&H2S~& 92$.$3^{o} && 134 pm}\\ \ce{&H2Se~& 91$.$0^{o} && 146pm}\\ \end{aligned}

The traditional textbook explanation would argue that the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the H-X-H angle to contract slightly. So instead of the H-O-H angle being the perfect tetrahedral angle (109.5$\ce{^{o}}$) it is slightly reduced to 104.5$\ce{^{o}}$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ are not hybridized. That is, The S-H and Se-H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X-H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S-H and Se-H bonds used pure p-orbitals we would expect an H-X-H interorbital angle of 90$\ce{^{o}}$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two X-H bonds the angle opens up a bit wider. This explanation would be consistent with the H-S-H angle being slightly larger than the corresponding H-Se-H angle. Since the H-Se bond is longer then the H-S bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the two O-H bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

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