Question
Explain the non linear shape of $H_2S$ and non planar shape of $PCl_3$ using valence shell electron pair repulsion theory.
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Answer
In the molecule $PCl_3$, The electronic configuration of P atom is $1s^2, 2s^2, 2p^6, 3s^2, 3p^3$. The valency of $P$ is $5$. In $PCl3$,
$P$ has $3$ single bonds and $1$ lone pair (pair of unshared electrons). This suggests us to use $sp^3$ hybrid orbitals on phosphorus. Each chlorine atom (valence shell configuration $3s^23p^5$) has one singly occupied $3p$ orbital. The P–Cl bonds are formed by the overlap of a phosphorus $sp3$ hybrid orbital with singly occupied chlorine $3p$ orbital. Each Cl atom holds three lone pairs.
In $H_2S$ , S is the central atom, which has 6 valence electrons, out of which 2 electron are used by two hydrogen forming two bonds and two remaining pairs remain as lone pairs around the atom, Thus due to lone pair lone pair repulsion offered by the two pairs of electrons, it has a bent shape.
$P$ has $3$ single bonds and $1$ lone pair (pair of unshared electrons). This suggests us to use $sp^3$ hybrid orbitals on phosphorus. Each chlorine atom (valence shell configuration $3s^23p^5$) has one singly occupied $3p$ orbital. The P–Cl bonds are formed by the overlap of a phosphorus $sp3$ hybrid orbital with singly occupied chlorine $3p$ orbital. Each Cl atom holds three lone pairs.
In $H_2S$ , S is the central atom, which has 6 valence electrons, out of which 2 electron are used by two hydrogen forming two bonds and two remaining pairs remain as lone pairs around the atom, Thus due to lone pair lone pair repulsion offered by the two pairs of electrons, it has a bent shape.
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Column I
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Column II
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i.
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$H_3O^+$ |
a.
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Linear |
|
ii.
|
$HC = CH$ |
b.
|
Angular |
|
iii.
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$\text{ClO}^{-}_{2}$ |
c.
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Tetrahedral |
|
iv.
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$\text{NH}^{+}_{4}$ |
d.
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Trigonal bipyramidal |
|
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e.
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Pyramidal |