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How many lone pair of electrons are present on the central atom of$\text{ C}{{\text{H}}_{\text{4}}}\text{ }$, $\text{ }{{\text{H}}_{\text{2}}}\text{O }$, $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$, $\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$ and $\text{ PC}{{\text{l}}_{5}}\text{ }$ molecules?

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Last updated date: 20th Jun 2024
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Answer
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Hint: The lone pairs are the valence electrons which do not take part in the bonding. Determine the valence electrons involved in the molecules and then subtract the total number of bonding electrons from the valence electrons to calculate the number of lone pairs.
\[\text{ Lone pairs = }\dfrac{1}{2}\left( \text{Valence }{{\text{e}}^{-}}\text{ in the molecule }-\text{ Bonding }{{\text{e}}^{-}}\text{ in the molecule} \right)\]
The lone pairs are a pair of valence electrons that are not shared by another atom in the covalent bond.it is also termed as the unshared pair or the non-bonding pair. The lone pairs are in the outermost shell of atoms. These pairs of electrons are not used in chemical bonding.
The lone pairs can find out by knowing the geometry of the molecule.
Step 1) Count all the number of valence electrons in the molecule.
Step 2) Count the total number of atoms that are bonded to the central atom and multiply it by 8 so that all the atoms complete the octet.
Step 3) Find the number of lone pairs by subtracting the valence electrons and bonded atoms from the total valence electrons.
Step 4) now we divide the lone pair electrons found in step 3) by 2 to get the number of lone pairs on the central atom.

Complete step by step solution:
Let's determine the number of lone pairs on the central atom.
A) Methane or$\text{ C}{{\text{H}}_{\text{4}}}\text{ }$:
The total number of valence electrons of carbon are:
$\text{ C = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{2}}}\text{ }$
The carbon has 4 valence electrons and each hydrogen has the 1 valence electron. Thus the total number of valence electrons in the $\text{ C}{{\text{H}}_{\text{4}}}\text{ }$are:
$\begin{align}
& \text{ Total Valence }{{\text{e}}^{-}}\text{ in C}{{\text{H}}_{\text{4}}}\text{ = V}\text{.E}\text{. in C + 4(V}\text{.E}\text{. in H) } \\
& \therefore \text{Total Valence }{{\text{e}}^{-}}\text{ in C}{{\text{H}}_{\text{4}}}\text{ = 4 + 4 = 8 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
There are four bonds on the carbon atom, each bond shares the two electrons in a bonding pair. Thus a total of bonding electrons,$\text{ = 4 }\times \text{ 2}{{\text{e}}^{-}}\text{ = 8}{{\text{e}}^{-}}$
Therefore, all the valence electrons are utilized in the bonding pairs so the central atom C has 0 lone pairs.
B) Water or$\text{ }{{\text{H}}_{\text{2}}}\text{O }$:
The total number of valence electrons of oxygen are:
$\text{ O = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{4}}\text{ }$
The oxygen has 6 valence electrons and each hydrogen has the 1 valence electron. Thus the total number of valence electrons in the $\text{ }{{\text{H}}_{\text{2}}}\text{O }$are:
$\begin{align}
& \text{ Total Valence }{{\text{e}}^{-}}\text{ in }{{\text{H}}_{\text{2}}}\text{O = V}\text{.E}\text{. in O + 2(V}\text{.E}\text{. in H) } \\
& \therefore \text{Total Valence }{{\text{e}}^{-}}\text{ in }{{\text{H}}_{\text{2}}}\text{O = 6 + 2 = 8 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
The oxygen forms two bonds with the two hydrogen atoms. Each $\text{ O}-\text{H }$ bond share two electrons .thus total of bonding pairs in the $\text{ }{{\text{H}}_{\text{2}}}\text{O }$molecules are $\text{ = 2 }\times \text{ 2}{{\text{e}}^{-}}\text{ = 4}{{\text{e}}^{-}}$
The total number of lone pairs would be equal to the bonding pairs subtracted from the valence electrons. Thus total lone pairs are:
$\begin{align}
& \text{ Lone pairs on O = V}\text{.E}\text{. in }{{\text{H}}_{\text{2}}}\text{O }-\text{ B}\text{.E}\text{.} \\
& \therefore \text{Lone pairs on O = 8 }-\text{ 4 = 4 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
The lone pairs are equal to,
$\text{ L}\text{.P}\text{. on O = }\dfrac{4{{e}^{-}}}{2}\text{ = 2 L}\text{.P}\text{. }$
Thus, the oxygen in the $\text{ }{{\text{H}}_{\text{2}}}\text{O }$molecules has 2 lone pairs.
C) Ammonia or $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$:
The total number of valence electrons of nitrogen are:
$\text{ N = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{3}}\text{ }$
The nitrogen has 5 valence electrons and each hydrogen has the 1 valence electron. Thus the total number of valence electrons in the $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$are:
$\begin{align}
& \text{ Total Valence }{{\text{e}}^{-}}\text{ in N}{{\text{H}}_{\text{3}}}\text{ = V}\text{.E}\text{. in N + 3(V}\text{.E}\text{. in H) } \\
& \therefore \text{Total Valence }{{\text{e}}^{-}}\text{ in N}{{\text{H}}_{\text{3}}}\text{ = 5 + 3 = 8 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
The nitrogen forms three bonds with the three hydrogen atoms. Each $\text{ N}-\text{H }$ bond share two electrons .thus total of boning pairs in the $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$molecules are $\text{ = 3 }\times \text{ 2}{{\text{e}}^{-}}\text{ = 6 }{{\text{e}}^{-}}$
The total number of lone pairs would be equal to the bonding pairs subtracted from the valence electrons. Thus total lone pairs are:
$\begin{align}
& \text{ Lone pairs on N = V}\text{.E}\text{. in N}{{\text{H}}_{\text{3}}}\text{ }-\text{ B}\text{.E}\text{.} \\
& \therefore \text{Lone pairs on N = 8 }-\text{ 6 = 2 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
The lone pairs are equal to,
$\text{ L}\text{.P}\text{. on N = }\dfrac{2{{e}^{-}}}{2}\text{ = 1 L}\text{.P}\text{. }$
Thus, the nitrogen in the $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$molecules has 1 lone pair.
D) Phosphorus trichloride or$\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$:
The total number of valence electrons of phosphorus are:
$\text{ P = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{3}}}\text{ }$
The phosphorus has 5 valence electrons and each chloride has the 1 valence electrons. Thus the total number of valence electrons in the $\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$are:
$\begin{align}
& \text{ Total Valence }{{\text{e}}^{-}}\text{ in PC}{{\text{l}}_{\text{3}}}\text{ = V}\text{.E}\text{. in P + 3(V}\text{.E}\text{. in Cl) } \\
& \therefore \text{Total Valence }{{\text{e}}^{-}}\text{ in PC}{{\text{l}}_{\text{3}}}\text{ = 5 + 3 = 8 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
The P forms three bonds with the three chlorine atoms. Each $\text{ P}-\text{Cl }$ bond share two electrons .thus total of boning pairs in the $\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$molecules are $\text{ = 3 }\times \text{ 2}{{\text{e}}^{-}}\text{ = 6 }{{\text{e}}^{-}}$
The total number of lone pairs would be equal to the bonding pairs subtracted from the valence electrons. Thus total lone pairs are:
\[\begin{align}
& \text{ Lone pairs on P = V}\text{.E}\text{. in PC}{{\text{l}}_{\text{3}}}\text{ }-\text{ B}\text{.E}\text{.} \\
& \therefore \text{Lone pairs on P = 8 }-\text{ 6 = 2 }{{\text{e}}^{-}}\text{ } \\
\end{align}\]
The lone pairs are equal to,
$\text{ L}\text{.P}\text{. on P = }\dfrac{2{{e}^{-}}}{2}\text{ = 1 L}\text{.P}\text{. }$
Thus, the phosphorus in the $\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$molecules have 1 lone pair.
E) Phosphorus pentachloride or$\text{ PC}{{\text{l}}_{5}}\text{ }$:
The total number of valence electrons of phosphorus are:
$\text{ P = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{3}}}\text{ }$
The phosphorus has 5 valence electrons and each chloride has the 1 valence electrons. Thus the total number of valence electrons in the $\text{ PC}{{\text{l}}_{5}}\text{ }$are:
$\begin{align}
& \text{ Total Valence }{{\text{e}}^{-}}\text{ in PC}{{\text{l}}_{5}}\text{ = V}\text{.E}\text{. in P + 5(V}\text{.E}\text{. in Cl) } \\
& \therefore \text{Total Valence }{{\text{e}}^{-}}\text{ in PC}{{\text{l}}_{5}}\text{ = 5 + 5 = 10 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
The P forms three bonds with the three chlorine atoms. Each $\text{ P}-\text{Cl }$ bond share two electrons .thus total of boning pairs in the $\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$molecules are $\text{ = 5 }\times \text{ 2}{{\text{e}}^{-}}\text{ = 10 }{{\text{e}}^{-}}$
The total number of lone pairs would be equal to the bonding pairs subtracted from the valence electrons. Thus total lone pairs are:
\[\begin{align}
& \text{ Lone pairs on P = V}\text{.E}\text{. in PC}{{\text{l}}_{\text{3}}}\text{ }-\text{ B}\text{.E}\text{.} \\
& \therefore \text{Lone pairs on P = 10 }-\text{ 10 = 0} \\
\end{align}\]
Thus, the phosphorus in the $\text{ PC}{{\text{l}}_{5}}\text{ }$molecules have 0 lone pair.
Therefore, the lone pairs on the molecules are listed as below:
MoleculeValence electronsBonding pairsLone pairs
$\text{ C}{{\text{H}}_{\text{4}}}\text{ }$$\text{8 }{{\text{e}}^{-}}$40
$\text{ }{{\text{H}}_{\text{2}}}\text{O }$$\text{8 }{{\text{e}}^{-}}$22
$\text{ N}{{\text{H}}_{\text{3}}}\text{ }$$\text{6 }{{\text{e}}^{-}}$31
$\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$$\text{8 }{{\text{e}}^{-}}$31
$\text{ PC}{{\text{l}}_{5}}\text{ }$$\text{10 }{{\text{e}}^{-}}$50


Note: We can determine the geometry of the molecules from the bonding pair, lone pairs by the $\text{ VSEPR }$ theory. The structure of the molecules are as follows:
MoleculeStructure /Geometry
$\text{ C}{{\text{H}}_{\text{4}}}\text{ }$Tetrahedral
$\text{ }{{\text{H}}_{\text{2}}}\text{O }$Bent V shape
$\text{ N}{{\text{H}}_{\text{3}}}\text{ }$Pyramidal
$\text{ PC}{{\text{l}}_{\text{3}}}\text{ }$Trigonal pyramidal
$\text{ PC}{{\text{l}}_{5}}\text{ }$Trigonal pyramidal.