
p-p overlapping will be observed in the molecule of
A. Hydrogen
B. Hydrogen bromide
C. Hydrogen chloride.
D. Chlorine.
Answer
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Hint: p-p overlapping in the simplest term can be defined as the penetration of the electron cloud of the p orbitals of the two atoms. So, we can say that p-p overlapping will be only possible in those molecules in which each atom must have at least one partially filled p orbital in it.
Complete step-by-step answer:To answer this type of question, we will see the atomic orbitals involved in the bonding in each molecule. For p-p overlapping in a binary molecule, both atoms should involve p atomic orbitals overlapping.
In other words, if in a binary molecule if even one atom will not have electrons in p atomic orbital then the p-p overlapping is not possible.
Now we will see option by option the molecules.
In the Option A. we are given with a hydrogen molecule $\left( {{H}_{2}} \right)$. This molecule is formed by the overlapping of atomic orbitals of two hydrogen atoms. We know the electronic configuration of hydrogen is $1{{s}^{2}}$. There are two electrons in s orbitals only and p orbitals have no electrons. hence, there will be s-s overlapping in the hydrogen molecule.
In the Option B. hydrogen bromide $\left( HBr \right)$ molecule is given. We know that the electronic configuration of bromine is $\left[ Ar \right]4{{s}^{2}}3{{d}^{10}}4{{p}^{5}}$. This means there are electrons in p orbitals present which can participate in overlapping. But hydrogen has no p orbitals, it has only s orbitals. So, there will be s-p overlapping and not p-p.
In the Option C. we are given with hydrogen chloride $\left( HCl \right)$. we know the electronic configuration of chlorine is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$. This shows that there are electrons in p orbitals in chlorine. But again, we know that hydrogen has electrons only in s orbital. Therefore, here also the overlapping will be s-p
In the Option D. we are given a chlorine molecule $\left( C{{l}_{2}} \right)$. We know that chlorine molecules are formed by the combination of two chlorine atoms. We also know that the configuration of chlorine is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$. Since, both atoms are chlorine. So, we can say that there will be p-p overlapping in the chlorine molecule.
Hence, option (D) is the correct option.
Note: p-p overlapping can be of two types. One head-on overlapping and another side wise overlapping. In head-on overlapping a sigma bond is formed whereas in sidewise overlapping of p-orbital a pi bond is formed. But always remember that a pi bond is only formed after there is at least one sigma bond in the compound.
Complete step-by-step answer:To answer this type of question, we will see the atomic orbitals involved in the bonding in each molecule. For p-p overlapping in a binary molecule, both atoms should involve p atomic orbitals overlapping.
In other words, if in a binary molecule if even one atom will not have electrons in p atomic orbital then the p-p overlapping is not possible.
Now we will see option by option the molecules.
In the Option A. we are given with a hydrogen molecule $\left( {{H}_{2}} \right)$. This molecule is formed by the overlapping of atomic orbitals of two hydrogen atoms. We know the electronic configuration of hydrogen is $1{{s}^{2}}$. There are two electrons in s orbitals only and p orbitals have no electrons. hence, there will be s-s overlapping in the hydrogen molecule.
In the Option B. hydrogen bromide $\left( HBr \right)$ molecule is given. We know that the electronic configuration of bromine is $\left[ Ar \right]4{{s}^{2}}3{{d}^{10}}4{{p}^{5}}$. This means there are electrons in p orbitals present which can participate in overlapping. But hydrogen has no p orbitals, it has only s orbitals. So, there will be s-p overlapping and not p-p.
In the Option C. we are given with hydrogen chloride $\left( HCl \right)$. we know the electronic configuration of chlorine is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$. This shows that there are electrons in p orbitals in chlorine. But again, we know that hydrogen has electrons only in s orbital. Therefore, here also the overlapping will be s-p
In the Option D. we are given a chlorine molecule $\left( C{{l}_{2}} \right)$. We know that chlorine molecules are formed by the combination of two chlorine atoms. We also know that the configuration of chlorine is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$. Since, both atoms are chlorine. So, we can say that there will be p-p overlapping in the chlorine molecule.
Hence, option (D) is the correct option.
Note: p-p overlapping can be of two types. One head-on overlapping and another side wise overlapping. In head-on overlapping a sigma bond is formed whereas in sidewise overlapping of p-orbital a pi bond is formed. But always remember that a pi bond is only formed after there is at least one sigma bond in the compound.
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