Question 14 Marks
In order to explain the characteristic geometrical shapes of polyatomic molecules, Pauling introduced the concept of hybridisation. The orbitals undergoing hybridisation should have nearly the same energy. There are various type of hybridisations involving s, p and d-type of orbitals. The type of hybridisation gives the characteristic shape of the molecule or ion.
1. Why all the orbitals in a set of hybridised orbitals have the same shape and energy?
2. Out of $XeF _2$ and $SF _2$ which molecule has the same shape as $NO _2^{+}$ion?
3. Out of $XeF _4$ and $XeF _2$ which molecule doesn't have the same type of hybridisation as P (Phosphorus) has in $PF _5$ ?
OR
Unsaturated compounds undergo additional reactions. Why?
1. Why all the orbitals in a set of hybridised orbitals have the same shape and energy?
2. Out of $XeF _2$ and $SF _2$ which molecule has the same shape as $NO _2^{+}$ion?
3. Out of $XeF _4$ and $XeF _2$ which molecule doesn't have the same type of hybridisation as P (Phosphorus) has in $PF _5$ ?
OR
Unsaturated compounds undergo additional reactions. Why?
Answer
View full question & answer→1. Hybrid orbitals are formed after combining atomic orbitals and have the equivalent shape and energy in the given set of hybridised orbitals.
2. $XeF _2$ molecule has the same shape as $NO _2^{+}$ion.
3. $XeF _4$ molecule doesn't have the same type of hybridisation as P (Phosphorus) has in $PF _5$.
OR
Unsaturated hydrocarbon molecules include two- or three-fold bonds of carbon. The $\pi$-bond is a multiple bond, which becomes unstable and hence adds across numerous bonds.
2. $XeF _2$ molecule has the same shape as $NO _2^{+}$ion.
3. $XeF _4$ molecule doesn't have the same type of hybridisation as P (Phosphorus) has in $PF _5$.
OR
Unsaturated hydrocarbon molecules include two- or three-fold bonds of carbon. The $\pi$-bond is a multiple bond, which becomes unstable and hence adds across numerous bonds.
