Question

What hybridization is expected on the central atom of each of the following molecules?
1. $Be{{H}_{2}}$
2. $C{{H}_{2}}B{{r}_{2}}$
3. $P{{F}_{6}}^{-}$
4. $B{{F}_{3}}$
a. $s{{p}^{2}},sp,s{{p}^{3}},s{{p}^{2}}$
b. $sp,s{{p}^{3}},s{{p}^{3}}d,s{{p}^{2}}$
c. $sp,s{{p}^{3}},s{{p}^{3}}{{d}^{2}},s{{p}^{2}}$
d. $s{{p}^{2}},sp,s{{p}^{2}},s{{p}^{3}}$

Answer 1

Hint: Hybridization of a molecule can also be calculated by a formula:
Hybridization = $\dfrac{1}{2}(V+M-C+A)$
where, V = valence electrons,
M = monovalent atom linked to central atom,
C = charge on cation,
A = charge on anion.
After getting the numerical value,
2 = $sp$
3 = $s{{p}^{2}}$
4 = $s{{p}^{3}}$
5 = $s{{p}^{3}}d$
6 = $s{{p}^{3}}{{d}^{2}}$

Complete step by step solution:
In $Be{{H}_{2}}$, the number of valence electrons in beryllium is two ($2{{s}^{2}}$), hydrogen needs two orbitals to form bonds with beryllium. So, one electron from 2s orbital jumps to 2p level and the orbitals hybridize to form hybrid orbitals. Thus, hybridization of one s and one p orbital takes place, the hybridization of $Be{{H}_{2}}$ will be sp and the shape is linear with an angle of 180°.
In $C{{H}_{2}}B{{r}_{2}}$, the number of valence electrons in carbon is 4 ($2{{s}^{2}}2{{p}^{2}}$), hydrogen and bromine both need two- two orbitals to form bonds with carbon. So, one electron from 2s orbitals jumps from 2s to 2p level and orbitals hybridize to form hybrid orbitals. Thus, hybridization of one s and three p orbitals takes place, the hybridization of $C{{H}_{2}}B{{r}_{2}}$ is $s{{p}^{3}}$. The shape will be tetrahedral with an angle of 109.5°.
In $P{{F}_{6}}^{-}$, the number of valence electrons in phosphorus is 6 $[Ne]\,3{{s}^{2}}3{{p}^{4}}$ as there is a negative charge on the compound. Now, the fluorine atom needs six orbitals to form bonds with phosphorus. So, one electron from 2s and one from 3p jumps to the 3d level and forms hybrid orbitals. There will be hybridization of one s, three p and two d orbitals. Thus, the hybridization is $s{{p}^{3}}{{d}^{2}}$. The shape of $P{{F}_{6}}^{-}$ is octahedral and the angle between the bonds is 90°.
Similarly, in $B{{F}_{3}}$, the number of valence electrons in boron is 3 ($2{{s}^{2}}2{{p}^{1}}$), fluorine needs 3 orbitals to form bonds with boron. So, one electron from 2s orbital jumps to 2p level and forms hybrid orbitals. Hybridization of one s and two orbitals takes place. Thus, the hybridization will be $s{{p}^{2}}$. The shape will be trigonal planar and the angle is 120°.

Therefore, the correct answer is (c).

Note: You can also find the hybridization by adding the number of sigma bonds and the lone pairs. dπ - pπ bonding is formed due to the sideways overlap of p and d orbitals. pπ-dπ bonds will be formed when π bonds are more than the number of unhybridized p orbitals left.