Question

In which of the following molecule back bonding is present,
A.$B{F}_3$
B.$H_{2}O$
C.$N{H}_3$
D.$HF$

Answer 1

Hint: We must remember that back bonding is seen between atoms in a compound in which one atom has a lone pair of electrons and another shows vacant orbital adjacent to each other.

Complete step by step answer:
We can call back bonding as a $\pi$-coordinate bond. Two types of compounds generally show back bonding.
Compounds in which the side atoms donate a pair of electrons to central atom
Compounds in which the central atoms donate a pair of electron to side atoms
The conditions for back bonding are,
Any one atom in the pair must contain one lone pair of electrons
At Least one atom should belong to 2nd period and other atom should belong to 3rd period
We know that the geometry of $B{F}_3$ is trigonal planar. All the three boron-fluorine bonds lie in the plane, and therefore, the p-orbital of boron and fluorine are said to be parallel.
Boron shows the presence of empty p-orbital and lone pair of electrons is present in the p-orbital of fluorine. Therefore, boron acts as Lewis acid and fluorine acts as Lewis base. A lone pair of electrons is donated by fluorine to boron atoms, and this type of bonding is called back bonding.
Therefore, option (A) is correct.
Water has a tetrahedral arrangement of molecules. This is due to the repulsion from the lone pair combination is greater than bond-pair repulsion. The existing pairs are not present in the same plane. One pair lies below the plane and the other one lies above. This bond geometry is called a distorted tetrahedron.
Therefore, option (B) is incorrect.
The molecular geometry of ammonia is trigonal pyramidal or distorted tetrahedral structure. This is because of the presence of a lone non-bonding pair that generally exerts more repulsion on the bonding orbitals. In ammonia, nitrogen is the central atom, three hydrogen atoms that are alike create the base and one pair of electrons forms the apex of the pyramid. Therefore, option (C) is incorrect.
The molecular geometry of $HF$ is tetrahedral and has a linear shape. It contains three lone pairs and one bond pair, and is $s{p}^3$ hybridized. Therefore, option (D) is incorrect.
Therefore option A is correct.


Note:
Back bonding permits the molecule to be stable as it satisfies its octet.
Back bonding results in decrease in bond length and increase in bond order.
In a pi-back bonding electron, it moves from an atomic orbital of one atom to anti-bonding orbital of another atom (or) ligand.
Nickel carbonyl and Zeise’s salt are examples of compounds that show pi-back bonding.