Inorganic Chemistry (B.A./B.Sc) 1st Sem Previous Year Solved Question Paper 2022

1. SnCl2 (Tin Dichloride):
 - Central Atom: In SnCl2, the central atom is tin (Sn).
 - Valence Electrons: Tin has 4 valence electrons (from Group 14) and each chlorine atom contributes one valence electron.
 - Total Valence Electrons: \(4 (Sn) + 2 (Cl) = 6\).

 To determine the hybridization, you can use the following formula:
 
L{Hybridization} = {Number of Valence Electrons on Central Atom} + {Number of Monovalent Atoms Attached} 

 For SnCl2:
 
 {Hybridization} = left( 6 + 2 – 0 \right) = 4 ]

 The result of 4 indicates that tin in SnCl2 undergoes sp3 hybridization, which means it forms four equivalent hybrid orbitals. These hybrid orbitals overlap with the chlorine atoms to form four sigma bonds.

2. BF4- (Tetrafluoroborate Ion):
 - Central Atom: In BF4-, the central atom is boron (B).
 - Valence Electrons: Boron has 3 valence electrons (from Group 13), and each fluorine atom contributes one valence electron.
 - Total Valence Electrons: \(3 (B) + 4 (F) = 7\).

 Using the same formula for hybridization:
 {Hybridization} ( 7 + 4 – 1 ]

 The -1 in the formula accounts for the negative charge on the tetrafluoroborate ion.

 The result of 4.5 indicates that boron in BF4- undergoes sp3d hybridization, which means it forms five equivalent hybrid orbitals. These hybrid orbitals overlap with the fluorine atoms to form sigma bonds.

In both cases, the hybridization of the central atom allows it to form the necessary sigma bonds with the surrounding atoms, which results in the stability and structure of the molecules.