Complete the following table. Molecule Type of Hybridisation Type of bonds Geometry Bond angle CH 4 - 4 C - H and 4 σ bonds Tetrahedral - NH 3 sp 3 3 N - H and 3 σ bonds 1 lone pair - - H 2 O - - Angular 104 . 5 ∘ BF 3 sp 2 - - 120 ∘ C 2 H 4 - - - 120 ∘ BeF 2 - 2 Be - F Linear - C 2 H 2 sp 3 σ + 2 π 1 C - C σ 2 C - H σ 2 C - C π - -

CH 4 : The 2 s and 2 p orbitals thus undergo hybridisation to form 4 sp 3 hybridised orbitals. Methane is a perfect tetrahedron with all H - C - H bond angles equal at 109 . 5 ° . NH 3 : There is one lone pair of electrons and three bond pairs, the resulting molecular geometry is trigonal pyramidal. The bond angle in a molecule of ammonia is 107 ° . H 2 O : Water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus, we say that the oxygen atom is sp 3 hybridised. The bond between O and 2 H atoms and so it will be a covalent bond. There are 2 O - H σ bonds, 2 lone pair . BF 3 : The geometry of the molecule of BF 3 is ‘Trigonal Planar. Boron and only three σ bonds are formed per Boron atom. C 2 H 4 : The molecular shape is predicted to be trigonal planar around each carbon atom. This is composed of a σ framework and a π -bond. In the ethene molecule, the carbon atoms are sp 2 hybridised. BeF 2 : In BeF 2 , Be atom is sp hybridised. The bond angle is 180 ° . C 2 H 2 : The carbon atom of the ethyne molecule is sp hybridised. The bond angle of sp hybridisation is 180 ° and hence it is a linear molecule.

E M B I B E

Chemistry>Chapter 5 - Chemical Bonding>Exercises>Q 5

MEDIUM

11th Maharashtra Board

IMPORTANT

Earn 100

Complete the flow chart.

Important Questions on Chemical Bonding

HARD

11th Maharashtra Board

IMPORTANT

Chemistry>Chapter 5 - Chemical Bonding>Exercises>Q 6

Complete the following table.

Molecule	Type of Hybridisation	Type of bonds	Geometry	Bond angle
${CH}_{4}$	-	$4 C - H and 4 σ$ bonds	Tetrahedral	-
${NH}_{3}$	${sp}^{3}$	$3 N - H and 3 σ bonds 1 lone pair$	-	-
$H_{2} O$	-	-	Angular	$104 . 5^{\circ}$
${BF}_{3}$	${sp}^{2}$	-	-	$120^{\circ}$
$C_{2} H_{4}$	-	-	-	$120^{\circ}$
${BeF}_{2}$	-	$2 Be - F$	Linear	-
$C_{2} H_{2}$	$sp$	$(3 σ + 2 π) 1 C - C σ 2 C - H σ 2 C - C π$	-	-