Name: Paramagnetic Nature of Liquid Oxygen
Uploaded: 2019-07-19
Duration: 4 min 21 s
Description: Understand the molecular orbital diagram for the oxygen molecule. Determine the magnetic character of the oxygen molecule based on molecular orbital theory.

Question

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	-
$C_{2} H_{2}$	$sp$	$(3 σ + 2 π) 1 C - C σ 2 C - H σ 2 C - C π$	-	-

Accepted Answer

${CH}_{4} :$

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 ${CH}_{4}$ , 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 ${NH}_{3}$ and hence it is a linear molecule.

Important Questions on Chemical Bonding

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