Test: Chemical Bonding- 2 - Chemistry MCQ

Isoelectronic species are those have same number of electrons.

• Allene has a linear arrangement of its carbon atoms, with the central carbon atom bonded to two terminal carbon atoms.
• The central carbon is involved in two double bonds, leading to its sp hybridization.
• The terminal carbon atoms are each connected to two hydrogen atoms and one carbon atom, which corresponds to sp2 hybridization.
• Thus, the hybridization types in allene are sp for the central carbon and sp2 for the terminal carbons.

Sulfur forms discrete polyatomic molecule as in S8

Carbon forms discrete polyatomic molecules as in diamond, etc.

The correct answer is Square Planar. The structure of XeF4 is Square Planar. This is because XeF4 follows the VSEPR theory, and the two lone pairs of electrons in XeF4 are placed in a plane that is perpendicular inside an octahedral setting. They face opposite to each other, which is at a distance of 180°. This is the sole reason for the square planar shape of xenon tetrafluoride.

In TeCl₄ out of 6 electrons:bonded electrons are 4,non-bonded electron pair is 1.
In ClF₃ out of 7 electrons:bonded electrons are 3,non-bonded electron pairs are 2.
In PCl₃ out of 5 electrons:bonded electrons are 3,non-bonded electron pair is 1.
In ICl₂− out of (7+1=8) electrons:bonded electrons are 2,non-bonded electron pairs are 3.

Both CO (6 + 8=14) and CN^– (6 + 7+1= 14) have the same electrons. so they are iso-electronic with each other.

The atoms in molecular covalent molecules are held together by strong covalent bonds. Although these bonds are strong, there are only weak forces of attraction between molecules. These weak attractive forces are called van der Waal’s forces and can be broken with little energy.

To determine the structure and hybridization of the molecule N(SiH3)3, we can follow these steps:

Step 1: Identify the central atom and its valence electrons
The central atom in this molecule is nitrogen (N). Nitrogen belongs to group 15 of the periodic table and has 5 valence electrons.

Step 2: Determine the bonding and lone pairs
In N(SiH3)3, nitrogen forms three sigma bonds with three silicon (Si) atoms from the three SiH3 groups. Since nitrogen has 5 valence electrons and it uses 3 of them to form bonds, it will have 2 electrons left, which will form a lone pair.

Step 3: Calculate the steric number
The steric number is calculated as the number of sigma bonds plus the number of lone pairs. Here, nitrogen has:
- 3 sigma bonds (to Si)
- 1 lone pair

Thus, the steric number is 3+1=4.

Step 4: Determine the hybridization
For a steric number of 4, the hybridization is sp3. However, since there is a lone pair, the molecular geometry will be affected.

Step 5: Identify the molecular geometry
With one lone pair and three bonding pairs, the molecular geometry will be trigonal pyramidal. This is because the lone pair occupies more space and pushes the bonding pairs down.

Step 6: Consider back bonding
In N(SiH3)3, there is a possibility of back bonding. Nitrogen can donate its lone pair to the empty d-orbitals of silicon, which can lead to a decrease in hybridization.

Step 7: Adjust the hybridization due to back bonding
Due to back bonding, the effective hybridization of nitrogen can be considered as sp2 instead of sp3. This means that the nitrogen will have a trigonal planar arrangement around it.

Final Answer
Thus, the structure of N(SiH3)3 is trigonal pyramidal with sp2 hybridization.

Pi bond electrons don’t contribute when we calculate the hybridization. So due to one sigma bond and one lone pair hybridization will be “sp”.

Hence D is correct.

The state of hybridization of boron and oxygen atom in boric acid (H₃​BO₃​) is sp²,sp³  respectively.

Boric acid ; has a network structure in which boron is trigonal having sp² and each oxygen atom is tetrahedral having sp³-hybridization with two lone pair of electrons on oxygen.

PCl₃ has sp³-hybridised phosphorus, with one lone pair. Therefore, molecule has pyramidal shape like ammonia.