Question

Assertion A-the hybridisation possessed by the oxygen atom in OF2 molecule. Reson R-sp2 hybrid structure have trigonal planar geometry. Both a and r true and R is correct explanation of A class 11 chemistry​

Answer 1

Answer:

Correct option is C)

In OF

2

, oxygen has 2 bond pairs and 2 lone pairs and hence hybridization is sp

3

.

Answer 2

Answer:

Option d. A is false but R is true.

Explanation:

• The hybridization of oxygen in $OF_{2}$  is $sp^{3}$.

• $OF_{2}$  or Oxygen difluoride is a chemical compound formed by the reaction $2F_{2} + 2NaOH$ → $OF_{2} + 2NaF+ H_{2} O$

• It is a colourless gaseous compound with a strong odour and often acts as an oxidizer.

• The electronic configuration of halogen Fluorine and Oxygen in their ground state reveals that,

$\tensor^{}_{9}{\mathrm{F=1s^{2}2s^{2}2p^{5} }}{} \end{document}$

$\tensor^{}_{8}{\mathrm{O= 1s^{2}2s^{2}2p^{4} }}{} \end{document}$

• After bonding with $F$ and filing the octet rule, the outermost shell of the central Oxygen atom has 4 hybridised orbitals i.e.$2s,2p_{x} ,2p_{y} ,2p_{z}$

• Here the steric number is 4 i.e. it has 2 bond pairs and 2 lone pairs.

• So the hybridization is $sp^{3}$. Therefore assertion is false.

• Hybridisation refers to the process by which the bonding orbitals possess the same energy and bond length.

• For $sp^{2}$ hybridized central atoms, the possible molecular geometry is trigonal planar.
• All the bonds remain in place showing the trigonal planar shape. When there found 3 bonds (3 atoms linked central atom) to another atom or lone pairs, it shows $sp^{2}$ hybridization. Therefore reason is true.

Hence option d. A is false but R is true is correct.