Carbon dioxide is therefore linear in electron-group geometry and in molecular geometry. The shape of CO2 is linear because there are no lone pairs affecting the orientation of the molecule. Therefore, the linear orientation minimizes the repulsion forces.

The difference between bond pair and lone pair is that a bond pair is composed of two electrons that are in a bond whereas a lone pair is composed of two electrons that are not in a bond.

The initial VSEPR shape for the CO2 molecule is Tetrahedral. For each multiple bond (double/triple bond), subtract one electron from the final total. The CO2 molecule has 2 double bonds so minus 2 electrons from the final total. So the overall total number of electrons should be 2, this is the electron region number.

tetrahedral

SO3 has 3 oxygen bound to it and no lone pairs since S is stable with 3 bonds. This means that is trigonal planar for electron and molecular geometries. O3 is linear like CO2. SiCl4 is tetrahedral because Si has 4 Cl bound to it.

SiCl4: The central atom has no lone pair and there are four bond pairs. The molecule has a grand total of 36 electrons: 3 lone pairs on each chlorine, 2 lone pairs on the central iodine, and 8 bonding electrons.

109.5 degrees

180 degrees

180o

120°

109.5 °

1 Answer. BF3 has trigonal plannar structure all the three B−F bonds lie in plane and thus p-orbitals of boron and fluorine become parallel. Fluorine donates its lone pair to boron and this bonding is called backbonding.

Nitrogen has lone pair (N atom is sp2 hybridised) and silicon has 2 d orbital so the lone pair of Nitrogen makes bond with empty d orbital of silicon therefore back bonding happens Silicon belongs to the group carbon . …

B in BF3 has an empty p orbitals which it utilises in back bonding. However, the electron deficient nature of BF is satisfied by a F. Now, BF4- doesnot have any vacant p orbital and therefore it cannot take part in back bonding.

Let us discuss how back bonding occurs in BF3 molecules. -BF3 molecule has 2p orbitals of each fluorine which has fully filled orbitals and one of the 2p orbital of boron atoms is vacant. Back bonding BF3 does not affect the bond angle, planarity and the geometry of the molecule. So, the correct answer is “Option C”.

This type of bonding occurs between atoms in a compound in which one atom has lone pair of electron and the other has vacant orbital placed adjacent to each other. A compound with back bonding has pi-bonding character since it results after formation of sigma bond.

Most typically it involves the breaking of a carbon-hydrogen bond and then using what were the C−H bonding electrons to stabilize an adjacent charge. In resonance structures, we can’t move nuclei, but we can move electrons.

Back bonding occurs between the same size of orbitals of small atoms. 3p orbitals of Al and Cl are comparatively larger. So, back bonding doesn’t occur in AlCl3.

(B) The back bonding in P(SiH3)3 isn’t significant like in N(SiH3)3 bcz P has its 3p orbital (large and not so efficient) whereas N has its 2p orbital, more efficient bonding. (C) Back bonding in N(SiH3)3 so N is sp2 hybridised, hence planar.