Answer:
1. A.
All statements are correct.
2.E
Explanation:
I. The hybridization of boron in BF3 is sp2
Hybridization is simply the mixing of orbitals to form new orbitals. In the above instance, one s orbital had mixed with 2 p orbitals.
Firstly, to get this hybridization, we draw the Lewis structure I.e using dots to represent the number of valence electrons on the atoms.
Boron has five valence electrons and hence it needs three more to complete an octet. This can be completed by sharing one electron from each of the fluorine atom.
Now, the hybridization is sp2 because a single pi bond is required for the double bond between the boron and only three sigma bonds are formed per boron atom.
II. The molecule XeF4 is non polar
It is non polar because it has no plane of symmetry.
III. The bond order of N2 is three.
Bond order is mathematically equal to = (number of electrons in the bonding orbital - number of electrons in the non bonding orbital)/2
The number of electrons in the bonding are 6 while the number of electrons in the nonbonding orbital is zero. Calculating this yields 6/2 = 3
IV. The molecule HCN has 2 pi bonds and 2 sigma bonds
2. Hybridization of I3- is sp3d
To know the number of hybrid bonds, we simply add the number of lone pairs with that of the neighbours and that's 3 + 2 = 5
Thus, the hybridization is sp3d
