This is a chem question that I would like help with

Question

Draw the best Lewis structures for NOF and SeCl4 and include formal charges on atoms with non-zero values. Concisely identify at least four steps of the procedure for drawing the Lewis structures.

Note: In the formula of the first molecule, atoms are in the order of their atomic numbers; the given formula does not necessarily reflect the actual arrangement of atoms in the structure.  

Answer 1

So, to begin with, we cannot easily draw on this Wyzant platform so I'll do my best to "describe" the drawing of the Lewis structures.

Steps to draw Lewis structure

(1) Find total number of valence electrons (ve)

(2) Place the least electronegative atom in the center

(3) Place the other atoms around the center atom

4) Distribute all the valence electrons around the outer elements to satisfy their octet and any electrons that are left over go on the center atom.

5) Check for octet for all atoms (note exceptions for period 3 and above) and rearrange if necessary

To find the molecular geometry refer to VSEPR rules or a table of such structures. The geometry will be determined by the number of bonds and free pairs around the central atom

NOF

step 1: N has 5 ve + O has 6 ve + F has 7 ve = 18 ve total

steps 2 and 3: N is least electronegative so you have O - N - F

step 4: Place 3 electron pairs around the O and the F, leaving one electron pair to go over the N. Can't draw but would look like O with 2 dots on top, to the left and on the bottom, N with 2 dots on top, and F with 2 dots on top, to the left and on the bottom.

step 5: The N in the center has only 6 electrons (not an octet), so move one lone pair from the O to form a double bond between O and N to now have O = N -F

To find the formal charge on each atom, take the number of valence electrons for that atom and subtract the # of nonbonding electrons and then subtract 1/2 the number of bonding electrons.

Formal charge for O = zero

Formal charge for N = zero

Formal charge for F = zero

Molecular geometry is linear

SeCl₄

step 1: Se has 6 ve + Cl has 7 ve (x4) = 28 ve = 34 ve total

step 2 and 3: Se in center with 4 Cl around it

step 4: Place 3 electron pairs around each Cl. This now has used up 32 ve leaving 2 ve to go on the center Se atom.

step 5: The 4 Cl atoms all have an octet and the Se has 10 electrons (an expanded octet). This is ok for Se because it is in period 4 so can violate the octet rule.

Formal charge on each Cl = zero

Formal charge on Se = zero

Molecular geometry is trigonal pyramidal