(I assume that the 4 directions north-south-east-west are meant with respect to the wire seen from the top.)
We can use the right-hand rule to understand the direction of the magnetic field generated by the wire. The thumb follows the direction of the current in the wire (upward), while the other fingers give the direction of the field in every point around the wire. Seen from the top, the field has an anti-clockwise direction. Therefore, if we take a point at east with respect to the wire, in this point the field has direction south.
OF2 -
<span>O has 6 electrons in outer shell and F has 7 in its outer shell </span>
<span>Therefore, you have to account for 20 electrons total in the </span>
<span>structure (7+7+6 = 20) </span>
<span>therefore draw it linear first. F ---- O-----F </span>
<span>The two bonds take care of 4 electrons now you have to add another 16. </span>
<span>Therefore 3 lone pairs on each F and 2 lone pair on O. </span>
<span>If you check for formal charges, all the atoms are neutral </span>
<span>F will have 3 lone pairs + 1 bond = 7 electrons (bond = 1/2 electron for formal charge distribution) therefore both the F's are neutral </span>
<span>Now look at the O: it should have 6.. it has two lone pair and 2 bonds = 4 electrons and 2 bonds = 1 electron each = 2 electrons from bonds = 6 total electrons for formal charge which is exactly the # it should have. There is no need for any double bond in this as there are no charges to be separated. </span>
<span>Now if u look at the # of domains around O you will see if you include the lone pairs it has a sp3 hybridization (4 domains) therefore a tetrahedron which has 2 lone pairs and 2 bonds.. since there are two lone pairs, the lone pair/bond pair repulsion is so high it is going to repel the two Fluorines and form a bent structure, looks a lot like H2O. </span>
change of an object relative to the position of another object.
