In N2O, nitrogen is the central atom and the oxygen atom is terminal. In OF2, however, oxygen is the central atom. Use formal charges to explain why.

Video transcript

Hey everyone. So here it says why does the best lewis structure of beryllium bromide contain two single B E B R bonds instead of two double B E B. R bonds. Alright, so we're going to say here that beryllium is in group two A. So it has two valence electrons. And we say normally if a halogen is not in the center, if the surrounding element it only makes single bonds. The Brahmins themselves are group seven a. So they have seven valence electrons. So this is the way it should be drawn here. They're talking about double bonds. The beryllium doesn't have enough electrons to do that. So it will borrow some electrons from the bro means okay and to be able to do that, the brawl means would share some electrons and make double bonds with the beryllium. Now again, the first structure is the correct way of drawing this. But let's look at the options are giving us here if be a beryllium bro, might contain two double bonds. Beryllium will be surrounded with more than eight electrons. That's not true Because we see that beryllium has 2468 electrons around it. It's still not violating the octet rule If beryllium bromine contained two double bonds. The summer formal charges would not be equal to zero. Alright, so remember formal charge equals group number and let's do it for beryllium equals group number minus the bonds the element is making plus each non bonding electron. So I'll be too -4 which is -2. And then formal charge of bro. Mean to formal charge of grown men. now Look at either bro mean so one of the Brahmins in group seven a. We see it making two bonds and it has four electrons not bonding. So here that would be plus one. That B plus one minus two plus one. Collectively the overall charge will be equal to zero. Okay, so here uh that's not true. The formal charges would still all add up to zero and beryllium boron might contain two double bonds. The formal charges of beryllium and bromine will be maximized instead of minimized as required by formal charge rules, if we did the formal charge of the original form, it would be 00 and zero. This is the best structure to drop because all the numbers are zero as they should be. So here see seems to be the best answer. Alright. If beryllium boron might contain two double bonds, both beryllium and browning will violate the octet rule. That's not true. We see that beryllium when it's forming two double bonds, it still wouldn't violate the octet rule and neither would the bro means each bromeliad would still have around it 2, 4, 6, 8 electrons. And then finally, if beryllium boron might contain two double bonds. Each of the chlorine atoms would be surrounded by more than eight electrons. First of all, chlorine is not part of this molecule. And even if we changed it to bromine atoms, it still wouldn't be true. We just saw that each bro mean even though they're making double bonds, would still have only eight electrons around them. So out of all the options on the options see is true.