What hybrid orbitals are used by the metal ion and how many unpaired electrons are present the complex ion [VCl₄]^- with tetrahedral geometry?

(a) sp³; 2 unpaired electrons

(b) sp³; 3 unpaired electrons

(c) sp³d²; 3 unpaired electrons

(d) sp³d²; 4 unpaired electrons

Video transcript

Hey, everyone. Let's look at our next question, identify the hybridization of the central metal ion in the complex. We've got in brackets an overall charge of negative two cobr four and determine the number of unpaired electrons complex adopts a tetrahedral geometry. And our choices are a SP 31 unpaired electron BS P 33 unpaired electrons. CD two sp 32 unpaired electrons or DD two sp 33 unpaired electrons. Well, before we even go any further, we can eliminate choices C and D with that D two sp three hybridization because that's a hybridization for an octahedral geometry, that's six total hybrid orbitals. So we have a tetrahedral geometry. We have only four ligands. So we're definitely going to be looking with a tetrahedral geometry and an sp three hybridization. We just know that right off the bat. So our answer choice has to be either A or B. So we need to look at how many D electrons we have and how they will fill in the orbitals along with the electrons participating in the bonds. So we start with our metal, which is Cobalt, Ceo and just elemental Cobalt has an electron configuration of argon as its nearest noble gas four S 2 3d 7. We need to find out what charge our cobalt will have. And we know our overall charge is negative two. Bromine is a charge of negative one and there's four of them. So we have a charge of negative four coming from the brom means overall charge of negative two. So our cobalt must be positive two. So we know our metal cobalt must be co two plus. So we've lost two electrons. So we write in argon and brackets no more four S electrons, we just have 3d 7. So to look at our orbital arrangement, we would draw our five D orbitals, 12345, we do have crystal field splitting. Um I haven't drawn the orbitals at a different energy level because we have a tetrahedral geometry and tetrahedral arrangements generally are we build with a small delta. So we'll fill in all the orbitals with one electron first before pairing anything up. So the energy level difference between them doesn't, doesn't really matter in terms of how we're gonna fill in our electrons. And then we know we'll have sp three hybridization. So I'll draw on my four S orbital and then my 34 P orbitals. So I have 70 orbitals on my cobalt. So again, small delta. So I fill in five electrons in the five D orbitals and then the last two get paired. So we have two paired electrons and three unpaired electrons. Now, let's look at the electrons coming from our ligand, our leggings. Excuse me, I have four Bromine Liggins, each has a pair of electrons that participates in bonding. So I have eight electrons from the Bromy. And it's important that I know they're in pairs. So they have to be paired in the hybrid orbitals because they're being shared in pairs from the bromine. So I start with all my 3d orbitals are full. So I start with my four S and put a pair on there. That's one pair. I know I have four pairs. Now, I've got three P orbitals left. So my other three pairs fill in the four P orbital. So I have four pairs of electrons shared in P three hybrid orbitals. And when I look at my overall complex, well, all my electrons being shared from the ligands are in pairs, but the electrons from Cobalt are not all in pairs. We have those three unpaired electrons from the Cobalt two plus. So we go over Taran choices and that gives us choice B sp three and three unpaired electrons. See you in the next video.