Problem

What are the electron-pair geometry and the molec…

11:02

Problem 92 Hard Difficulty

Identify the electron pair geometry and the molecular structure of each of the following molecules or ions:
(a) $\mathrm{IF}_{6}+$
(b) $\mathrm{CF}_{4}$
(c) $\mathrm{BF}_{3}$
(d) $\mathrm{SiF}_{5}-$
(e) $\mathrm{BeCl}_{2}$

Answer

(a)
The central atom of the molecule $\mathrm{IF}_{6}^{+},$ iodine has only 6 valence electrons due to the positive
sign. Because of the presence of 6 fluoride ions, there will be no lone pair present on the iodine
atom. Hence the electron-pair geometry will be the same as the molecular structure. It will be
Octahedral.
(b)
The central atom of $\mathrm{CF}_{4}$ , carbon has no lone pair of electrons because all the 4 valence shell
electrons of carbon are bonded to fluorine atoms. Hence the electron-pair geometry will be the
same as the molecular structure. It will be tetrahedral.
(c)
The central atom of $\mathrm{BF}_{3}$ , boron has no lone pair of electrons because all the 3 valence shell
electrons of boron are bonded to fluorine atoms. Hence the electron-pair geometry will be the
same as the molecular structure. It will be trigonal planar.
(d)
The central atom of $\mathrm{SiF}_{5}^{-},$ silicon has no lone pair of electrons because all the 4 valence shell
electrons are bonded to four fluorine atoms and the electron due to the presence of negative
charge is bonded with the fifth fluorine atom. Hence the electron-pair geometry will be the same
as the molecular structure. It will be trigonal bipyramidal.
(e)
The central atom of BeCl, beryllium has no lone pair of electrons because both the valence
shell electrons are bonded to chlorine atoms. Hence the electron-pair geometry will be the same
as the molecular structure. It will be Linear.

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Video Transcript

to determine the electron pair geometry and the molecular structure of the molecules or Polly Atomic ions that are given in this question. We must first draw the Louis structure, then, from the Lewis structure, used the Vesper model Valence Shell Electron Pair propulsion in order to determine the geometry and structure of the molecule. The first one is I F six. Plus, there are seven Valence electrons from My A dying and then six times seven from floor een minus one. Because of the plus one charge, this sums up to a total of 48 valence electrons. We'll put iodine in the middle, surrounded with six florins that used up 12 valence electrons. So we have 36 left over which we can put around all of the floor. Ain's There are no lone pairs. There are six pairs of electrons surrounding iodine. Six pairs of electrons results in an electron pair, geometry or an electron group. I will call it because sometimes it's not a pair when we have double and triple bonds, so I'll call it an electron group geometry of Octa. He'd RL because there are no lone pairs. The molecular structure is also Octa he'd RL. The next one is CF four Carbon tetra fluoride Carbon has four valence electrons. Flooring has seven. There are four of them for a total of 32 Valence electrons will surround carbon with the four floor Eanes that used up eight valence electrons. We have 24 left over which we can surround the floor Eanes so that they have an octet Carbon. Now, already having an octet. Carbon has four electron groups surrounding it. So it's electron group. Geometry will be Tetra hydro all because there are no lone pairs. The molecular structure is also Tetra hydro. The next one is BF three. Boron has three valence electrons. Florian has seven. There are three of them, giving us a total of 24 valence electrons. We'll put boron in the middle, surrounded with three floor Eanes that used up six valence electrons. So we now have 18 left over, which is just enough to given octet all the floor. Eanes. But what about boron? It doesn't have an octet. Should we form a double bond of one of the floor Eanes with boron? Well, it turns out flooring never forms a double bond, and boron is one of the exceptions to the octet rule, where in some molecules such as this one, it's just fine with six valence electrons. So there are three electron groups. They're all bonding groups surrounded. The boron three electron groups gives us an electron group geometry of tribunal plainer because all the electron groups air bonding, there are no lone pairs. The molecular structure is also a tribunal plainer for the next one. We have s I f five minus are four valence electrons from silicone seven from floor een there five of them, plus one more because of the minus one charge. This gives us a total of 40 valence electrons. We'll put silicone in the middle, put five floor Eames around it that used up 10 valence electrons. We now have 30 left over 30 is just enough to give all of the floor Eanes and octet when all of the Florian's have an octet. We've used up all of our valence electrons, so we'll put it in brackets and put a negative charge on the outside. You'll notice now that there are five electron groups surrounding silicone. Five electron groups results in a tribunal by parameter all electron group geometry and because all of them are bonding pairs, not lone pairs than the molecular structure is equivalent at tribunal by parameter all for the last one, we have BCL two. Beryllium has to valence electrons. Chlorine has seven there, two of them giving us a total of 16 valence electrons. We'll put beryllium in the middle. Put to chlorine is around it that used up four valence electrons, so we have 12 left over, which is just enough to given octet to the chlorine. But beryllium doesn't have an octet. So should we form double bonds here and here. The answer is no, because halogen is rarely form double bonds. And beryllium is one of the exceptions where in some molecules such as this one, it is happy with four valence electrons rather than 18 rather than eight. Because there are two electron groups surrounding beryllium and they're both bonding than the electron group. Geometry is linear and with no lone pairs on beryllium, the molecular structure is also linear

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