Onceyou have ascertained the Lewis dot structure of something, what do youdo with it? In other parts of this web page, we discuss how the geometryaround a central atom is determined by the number of objects surroundingit. (Objects can be bonds or lone pairs. Single, double, andtriple bonds all count as a single "object.") Molecular geometryis determined by how the electrons about an atom are distributed.Lewis structures get us started; the VSEPR model takes us a step further.

You are watching: What is the value of the bond angle in icl2−

Beforewe get started, let"s have a big printout of a major table. Thisis for the convenience of web users and, at the same time, should infuriatethose wasting time, paper, and ink printing these notes!  Thereis enough information in the answers below to enable you to draw Lewisstructures; you are urged to do so!Problem7.67: What shape do you expect for molecules that meet thefollowing descriptions?
 (a) A central atom with two lone pairs and three bonds toother atoms.This would be T-shaped. Referto table above. (b) A central atom with two lone pairs and two bonds to otheratoms.This would be bent (e.g.,water). (c) A central atom with two lone pairs and four bonds toother atoms.Referring to the table above, yousee that his would be square planar.
Problem7.70:
What shape do you expect for each of the followingmolecules?
 (a) H2Se 2 bonded atoms + 2 lone pairs:Bent (b) TiCl4 4 bonded atoms + 0 lone pairs:Tetrahedral (c) O3 2 bonded atoms + 1 lone pair:Bent (d) GaH3 3 bonded atoms + 0 lone pairs:Trigonal planar
Problem7.71: What s hapedo you expect for each of the following molecules?
 (a) XeO4 4 bonded atoms + 0 lone pairs:Tetrahedral (b) SO2Cl2 4 bonded atoms + 0 lone pairs:Tetrahedral (c) OsO4 4 bonded atoms + 0 lone pairs:Tetrahedral (d) SeO2 2 bonded atoms + 1 lone pair:Bent
Problem7.73: Predict the shape of each of the following ions:
 (a) NO3- 3 bonded atoms + 0 lone pairs:Trigonal Planar (b) NO2+ 2 bonded atoms + 0 lone pairs:Linear (c) NO2- 2 bonded atoms + 1 lone pair:Bent
Problem7.74: What shape do you expect for each of the followinganions?
 (a) PO43- 4 bonded atoms + 0 lone pairs:Tetrahedral (b) MnO4- 4 bonded atoms + 0 lone pairs:Tetrahedral (c) SO42- 4 bonded atoms + 0 lone pairs:Tetrahedral (d) SO32- 3 bonded atoms + 1 lone pair:Trigonal pyrimidal (e) ClO4- 4 bonded atoms + 0 lone pair:Tetrahedral (f) SCN- 2 bonded atoms + 0 lone pair:Linear
Problem7.76: What bond angles do you expect for each of the following?
 (a) The F-S-F bond angle in SF2 The S is bound to two fluorinesand has two lone pairs of electrons. Thus the bond angle would beexpected to be about 109o. (b) The H-N-N angle in N2H2 N bound two atoms. Each Nhas a single lone pair. Thus, would expect bond angle of about 120o. (c) The F-Kr-F angle in KrF4 Kr bound to four fluorines andcontains two extra pairs of electrons. This molecule would be squareplanar and the bond angle would be 90o. (d) The Cl-N-O angle in NOCl N bound to two atoms (single bondto Cl and double bond to O). Has a lone pair. Thus, bent atabout 120o.
Problem7.77:
What bond angles do you expect for each of the following?
 (a) The Cl-P-Cl anglein PCl6- Phosphorus is bound to 6chlorines and, in the Lewis sstructure, you see no lone pairs. Hence,this is octahedral and the bond angle is 90o. (b) The Cl-I-Cl anglein ICl2- I is bound to Cl and has 3 lonepairs in addition. Thus, it is a trigonal bipyramidal structure;with just three atoms, they would be on the axis and the bond angle wouldbe linear, namely, 180o. (c) The O-S-O angle inSO42- This molecule has found atoms boundto the central sulfur and no extra lone pairs. Thus it is tetrahedraland the angle is approximately 109.47122o.Useful hint: This iscos-1(-1/3)on your calculator. (d) The O-B-O angle inBO33- Boron is bound to 3 oxygens andthere are no lone pairs. Thus, this is trigonal planar and the bondangle is 120o.
Here are the chartsagain for those of you who prefer to scroll down (from above) or don"twant to do too much scrolling for the two remaining problems. Again,if you are printing these, well, ... .  Problem7.78: Acrylonitrile is used as the starting material formanufacturing acrylic fibers. Predict values for all bond anglesin acrylonitrile. Let us label the carbonsfrom left to right as a, b, and c. Thefirst two carbons have three objects about them and the third has two.None of the carbons has any spare lone pairs. (Later on, it willseem quite straightforward to view these as sp2,sp2, and sphybridization, respectively--but we are not that far along yet!)Anyway, here are the various bondangles. Except for a missing lone pair at the right of the terminalN, teh structure as drawn is the Lewis structure.
 Atoms in Bond Approx. Bond Angle H-Ca-H 120o H-Ca-Cb 120o Ca-Cb-Cc 120o Cb-Cc-N 180o Ca-Cb-H 120o H-Cb-Cc 120o
Nothing much to this!Problem7.80: Explain why cyclohexane, a substance that containsa six-membered ring of carbon atoms, is not flat but instead has a puckered,nonplanar, shape. Predict the values of the C-C-C bond angles.

See more: How Much Is A 1943 Silver Quarter Worth, 1943 Quarter All the carbons are surroundby four objects. Thus, the ring would be expected to pucker sinceflatness can occur only if C has three objects around it. The surroundingshere are tetrahedral and one would expect the C-C-C typical bond angleto be approximately 109.47122o.