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What is the electron sea model for bonding in metals?

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 33E Chapter 9

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 33E

Problem 33E

What is the electron sea model for bonding in metals?

Step-by-Step Solution:
Step 1 of 3

Solution:Step-1 (a) BrF i4square planar where as BF is tetrahe4alBr has 7 valence electrons in its ground state electronic configuration whereas B only has three.Explanation:If you look at the positions of Br and B on the periodic table, they have ground state electronconfigurations of the following:For Br,[Ar]4s23d104p5For B,[He]2s22p1Looking at these electron configurations, it can be seen that the valence shell of bromine (n=4)has 7 electrons, 2 from the s subshell, and 5 in the p subshell.Since Br is the central atom, and is bonded to four F atoms (who also have 7 valence electrons),4 of the 7 electrons in the valence shell of Br will be bonded to an F atom. This means that 3electrons will not be in a bond. Since the compound is negatively charged, this means theadditional electron (carrying the negative charge) will be paired with one of the 3 lone electrons.Drawing this structure out, you will see that the four F's will align in a square plane, and lonepairs will be on either side of the plane. Hence, square planar, as shown below where green dotscame from Br, blue from F, and red is the extra electron providing the overall negative charge.Looking at BF , B only provides 3 valence electrons, so 3 of the F's will bond directly to these, 4and the fourth F will occupy the open p-orbital of B.Again, an extra electron shown in red that is not present in the ground state configurations willprovide the negative charge (shown in red)\nHence in BF , th4e are 4 bonded pairs on the central atom, and there is no lone pair of electronson it, so that the shape of the molecule is tetrahedral. But in BrF , there are 4 bonded pair on 4the central atom along with 2 lone pairs, which causes repulsion between the bond pairs and lonepairs resulting into square planar geometry of the molecule.Step-2(b) The H-X-H bond angles of H O, H S and H2Se wil2vary depend2n the electronegativity ofthe central atoms. The Lewis structure of the above molecules are,Electronegativity is defined as a measure of the tendency of an atom to attract a bonding pair ofelectrons towards itself. As we move down a group the electronegativity decreases whereasalong a period (left-right) the electronegativity increases.Among H O, H2S and 2 Se , O, S2nd Se all are present in same group (group 16) in periodictable. O has higher electronegativity , than second one is S and Se has least electronegativityamong all the three atoms. So O can attract the electron pair more strongly as compared to theother. So the tendency of the bond pairs to retain in maximum repulsion will be higher in case ofH 2molecule , which makes the bond angle greater. Similarly due to low electronegativity of Sthen O, its bond angle will less then H O and the ele2ronegativity of Se is lesser than S, so itsbond angle is too small as compared to the two other molecules.Hence the decreasing order of bond angles will be H2 >H S 2 Se 2

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Chapter 9, Problem 33E is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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What is the electron sea model for bonding in metals?