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Chapter 8: Problem 39

Using only the periodic table as your guide, select the most electronegative atom in each of the following sets: (a) Se, Te, $\mathrm{Br}, \mathrm{I} ;(\mathbf{b}) \mathrm{Be}, \mathrm{Mg}, \mathrm{C}, \mathrm{Si} ;(\mathbf{c}) \mathrm{Al}, \mathrm{Si}, \mathrm{P}, \mathrm{S} ;(\mathbf{d}) \mathrm{O}, \mathrm{P}, \mathrm{Ge}, \mathrm{In}$.

Short Answer

Expert verified
The most electronegative elements in each set are: (a) \(Br\) (b) \(C\) (c) \(S\) (d) \(O\)

Step by step solution

01

Set (a): Se, Te, Br, I

Referring to the periodic table, we can see that Se, Te, Br, and I are in the same period with increasing atomic number. As electronegativity increases from left to right in a period, the most electronegative element in this set is \(Br\).
02

Set (b): Be, Mg, C, Si

In this set, Be and Mg belong to Group 2 and C and Si belong to Group 14. Since electronegativity increases from left to right in a period, C and Si should be more electronegative than Be and Mg. Now, between C and Si, Si is below C in the periodic table, and electronegativity generally decreases down a group. Therefore, the most electronegative element in this set is \(C\).
03

Set (c): Al, Si, P, S

In this set, Al, Si, P, and S are in the same period with increasing atomic number. Electronegativity increases from left to right in a period. Therefore, the most electronegative element in this set is \(S\).
04

Set (d): O, P, Ge, In

In this set, O and P belong to Group 16 and 15 respectively, while Ge and In belong to Group 14 and 13 respectively. Electronegativity increases from left to right in a period, so O and P should be more electronegative than Ge and In. Between O and P, P is below O in the periodic table, and electronegativity generally decreases down a group. Therefore, the most electronegative element in this set is \(O\). To summarize the most electronegative elements in each set: (a) \(Br\) (b) \(C\) (c) \(S\) (d) \(O\)

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Most popular questions from this chapter

In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance structures are possible that satisfy the octet rule? (c) On the basis of the formal charges, which Lewis structure is expected to be dominant for \(\mathrm{BeCl}_{2}\) ?

(a) Which of these compounds is an exception to the octet rule: carbon dioxide, water, ammonia, phosphorus trifluoride, or arsenic pentafluoride? (b) Which of these compounds or ions is an exception to the octet rule: borohydride \(\left(\mathrm{BH}_{4}^{-}\right),\) borazine $\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\right.$ which is analogous to benzene with alternating \(\mathrm{B}\) and \(\mathrm{N}\) in the ring \(),\) or boron trichloride?

Write the Lewis symbol for atoms of each of the following elements: $(\mathbf{a}) \mathrm{Te},(\mathbf{b}) \mathrm{Si},(\mathbf{c}) \mathrm{Kr},(\mathbf{d}) {\mathrm{P}}$.

A new compound is made that has a \(\mathrm{C}-\mathrm{O}\) bond length of $120 \mathrm{pm}$. Is this bond likely to be a single, double, or triple C-O bond?

(a) Draw the best Lewis structure(s) for the nitrite ion, \(\mathrm{NO}_{2}^{-}\). (b) With what allotrope of oxygen is it isoelectronic? (c) What would you predict for the lengths of the bonds in \(\mathrm{NO}_{2}^{-}\) relative to \(\mathrm{N}-\mathrm{O}\) single bonds and double bonds?
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