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Chapter 22: Problem 13

Consider the elements \(\mathrm{O}, \mathrm{Ba}, \mathrm{Co}, \mathrm{Be}, \mathrm{Br}\) , and Se. From this list, select the element that (a) is most electronegative, (b) exhibits a maximum oxidation state of \(+7,(\mathrm{c})\) loses an electron most readily, (\mathbf{d} ) ~ f o r m s ~ \(\pi\) bonds most readily, (e) is a transition metal, \((\mathbf{f})\) is a liquid at room temperature and pressure.

Short Answer

Expert verified
(a) Most electronegative element: Oxygen (O) (b) Maximum oxidation state of +7: Bromine (Br) (c) Loses an electron most readily: Barium (Ba) (d) Forms π bonds most readily: Oxygen (O) (e) Is a transition metal: Cobalt (Co) (f) Is a liquid at room temperature and pressure: Bromine (Br)

Step by step solution

01

(a) Most electronegative element

To find the most electronegative element, we need to look at the periodic table and find the element with the highest electronegativity value. In the given list of elements, the electronegativity values are as follows: O: 3.44 Ba: 0.89 Co: 1.88 Be: 1.57 Br: 2.96 Se: 2.55 From these values, oxygen (O) has the highest electronegativity value, 3.44. Therefore, oxygen is the most electronegative element among the given elements.
02

(b) Maximum oxidation state of +7

To find the element that exhibits a maximum oxidation state of +7, we need to consider the elements' electron configurations and the number of valence electrons. Among the given elements, bromine (Br) is the one that can exhibit a maximum oxidation state of +7 due to its electron configuration [Ar] 3d10 4s2 4p5. In its maximum oxidation state, it loses all seven valence electrons to achieve a more stable electronic configuration.
03

(c) Loses an electron most readily

In order to determine which element loses an electron most readily, we should look for the element with the lowest ionization energy, as it represents the energy required to remove an electron from the atom. Among the given elements, the ionization energies are: O: 1314 kJ/mol Ba: 502.9 kJ/mol Co: 760 kJ/mol Be: 899 kJ/mol Br: 1139.9 kJ/mol Se: 941 kJ/mol From these values, barium (Ba) has the lowest ionization energy. Therefore, Ba loses an electron most readily among the given elements.
04

(d) Forms π bonds most readily

To determine which element forms π bonds most readily, we need to consider the elements' electron configurations and their ability to form double or triple bonds. Elements with unpaired electrons in the p orbitals are most likely to form π bonds. In the given list of elements, oxygen (O) has the electron configuration [He] 2s2 2p4, which has two unpaired electrons in the p orbital. Therefore, oxygen forms π bonds most readily.
05

(e) Is a transition metal

Transition metals are elements with partially filled d orbitals. Among the given elements, only cobalt (Co) is a transition metal with an electron configuration of [Ar] 3d7 4s2.
06

(f) Is a liquid at room temperature and pressure

To determine which element is a liquid at room temperature and pressure, we need to look at the physical state of each element at standard conditions. Among the given elements, bromine (Br) is a liquid at room temperature and pressure.

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Most popular questions from this chapter

The dissolved oxygen present in any highly pressurized, high-temperature steam boiler can be extremely corrosive to its metal parts. Hydrazine, which is completely miscible with water, can be added to remove oxygen by reacting with it to form nitrogen and water. (a) Write the balanced equation for the reaction between gaseous hydrazine and oxygen.(b) Calculate the enthalpy change accompanying this reaction. (c) Oxygen in air dissolves in water to the extent of 9.1 ppm at \(20^{\circ} \mathrm{C}\) at sea level. How many grams of hydrazine are required to react with all the oxygen in 3.0 \(\times 10^{4} \mathrm{L}\) (the volume of a small swimming pool) under these conditions?

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