Chemistry & Chemical Reactivity, Hybrid Edition (with OWLv2 24-Months Printed Access Card)
Chemistry & Chemical Reactivity, Hybrid Edition (with OWLv2 24-Months Printed Access Card)
9th Edition
ISBN: 9781285462530
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 21.2, Problem 3CYU

(a)

Interpretation Introduction

Interpretation: The compound or ion of a second period element, that has formula and lewis structure analogous to given formula PH4+  has to be predicted.

Concept introduction: The elements belonging to the same group of periodic table have same number of electrons in their valence shell. Thus, their physical and chemical properties are also similar.

These elements can exist in similar oxidation state due to same number of valence electrons in the outermost shell. Hence, the elements of same group form compounds with same formula.

If two elements belongs to nA group of periodic table, then both the elements have +n as highest oxidation number. This is because the highest oxidation number is equal to the group number. So both the elements of same group  have same oxidation number and forms analogous compound of the same chemical formula. The other possible oxidation numbers apart from the highest oxidation number are as per the number of valence electrons in the outermost shell.

However, the lighter elements of the group do not expand their octet while the heavier elements can expand their octet and form compounds with different oxidation numbers. So if the elements belong to group 5A of periodic table then the lighter element only has oxidation number of +3, which is equal to number of electrons in outermost shell. But the heavier element can have both oxidation number of +3 and +5.

So, the same number of valence electrons in outermost shell of the elements in same group is responsible for the similar chemical properties of these elements. Thus they also form compounds with same formula.

(b)

Interpretation Introduction

Interpretation: The compound or ion of a second period element, that has formula and lewis structure analogous to given formula S22  has to be predicted.

Concept introduction: The elements belonging to the same group of periodic table have same number of electrons in their valence shell. Thus, their physical and chemical properties are also similar.

These elements can exist in similar oxidation state due to same number of valence electrons in the outermost shell. Hence, the elements of same group form compounds with same formula.

If two elements belongs to nA group of periodic table, then both the elements have +n as highest oxidation number. This is because the highest oxidation number is equal to the group number. So both the elements of same group  have same oxidation number and forms analogous compound of the same chemical formula. The other possible oxidation numbers apart from the highest oxidation number are as per the number of valence electrons in the outermost shell.

However, the lighter elements of the group do not expand their octet while the heavier elements can expand their octet and form compounds with different oxidation numbers. So if the elements belong to group 5A of periodic table then the lighter element only has oxidation number of +3, which is equal to number of electrons in outermost shell. But the heavier element can have both oxidation number of +3 and +5.

So, the same number of valence electrons in outermost shell of the elements in same group is responsible for the similar chemical properties of these elements. Thus they also form compounds with same formula.

(c)

Interpretation Introduction

Interpretation: The compound or ion of a second period element, that has formula and lewis structure analogous to given formula P2H4  has to be predicted.

Concept introduction: The elements belonging to the same group of periodic table have same number of electrons in their valence shell. Thus, their physical and chemical properties are also similar.

These elements can exist in similar oxidation state due to same number of valence electrons in the outermost shell. Hence, the elements of same group form compounds with same formula.

If two elements belongs to nA group of periodic table, then both the elements have +n as highest oxidation number. This is because the highest oxidation number is equal to the group number. So both the elements of same group  have same oxidation number and forms analogous compound of the same chemical formula. The other possible oxidation numbers apart from the highest oxidation number are as per the number of valence electrons in the outermost shell.

However, the lighter elements of the group do not expand their octet while the heavier elements can expand their octet and form compounds with different oxidation numbers. So if the elements belong to group 5A of periodic table then the lighter element only has oxidation number of +3, which is equal to number of electrons in outermost shell. But the heavier element can have both oxidation number of +3 and +5.

So, the same number of valence electrons in outermost shell of the elements in same group is responsible for the similar chemical properties of these elements. Thus they also form compounds with same formula.

(d)

Interpretation Introduction

Interpretation: The compound or ion of a second period element, that has formula and lewis structure analogous to given formula PF3  has to be predicted.

Concept introduction: The elements belonging to the same group of periodic table have same number of electrons in their valence shell. Thus, their physical and chemical properties are also similar.

These elements can exist in similar oxidation state due to same number of valence electrons in the outermost shell. Hence, the elements of same group form compounds with same formula.

If two elements belongs to nA group of periodic table, then both the elements have +n as highest oxidation number. This is because the highest oxidation number is equal to the group number. So both the elements of same group  have same oxidation number and forms analogous compound of the same chemical formula. The other possible oxidation numbers apart from the highest oxidation number are as per the number of valence electrons in the outermost shell.

However, the lighter elements of the group do not expand their octet while the heavier elements can expand their octet and form compounds with different oxidation numbers. So if the elements belong to group 5A of periodic table then the lighter element only has oxidation number of +3, which is equal to number of electrons in outermost shell. But the heavier element can have both oxidation number of +3 and +5.

So, the same number of valence electrons in outermost shell of the elements in same group is responsible for the similar chemical properties of these elements. Thus they also form compounds with same formula.

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Chapter 21 Solutions

Chemistry & Chemical Reactivity, Hybrid Edition (with OWLv2 24-Months Printed Access Card)

Ch. 21.8 - Prob. 4QCh. 21.8 - Prob. 3RCCh. 21.11 - Prob. 1QCh. 21.11 - Prob. 2QCh. 21 - Give examples of two basic oxides. Write equations...Ch. 21 - Prob. 2PSCh. 21 - Prob. 3PSCh. 21 - Prob. 4PSCh. 21 - Prob. 5PSCh. 21 - Prob. 6PSCh. 21 - For the product of the reaction you selected in...Ch. 21 - For the product of the reaction you selected in...Ch. 21 - Prob. 9PSCh. 21 - Prob. 10PSCh. 21 - Place the following oxides in order of increasing...Ch. 21 - Place the following oxides in order of increasing...Ch. 21 - Prob. 13PSCh. 21 - Prob. 14PSCh. 21 - Prob. 15PSCh. 21 - Prob. 16PSCh. 21 - Prob. 17PSCh. 21 - Prob. 18PSCh. 21 - Prob. 19PSCh. 21 - Prob. 20PSCh. 21 - Prob. 21PSCh. 21 - Write balanced equations for the reaction of...Ch. 21 - Prob. 23PSCh. 21 - (a) Write equations for the half-reactions that...Ch. 21 - When magnesium bums in air, it forms both an oxide...Ch. 21 - Prob. 26PSCh. 21 - Prob. 27PSCh. 21 - Prob. 28PSCh. 21 - Calcium oxide, CaO, is used to remove SO2 from...Ch. 21 - Prob. 30PSCh. 21 - Prob. 31PSCh. 21 - The boron trihalides (except BF3) hydrolyze...Ch. 21 - When boron hydrides burn in air, the reactions are...Ch. 21 - Prob. 34PSCh. 21 - Write balanced equations for the reactions of...Ch. 21 - Prob. 36PSCh. 21 - Prob. 37PSCh. 21 - Alumina, Al2O3, is amphoteric. Among examples of...Ch. 21 - Prob. 39PSCh. 21 - Prob. 40PSCh. 21 - Describe the structure of pyroxenes (see page...Ch. 21 - Describe how ultrapure silicon can be produced...Ch. 21 - Prob. 43PSCh. 21 - Prob. 44PSCh. 21 - Prob. 45PSCh. 21 - Prob. 46PSCh. 21 - Prob. 47PSCh. 21 - The overall reaction involved in the industrial...Ch. 21 - Prob. 49PSCh. 21 - Prob. 50PSCh. 21 - Prob. 51PSCh. 21 - Prob. 52PSCh. 21 - Prob. 53PSCh. 21 - Prob. 54PSCh. 21 - Prob. 55PSCh. 21 - Sulfur forms a range of compounds with fluorine....Ch. 21 - The halogen oxides and oxoanions are good...Ch. 21 - Prob. 58PSCh. 21 - Bromine is obtained from brine wells. The process...Ch. 21 - Prob. 60PSCh. 21 - Prob. 61PSCh. 21 - Halogens combine with one another to produce...Ch. 21 - The standard enthalpy of formation of XeF4 is 218...Ch. 21 - Draw the Lewis electron dot structure for XeO3F2....Ch. 21 - Prob. 65PSCh. 21 - Prob. 66PSCh. 21 - Prob. 67GQCh. 21 - Prob. 68GQCh. 21 - Consider the chemistries of the elements...Ch. 21 - When BCl3 gas is passed through an electric...Ch. 21 - Prob. 71GQCh. 21 - Prob. 72GQCh. 21 - Prob. 73GQCh. 21 - Prob. 74GQCh. 21 - Prob. 75GQCh. 21 - Prob. 76GQCh. 21 - Prob. 77GQCh. 21 - Prob. 78GQCh. 21 - Prob. 79GQCh. 21 - Prob. 80GQCh. 21 - Prob. 81GQCh. 21 - Prob. 83GQCh. 21 - Prob. 84GQCh. 21 - A Boron and hydrogen form an extensive family of...Ch. 21 - In 1774, C. Scheele obtained a gas by reacting...Ch. 21 - What current must be used in a Downs cell...Ch. 21 - The chemistry of gallium: (a) Gallium hydroxide,...Ch. 21 - Prob. 89GQCh. 21 - Prob. 90GQCh. 21 - Prob. 91GQCh. 21 - Prob. 92GQCh. 21 - Prob. 93ILCh. 21 - Prob. 94ILCh. 21 - Prob. 95ILCh. 21 - Prob. 96ILCh. 21 - Prob. 97ILCh. 21 - Prob. 98ILCh. 21 - Prob. 99SCQCh. 21 - Prob. 100SCQCh. 21 - Prob. 101SCQCh. 21 - Prob. 102SCQCh. 21 - Prob. 103SCQCh. 21 - Prob. 104SCQCh. 21 - Prob. 105SCQCh. 21 - Prob. 106SCQCh. 21 - Prob. 107SCQCh. 21 - Prob. 108SCQCh. 21 - Prob. 109SCQCh. 21 - Prob. 110SCQCh. 21 - Comparing the chemistry of carbon and silicon. (a)...Ch. 21 - Prob. 112SCQCh. 21 - Xenon trioxide, XeO3, reacts with aqueous base to...
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