Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.61P

(a)

Interpretation Introduction

Interpretation:

The bond strength, bond length, and bond order of NN bond in N2H4, N2H2 and N2 is to be compared.

Concept introduction:

To draw the Lewis structure of the molecule there are following steps:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

Bond strength is directly proportional to the bond order. While the bond length is inversely proportional to the bond order thus higher bond order means smaller bond length.

A single bond is weaker and longer than a double bond and a triple bond is most stronger and shorter.

(a)

Expert Solution
Check Mark

Answer to Problem 10.61P

The bond order of the NN bond in N2H4, N2H2 and N2 are 1,2 and 3 respectively.

Therefore the order of NN bond strength is as follows:

  N2H4<N2H2<N2

And the order of NN bond length is as follows:

  N2H4>N2H2>N2

Explanation of Solution

Both N atoms are central atoms and other atoms are evenly distributed.

The total number of valence electrons of N2H4 is calculated as,

  Total valence electrons(TVE)=[[(Total number of N atom)(Valence electrons of N)]+[(Total number of H atom)(Valence electrons of H)]]        (1)

Substitute 2 for the total number of N atom, 5 for valence electrons of N, 4 for the total number of H and 1 for the valence electrons of H in equation (1).

  Total valence electrons(TVE)=[(2)(5)+(4)(1)]=14

N2H4 has 14 valence electrons, from which ten are used in the formation of all single bonds. The remaining 4 electrons are used in completing the octets of both the N atoms.

Lewis structure of N2H4,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.61P , additional homework tip  1

Therefore the bond order of the NN bond in N2H4 is 1.

The total number of valence electrons of N2H2 is calculated as follows:

Substitute 2 for the total number of N atom, 5 valence electrons of N, 2 for the total number of H and 1 for the valence electrons of H in equation (1).

  Total valence electrons(TVE)=[(2)(5)+(2)(1)]=12

N2H2 has 12 valence electrons, from which six are used in the formation of all single bonds. From the 12 valence electrons remaining 6 electrons are not able to fulfill the octets of both N atoms due to which formation of NN double bonds takes place.

Lewis structure of N2H2,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.61P , additional homework tip  2

Therefore the bond order of the NN bond in N2H2 is 2

The total number of valence electrons of N2 is calculated as,

  Total valence electrons(TVE)=[(Total number of N atom)(Valence electrons of N)]        (2)

Substitute 2 for the total number of N atom, and 5 valence electrons of in equation (2).

  Total valence electrons(TVE)=[(2)(5)]=10

N2 has 10 valence electrons, from which two electrons are used to place a single bond between both N atoms. Remaining four pairs of electrons are not enough to complete the octets of both N atoms the requirement is of six electrons. Hence two lone pairs become bonding pair and a triple bond will form.

Lewis structure of N2 is,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.61P , additional homework tip  3

Therefore the bond order of the NN bond in N2 is 3

In Hydrazine (N2H4), both N atoms are attached with single bonds and a single bond is weaker and longer. In Diazene (N2H2) both N atoms are connected with a double bond and its strength is more and bond length is shorter than the Hydrazine. Nitrogen has a triple bond. The triple bond is most stronger and shorter among all three.

Thus the order of NN bond strength is N2H4<N2H2<N2.

And the order of NN bond length is N2H4>N2H2>N2.

Conclusion

Bond strength is directly proportional to the bond order and bond length is inversely proportional to bond order. Hydrazine has a weakest and longest NN bond and nitrogen has the strongest and NN weakest bond.

Interpretation Introduction

Interpretation:

The Lewis structure for tetrazene (H2NNNNH2) is to be drawn. Also ΔHrxn° for the decomposition of tetrazene is to be calculated.

Concept introduction:

To draw the Lewis structure of the molecule following steps are used.

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

The heat of the reaction (ΔHrxn°) is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. ΔHrxn° is negative for exothermic reaction and ΔHrxn° is positive for an endothermic reaction.

The formula to calculate ΔHrxn° of reaction is as follows:

  ΔHrxn°=ΔHreactant bond broken°+ΔHproduct bond formed°

Or,

  ΔHrxn°=BEreactant bond brokenBEproduct bond formed

The bond energy of reactants is positive and the bond energy of products is negative.

Expert Solution
Check Mark

Answer to Problem 10.61P

The Lewis structure for Tetrazene is

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.61P , additional homework tip  4

The heat of reaction (ΔHrxn°) for the decomposition of Tetrazene is  –367 kJ.

Explanation of Solution

The total number of valence electrons of N4H4 is calculated as follows:

  Total valence electrons(TVE)=[[(Total number of N atom)(Valence electrons of N)]+[(Total number of H atom)(Valence electrons of H)]]        (3)

Substitute 4 for the total number of N atom, 5 for valence electrons of N, 4 for the total number of H and 1 for the valence electrons of H in equation (3).

  Total valence electrons(TVE)=[(4)(5)+(4)(1)]=24

N4H4 has 24 total number of valence electrons. 7 pairs of electrons are used in the formation of single bonds between the atoms. Remaining 7 pairs are used to complete the octets of all the N atoms but one N remain short of two electrons due to which one lone pair is used to make a double bond.

The Lewis structure of Tetrazene is,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.61P , additional homework tip  5

The given chemical equation for the decomposition of Tetrazene is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.61P , additional homework tip  6

The number of broken bonds is 4 NH bonds, 2 NN bonds, and 1 N=N bond.

The number of bonds formed is 4 NH bonds, 1 NN bond, and 1 NN bond.

The formula to calculate the enthalpy of the given reaction is as follows:

  ΔHrxn°=(4BENH+2BENN+1BEN=N)(4BENH+1BENN+1BENN)        (4)

Substitute 391 kJ/mol for BENH, 160kJ/mol for BENN, 418kJ/mol for BEN=N and 965kJ/mol for BENN in the equation (4).

  ΔHrxn°=[((4 mol)(391kJ/mol)+(2 mol)(160kJ/mol)+(1 mol)(418kJ/mol))(4 mol)(391kJ/mol)+(1 mol)(160kJ/mol)+(1 mol)(965kJ/mol)]=367kJ

Conclusion

Only one Lewis structure is possible for tetrazene.

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Chapter 10 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 10.2 - Prob. 10.6AFPCh. 10.2 - Prob. 10.6BFPCh. 10.2 - Prob. 10.7AFPCh. 10.2 - Prob. 10.7BFPCh. 10.2 - Prob. 10.8AFPCh. 10.2 - Prob. 10.8BFPCh. 10.3 - Prob. 10.9AFPCh. 10.3 - Prob. 10.9BFPCh. 10.3 - Prob. B10.1PCh. 10.3 - Prob. B10.2PCh. 10 - Prob. 10.1PCh. 10 - When is a resonance hybrid needed to adequately...Ch. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Draw a Lewis structure for (a) SiF4; (b) SeCl2;...Ch. 10 - Draw a Lewis structure for (a) ; (b) C2F4; (c)...Ch. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.11PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - Molten beryllium chloride reacts with chloride ion...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Phosgene is a colorless, highly toxic gas that was...Ch. 10 - If you know the formula of a molecule or ion, what...Ch. 10 - In what situation is the name of the molecular...Ch. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Consider the following molecular shapes. (a) Which...Ch. 10 - Use wedge-bond perspective drawings (if necessary)...Ch. 10 - Prob. 10.33PCh. 10 - Determine the electron-group arrangement,...Ch. 10 - Determine the electron-group arrangement,...Ch. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Determine the shape, ideal bond angle(s), and the...Ch. 10 - Prob. 10.41PCh. 10 - Determine the shape around each central atom in...Ch. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Arrange the following ACln species in order of...Ch. 10 - State an ideal value for each of the bond angles...Ch. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - How can a molecule with polar covalent bonds not...Ch. 10 - Prob. 10.54PCh. 10 - Consider the molecules SCl2, F2, CS2, CF4, and...Ch. 10 - Consider the molecules BF3, PF3, BrF3, SF4, and...Ch. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - When SO3 gains two electrons, forms. (a) Which...Ch. 10 - The actual bond angle in NO2 is 134.3°, and in it...Ch. 10 - Prob. 10.69PCh. 10 - Propylene oxide is used to make many products,...Ch. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - A gaseous compound has a composition by mass of...Ch. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Pure HN3 (atom sequence HNNN) is explosive. In...Ch. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Oxalic acid (H2C2O4) is found in toxic...Ch. 10 - Prob. 10.87PCh. 10 - Hydrazine (N2H4) is used as a rocket fuel because...Ch. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Consider the following molecular shapes: Match...Ch. 10 - Prob. 10.93PCh. 10 - Prob. 10.94PCh. 10 - Prob. 10.95PCh. 10 - Phosphorus pentachloride, a key industrial...
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