Chapter 18, Problem 18.18CP

(a)

Interpretation Introduction

Interpretation:

The binary hydrides shown in the Figure 1 has to be identified as ionic, covalent and interstitial.

Concept Introduction:

The compound containing hydrogen and just one other element is called as binary hydride.  Type of hydride formation depends upon the element present in the group.  Types of binary compounds are, ionic hydride, metallic hydride and covalent hydride.

Ionic hydride: Ionic hydrides are formed by alkali metals and heavier alkaline earth metals.  They contain cations and ${\text{H}}^{\text{-}}$ anions in face centered cubic crystal.

Covalent hydride: Covalent hydrides are formed by non-metals.  These compound contains hydrogen which is bonded to another element by covalent bond.  Most of the covalent hydrides consists of separate, small molecule have relatively weak intermolecular force of attraction, so they are gas or volatile liquid at normal temperature.

Metallic hydride: Metallic hydrides are formed by transition metals, lanthanides and actinide metals in which hydrogen will be present in variable amount.  Metallic hydride has general formula of ${\text{AH}}_{\text{x}}$.  They are often called as interstitial hydride because they consists of crystal lattice of metals with smaller hydrogen occupying holes or interstitial.

Answer to Problem 18.18CP

In Figure 1,

$\left(1\right)$ is found as covalent hydride

$\left(2\right)$ is found as ionic hydride

$\left(3\right)$ is found as covalent hydride

$\left(4\right)$ is found as interstitial hydride

Explanation of Solution

Given:

Given representation is shown in Figure 1.

Figure 1

In this picture, ivory spheres denotes Hydrogen atoms or ions and purple spheres denotes an atom or ion of the element.

Picture (1) and (3) shows that hydrides (smaller atoms) are attached to the central atom, smaller molecules have relatively weak intermolecular force of attraction.  This type of hydride are known as covalent hydride.

Picture (4) shows interstitial arrangement of hydrogen atom.  They are often called as interstitial hydride because they consists of crystal lattice of metals with smaller hydrogen occupying holes or interstitial.

Picture (2) shows that it is slice of face centred arrangement.  Ionic hydride only gives face centred arrangement.  Thus it is concluded as ionic hydride formed by alkali metals.

So, picture $\left(1\right)$ and (3) are covalent hydride, $\left(2\right)$ is ionic hydride and $\left(4\right)$ is interstitial hydride.

(b)

Interpretation Introduction

Interpretation:

The oxidation state of hydrogen and other element in compound (1), (2) and (3) has to be identified.

Concept Introduction:

Oxidation number: Oxidation number is defined as degree of oxidation (that is loss of an electrons) of an atom within the chemical compound.

There is a slight difference between oxidation state and oxidation number.  In oxidation state, the electronegativity of an atom in a bond should be considered.  However, while describing oxidation number, electronegativity will not be considered.

Answer to Problem 18.18CP

In compound (1), oxidation state of hydride is $+1$ and oxidation state of other elements is $-3$

In compound (2), oxidation state of hydride is $-1$ and oxidation state of other elements is $+1$

In compound (3), oxidation state of hydride is $+1$ and oxidation state of other elements is $-2$

Explanation of Solution

Given:

Given representation is shown in Figure 1.

Figure 1

In this picture, ivory spheres denotes Hydrogen atoms or ions and purple spheres denotes an atom or ion of the element.

Picture (1) illustrate a slice of the structure of nonstoichiometric interstitial hydride.

Calculation of oxidation state of hydride and other element in each compound as follows,

Compound (1):

Generally oxidation state of hydride in covalent compound is $+1$ and other element is calculated as follows,

Consider x as oxidation state of other element in the compound.

$\begin{array}{l}\text{x + 3}\left(\text{+1}\right)\text{= 0}\\ \text{x = -3}\end{array}$

Therefore, the oxidation state of hydride in compound (1) is $+1$ and other element is $-3$

Compound (2):

Generally oxidation state of hydride in ionic compound is $-1$ and other element is calculated as follows,

Consider x as oxidation state of other element in the compound.

$\begin{array}{l}\text{x + }\left(\text{-1}\right)\text{= 0}\\ \text{x = +1}\end{array}$

Therefore, the oxidation state of hydride in compound (2) is $-1$ and other element is $+1$

Compound (3):

Generally oxidation state of hydride in covalent compound is $+1$ and other element is calculated as follows,

Consider x as oxidation state of other element in the compound.

$\begin{array}{l}\text{x + 2}\left(\text{+1}\right)\text{= 0}\\ \text{x = -2}\end{array}$

Therefore, the oxidation state of hydride in compound (3) is $+1$ and other element is $-2$