In the previous section, we saw the properties of dihydrogen. In this section, we will see hydrides.
Hydrides
• Dihydrogen reacts with almost all other elements except noble gases.
• After such reactions, binary compounds are formed. (A Binary compound is a compound which has exactly two different elements)
• The binary compounds thus formed are called hydrides.
• Let us denote the element (with which dihydrogen reacts) by the letter ‘E’.
♦ Then the hydride can be expressed as EHx.
✰ Example: MgH2
♦ Or it can be expressed as EmHn
✰ Example: B2H6
◼ Hydrides are classified into three categories:
1. Ionic hydrides
♦ They are known as saline hydrides also.
♦ They are known as saltlike hydrides also.
2. Covalent hydrides
♦ They are known as molecular hydrides also.
3. Metallic hydrides
♦ They are known as non-stoichiometric hydrides also.
♦ They are known as interstitial hydrides also.
Let us see them in detail:
Ionic hydrides
Details can be written in 9 steps:
1. Ionic hydrides are compounds formed when dihydrogen reacts with any of the s-block elements.
2. We know that s-block elements are metals. Metals donate electrons and become positive ions.
• The H atoms are forced to accept those electrons. (These metals are above H in the reactivity series)
• Thus H atoms become H- ions.
3. Positive ions of the metal and negative H- ions combine to form ionic molecules. Let us see some examples:
♦ NaH (sodium hydride) is actually Na+H-
♦ MgH2 (magnesium hydride) is actually Mg2+H2-
✰ Note that, each Mg atom donates two electrons.
✰ But one H atom can accept only one electron.
✰ So two H atoms are required for each Mg atom.
4. Some ionic hydrides show significant covalent character. (We have discussed about 'covalent character in ionic compounds' in an earlier section 4.12)
• LiH, BeH2 and MgH2 are the main hydrides which show covalent character. Li and H are light elements. They have very small size and mass.
• BeH2 and MgH2 have a polymeric structure. This can be explained in steps:
(i) Consider a sample of MgH2.
(ii) In that sample, individual MgH2 molecules will not be present. Instead, long chains will be present.
(iii) Each chain will contain MgH2 molecules attached together.
(iv) The chain is called polymer and single molecule is called monomer.
5. Ionic hydrides are crystalline in nature.
• They are also non-volatile. That means, they do not vaporize at room temperature and pressure.
6. Ionic compounds do not conduct electricity. But in the molten state, they do conduct electricity.
• Electrolysis of a molten ionic hydride will yield dihydrogen gas at the anode.
• This is because, the H- ions in the electrolyte moves towards the anode and get oxidized to H atoms. These H atoms combine to give H2 gas.
2H- (melt) → H2 (g) + 2e-
7. Ionic hydrides react violently with water.
• NaH is an example:
NaH (s) + H2O (l) → NaOH (aq) + H2 (g)
• This reaction can be explained in 2 steps:
(i) Initially, Na+ is stable when it is with H-. So NaH is stable as a whole.
(ii) But when water is available, the Na+ attacks the H2O molecule to obtain OH- ions.
♦ The combination of Na+ and OH-
♦ is more stable than
♦ the combination of Na+ and H-.
8. LiH do not react violently. So it is used in the synthesis of other hydrides.
8LiH + Al2Cl6 → 2LiAlH4 + 6LiCl
2LiH + B2H6 → 2LiBH4
9. Ionic hydrides are stoichiometric compounds. This can be explained in steps:
(i) Consider any one ionic hydride.
(ii) Take different samples of that hydride from different sources.
(iii) What ever be the source, the constituent elements will be present in the same ratio.
• For example:
♦ The ratio between masses of Na and H in all samples of NaH will be 1 : 1
♦ The ratio between masses of Mg and H in all samples of MgH2 will be 1 : 2
Covalent hydrides
This can be written in 5 steps:
1. Dihydrogen forms covalent compounds with most of the p-block elements.
2. We know that, p-block elements are mostly non-metals. We saw that, compounds of dihydrogen with s-block elements (which are metals) are called hydrides. For convenience, the compounds of dihydrogen with non-metals are also called hydrides.
3. Many hydrides of non-metals are familiar to us. CH4, HCl, HF, NH3 are examples.
4. Covalent compounds are more volatile than ionic compounds. The reason can be explained in 2 steps:
(i) In ionic compounds, there is strong inter molecular forces. The ions are arranged in a lattice structure. Greater energy is required to break the lattice structure. So they do not vaporize easily.
(ii) In covalent compounds, the inter molecular forces are not so strong. So they vaporize easily.
5. Covalent hydrides are further classified as:
A. Electron-deficient covalent hydrides
B. Electron-precise covalent hydrides
C. Electron-rich covalent hydrides
We will now see each one in detail:
A. Electron-deficient covalent hydrides
This can be explained using an example. It can be written in 7 steps:
(i) Consider the hydride B2H6 (diborane)
♦ We want to draw the Lewis structure of this molecule.
(ii) The subshell configuration of B is 1s22s22p1
♦ So it has 3 electrons in the outer most shell.
(iii) We know that the H atom has one electron in the outer most shell.
• So if we try to draw the conventional Lewis structure, it will be as shown in the fig.9.1 below:
 |
| Fig.9.1 |
♦ The red dots are the outer most electrons of B.
♦ The green dots are the outer most electrons of H.
(iv) We see that enough number of electrons are not available. This is indicated by the hollow red circles of B.
• Such covalent hydrides are known as electron-deficient covalent hydrides. (We will see the actual structure of B2H6 in higher classes)
(v) Boron belongs to the group 13 of the periodic table. All elements of group 13 form electron-deficient covalent hydrides.
(vi) We can write:
Covalent hydrides which do not have enough electrons to write the conventional Lewis structure are known as electron-deficient covalent hydrides.
(vii) These hydrides are very unstable. They will be trying to obtain electrons from other atoms. So they are Lewis acids. (Details here)
B. Electron-precise covalent hydrides
• Covalent hydrides which have the required electrons to write the conventional Lewis structure are known as electron-precise covalent hydrides.
• All elements of group 14 form such hydrides.
♦ CH4 is an example.
C. Electron-rich covalent hydrides
This can be explained in 6 steps:
(i) These covalent hydrides have excess electrons in the form of lone pairs. We know that lone pairs do not take part in bonding. But they play important roles in determining the shapes of the molecules.
(ii) Elements of group 15 form covalent hydrides which have one lone pair.
♦ NH3 is an example (See fig.4.107 in section 4.18)
(iii) Elements of group 16 form covalent hydrides which have two lone pair.
♦ H2O is an example (See fig.4.104 in section 4.17)
(iv) Elements of group 17 form covalent hydrides which have three lone pairs.
♦ HF is an example
(v) These hydrides will donate electrons to other atoms. So they are Lewis bases. (Details here)
(vi) In these hydrides, the N, O and F atoms carry lone pairs.
• Also, N, O and F are highly electronegative. They pull the shared pair away from H, making the H partially positive charged (δ+).
• So the H with δ+ charge will get attracted to the lone pairs in other molecules. This will create hydrogen bonding. (Details here)
• Hydrogen bonding will cause the molecules to accumulate together, thus raising the boiling point of the sample.
The following solved example will illustrate the effect of hydrogen bonding.
Solved example 9.1
Would you expect the hydrides of N, O and F to have lower boiling points than the hydrides of their subsequent group members? Give reason?
Solution:
1. Here we have to make three comparisons:
(i) Comparison between the following two boiling points:
♦ Boiling point of the hydride of N.
✰ Hydride of N is NH3
♦ Boiling point of the hydride of the element below N in the same group.
✰ Element below N in the same group is P.
✰ Hydride of P is PH3
(ii) Comparison between the following two boiling points:
♦ Boiling point of the hydride of O.
✰ Hydride of O is H2O
♦ Boiling point of the hydride of the element below O in the same group.
✰ Element below O in the same group is S.
✰ Hydride of S is H2S
(iii) Comparison between the following two boiling points:
♦ Boiling point of the hydride of F.
✰ Hydride of F is HF
♦ Boiling point of the hydride of the element below F in the same group.
✰ Element below F in the same group is Cl.
✰ Hydride of Cl is HCl
2. In each case, we are comparing with the hydride of the element which comes below in the group.
• The element below will be larger.
• So it’s hydride will have more electrons. More electrons give rise to greater Van der Walls forces of attraction between molecules.
• Consequently, they will have higher boiling points.
• Based on this, we are tempted to conclude that:
♦ PH3 has higher boiling point than NH3
♦ H2O has higher boiling point than H2S
♦ HF has higher boiling point than HCl
3. But such a conclusion will be wrong. The reason can be written in 3 steps:
(i) N, O and F are highly electronegative atoms.
(ii) They will pull electrons from the H atoms.
(iii) This will give rise to hydrogen bonding between molecules. Thus the boiling points will be higher.
4. So we can write:
♦ Boiling point of NH3 will be higher than that of PH3.
♦ Boiling point of H2O will be higher than that of H2S.
♦ Boiling point of HF will be higher than that of HCl.
Metallic Hydrides
This can be written in 6 steps:
1. Metallic hydrides are formed by some d-block elements and some f-block elements.
2. We know that d-block elements starts from group 3 and ends in group 12.
♦ All these groups do not give hydrides.
♦ Groups 7, 8 and 9 do not form hydrides at all.
♦ In group 6, only Cr forms hydride. It is CrH.
3. We know that metals are good conductors of heat and electricity.
• But when the metals in the d and f blocks form hydrides, those hydrides do not conduct heat and electricity as good as the parent metals.
4. Metallic hydrides are non-stoichiometric. This can be explained in 4 steps using some examples:
(i) First we will see the stoichiometric compound. It can be explained using an example:
• Consider different samples of MgH2 obtained from different sources. We can take a molecule of MgH2 from any of those samples. That molecule will contain one atom of Mg and two atoms of H.
• Again, consider different samples of MgH2 obtained from different sources. Let each of those samples weigh exact 50 grams.
• Find the percentage of the mass of Mg in each sample. It will be same for all the samples.
• Find the percentage of the mass of H in each sample. It will be same for all the samples.
• That means, MgH2 obeys the law of constant proportion.
• The ratio of the two percentages will be integers.
• Such compounds are called stoichiometric compounds.
(Recall how we determined empirical formula and molecular formula. Details here)
(ii) Now we will see non-stoichiometric compounds.
• These compounds do not obey the law of constant proportion. The ratio of percentages cannot be written as integers.
• ZrH1.3-1.75 (Zirconium hydride) is an example.
♦ Zirconium hydride taken from some sources will have the mass ratio of 1:1.3.
♦ Zirconium hydride taken from some other sources will have the mass ratio of 1:1.75.
♦ Intermediate ratios are also possible. That is why we write 1.3-1.75
• Such compounds are called non-stoichiometric compounds.
(iii) Non-stoichiometry can arise due to the following two reasons:
♦ Some atoms of one of the elements is missing from the lattice.
♦ There are too many atoms of one of the elements.
✰ These extra atoms occupy interstitial spaces in the lattice.
(iv) In the case of Lanthanum hydride (LaH2.87), some H atoms are missing from the lattice.
• The subscript should have been greater than '2.87'
• So we can write:
LaH2.87 is deficient in hydrogen.
• In fact, all metallic hydrides are deficient in hydrogen.
5. Some other examples of metallic hydrides are:
♦ YbH2.55 (Ytterbium hydride),
♦ TiH1.5-1.8 (Titanium hydride)
✰ Here, different samples will have different mass ratios of hydrogen.
✰ Smallest possible mass ratio is 1:1.5
✰ Largest possible mass ratio is 1:1.8
✰ Intermediate ratios are also possible.
♦ ZrH1.3-1.75 (Zirconium hydride)
✰ Here, different samples will have different mass ratios of hydrogen.
✰ Smallest possible mass ratio is 1:1.3
✰ Largest possible mass ratio is 1:1.75
✰ Intermediate ratios are also possible.
♦ VH0.56 (Vanadium hydride),
♦ NiH0.6-0.7 (Nickel hydride)
✰ Here, different samples will have different mass ratios of hydrogen.
✰ Smallest possible mass ratio is 1:0.6
✰ Largest possible mass ratio is 1:0.7
✰ Intermediate ratios are also possible.
6. Metallic hydrides can absorb large amounts of hydrogen in the interstices of the metal lattice.
• The hydrogen atoms in the interstices are not chemically bonded to the metal atoms. They are readily available for reactions.
♦ So many metallic hydrides are used in catalytic reduction reactions.
♦ They are also used in hydrogenation reactions.
• Metallic hydrides are also used as storage media for hydrogen.
• They are also a valuable source of energy.
• We have completed a discussion on hydrides. Based on the discussion, we can pictorially show the classification of hydrides. It is shown in the fig.9.2 below:
 |
| Fig.9.2 |
In the next section, we
will see water.
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment