Chemistry Hydrides and it's types

Topics to be covered

`=>` Hydrides
`=>` Ionic or saline hydrides
`=>` Covalent or molecular hydrides
`=>` Metallic or non-stoichiometric hydrides


Dihydrogen, under certain reaction conditions, combines with almost all elements, except noble gases, to form binary compounds, called hydrides. If `color{red}(‘E’)` is the symbol of an element then hydride can be expressed as `color{red}(EH_x)` (e.g., `color{red}(MgH_2)`) or` color{red}(E_mH_n)` (e.g.,`color{red}(B_2H_6)`).

`color{green}("The hydrides are classified into three categories :")`

(i) Ionic or saline or saltlike hydrides.

(ii) Covalent or molecular hydrides.

(iii) Metallic or non-stoichiometric hydrides.

Ionic or Saline Hydrides

`color{green}(" ★)` These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character.

`color{green}(" ★)` Significant covalent character is found in the lighter metal hydrides such as `color{red}(LiH, BeH_2)` and `color{red}(MgH_2)`. In fact `color{red}(BeH_2)` and `color{red}(MgH_2)` are polymeric in structure.

`color{green}(" ★)` The ionic hydrides are crystalline, non-volatile and nonconducting in solid state. However, their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode, which confirms the existence of `color{red}(H^–)` ion.

`color{red}(2H^(-) (text(melt)) oversettext(anode)→ H_2 (g) +2e^(-))`

`color{green}(" ★)` Saline hydrides react violently with water producing dihydrogen gas.

`color{red}(NaH(s) +H_2O(aq) → NaOH(aq) +H_2(g))`

`color{green}(" ★)` Lithium hydride is rather unreactive at moderate temperatures with `color{red}(O_2) `or `color{red}(Cl_2)`. It is,
therefore, used in the synthesis of other useful hydrides, e.g.,

`color{red}(8LiH + Al_2Cl_6 → 2LiAlH_4 + 6LiCl)`

`color{red}(2LiH + B_2H_6 → 2LiBH_4)`

Covalent or Molecular Hydride

`color{green}( ★)` Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar examples are `color{red}(CH_4, NH_3, H_2O)` and `color{red}(HF)`.

`color{green}( ★)` For convenience hydrogen compounds of nonmetals have also been considered as hydrides. Being covalent, they are volatile compounds.

`color{green}( ★)` Molecular hydrides are further classified according to the relative numbers of electrons and bonds in their Lewis structure into :
(i) electron-deficient, (ii) electron-precise, and (iii) electron-rich hydrides.

`color{green}( ★)` An electron-deficient hydride, as the name suggests, has too few electrons for writing its conventional Lewis structure. Diborane `color{red}((B_2H_6))` is an example. In fact all elements of group 13 will form electron-deficient compounds.They act as Lewis acids i.e., electron acceptors.

`color{green}( ★)` Electron-precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 form such compounds (e.g., `color{red}(CH_4)`) which are tetrahedral in geometry.

`color{green}( ★)` Electron-rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 form such compounds. (`color{red}(NH_3)` has 1- lone pair, `color{red}(H_2O – 2))` and `color{red}(HF –3))` lone pairs). They will behave as Lewis bases i.e., electron donors. The presence of lone pairs on highly electronegative atoms like `color{red}(N, O)` and `color{red}(F)` in hydrides results in hydrogen bond formation between the molecules. This leads to the association of molecules.

Q 3141291123

Would you expect the hydrides of `N, O` and `F` to have lower boiling points than the hydrides of their subsequent group
members ? Give reasons.


On the basis of molecular masses of `NH_3, H_2O` and `HF`, their boiling points are expected to be lower than those of the subsequent group member hydrides. However, due to higher electronegativity of `N, O` and `F`, the magnitude of hydrogen bonding in their hydrides will be quite appreciable. Hence, the boiling points `NH_3, H_2O` and `HF` will be higher than the hydrides of their subsequent group members.

Metallic or Non-stoichiometric (or Interstitial ) Hydrides

`color{green}( ★)` These are formed by many d-block and f-block elements.

`color{green}(★)` However, the metals of group 7, 8 and 9 do not form hydride.

`color{green}( ★)` Even from group 6, only chromium forms `color{red}(CrH)`.

`color{green}( ★)` These hydrides conduct heat and electricity though not as efficiently as their parent metals do.

`color{green}(★)` Unlike saline hydrides, they are almost always nonstoichiometric, being deficient in hydrogen. For example, `color{red}(LaH_(2.87), YbH_(2.55), TiH_(1.5–1.8), ZrH_(1.3–1.75), VH_(0.56), NiH_(0.6–0.7), PdH_(0.6–0.8))` etc. In such hydrides, the law of constant composition does not hold good.

`color{green}(★)` Earlier it was thought that in these hydrides, hydrogen occupies interstices in the metal lattice producing distortion without any change in its type. Consequently, they were termed as interstitial hydrides. However, recent studies have shown that except for hydrides of `color{red}(Ni, Pd, Ce)` and `color{red}(Ac)`, other hydrides of this class have lattice different from that of the parent metal.

`color{green}(★)` The property of absorption of hydrogen on transition metals is widely used in catalytic reduction / hydrogenation reactions for the preparation of large number of compounds. Some of the metals (e.g., Pd, Pt) can accommodate a very large volume of hydrogen and, therefore, can be used as its storage media. This property has high potential for hydrogen storage and as a source of energy.
Q 3111391220

Can phosphorus with outer electronic configuration `3s^2 3p^3` form `PH_5 ?`


Although phosphorus exhibits +3 and +5 oxidation states, it cannot form `PH_5`. Besides some other considerations, high `Δ_a H` value of dihydrogen and `Δ_(eg)H` value of hydrogen do not favour to exhibit the highest oxidation state of `P`, and consequently the formation of `PH_5`.