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What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water.

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Last updated date: 17th Jun 2024
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Answer
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Hint: Based on the availability of electrons in a compound containing hydrogen, it is classified into the respective hydrides. Try to figure out which hydride the compounds are, based on the number of electrons. Reactivity towards water decreases down the group.

Complete Solution :
In order to answer our question, we need to know about hydrides. Hydrogen combines with a large number of other elements including metal and non-metals, except noble gases to form binary compounds called hydrides. The types of hydrides are:
Ionic or saline hydrides: The ionic hydrides are stoichiometric which are formed when hydrogen combines with elements of s-block elements except Be. Ionic hydrides are formed by transfer of electrons from metal to hydrogen atoms and contain hydrogen as ${{H}^{+}}$ on e g, sodium hydride (NaH ,calcium hydride etc.
 Some of the properties of saline hydrides are as follows :
i.These are white crystalline solids, have ionic lattices and behave like salts.
ii. They have high melting and boiling points
iii. These are non-volatile and non-conducting in solid state, however they conduct electricity in fused state liberating hydrogen at anode which confirms the existence of ${{H}^{-}}$ ion.
Covalent or molecular hydrides: Covalent or molecular hydrides are the compounds of hydrogen which are associated with elements of p-block. They are of 3 types:

- Electron deficient: These hydrides have a lesser number of electrons than that required for writing the conventional Lewis structure. All the hydrides of group 13 form electron deficient compounds like BH, which polymerise to form ${{B}_{2}}{{H}_{6}}$ As these are electron deficient, so they have a tendency to accept the electrons, therefore, act as Lewis acids.

- Electron precise: Such compounds have required a number of electrons to write their conventional Lewis structures. The elements of group 14 form such hydrides. The hydrides have tetrahedral geometry

- Electron rich: Electron-rich hydrides have excess electrons in the form of one or more lone pairs of electrons around the central electronegative element. Therefore, these hydrides due to the presence of lone pairs of electrons form electron-rich hydrides intact the presence of one pair of electrons on electronegative elements results in the formation of molecules due to hydrogen bonding. These types of hydrides are formed by group 15-17 elements.

- Metallic hydrides: The elements of group 3, 4. 5 (d-block) and f-block elements form metallic hydrides In group B only chromium forms hydride (CrH). Metals of group 7, 8.9 do not form hydrides. These hydrides are known as metallic hydrides because they conduct electricity Moreover these hydrides are non-stoichiometric with variable composition being deficient in hydrogen.
Now, let us come to our question. We know that elements 15 ,19 ,23 and 44 represent nitrogen, potassium, vanadium and ruthenium.

- Ammonia will form electron rich hydride as it has a lone pair of electrons. Potassium is highly electropositive in nature, so it forms ionic hydride, whereas vanadium and ruthenium form metallic hydrides. Ammonia is a lewis base, potassium hydride reacts well with water, but vanadium and ruthenium do not react. The equations are:
\[\begin{align}
 & KH+{{H}_{2}}O\to KOH+{{H}_{2}} \\
 & N{{H}_{3}}+{{H}_{2}}O\to O{{H}^{-}}+N{{H}_{4}}^{+} \\
\end{align}\]
We obtain the order of the reactivity of hydrides towards water as: Ru < KH < N{{H}_{3}} < KH, which gives our answer.

Note: The metallic hydrides have properties similar to their parent metal, for example they are lustrous, conduct electricity , are hard and also have magnetic properties. However, their density is less.