Animated Solution for Chemistry - s and p-Block Elements: The correct statements among (a) to (d) are : 1. Saline hydrides produce ${\text{H}}_2$ gas when reacted with ${\text{H}}_2\text{O}$. 2. Reaction of ${\text{LiAlH}}_4$ with ${\text{BF}}_3$ leads to ${\text{B}}_2{\text{H}}_6$. 3. ${\text{PH}}_3$ and ${\text{CH}}_4$ are electron rich and electron precise hydrides, respectively. 4. $\text{HF}$ and ${\text{CH}}_4$ are called as molecular hydrides.

A Comprehensive Review

Hydrogen, the simplest and most abundant element in the universe, forms a fascinating array of compounds known as hydrides when it reacts with other elements. This problem takes us on a conceptual journey through the different classes of hydrides, testing our understanding of their chemical properties, synthesis, and structural classifications. Let's break down each statement to see why they all hold true.

Saline Hydrides

Let's evaluate the first statement regarding saline hydrides. Saline, or ionic hydrides, are formed when hydrogen reacts with highly electropositive s-block elements (like alkali and alkaline earth metals). These compounds contain the highly reactive hydride ion, ${\text{H}}^{-}$.

When these ionic hydrides come into contact with water, a violent and highly exothermic reaction occurs. The hydride ion acts as an exceptionally strong base, aggressively pulling a proton (${\text{H}}^{+}$) from the water molecule. This acid-base reaction releases copious amounts of dihydrogen gas:

Because of this vigorous reaction, statement one is absolutely correct.

Synthesis of Diborane

Moving to statement two, we delve into the synthesis of diborane (${\text{B}}_2{\text{H}}_6$), a crucial electron-deficient hydride. How do we prepare it in the laboratory?

A classic and highly efficient method involves the reduction of boron trifluoride (${\text{BF}}_3$) using a powerful reducing agent, lithium aluminium hydride (${\text{LiAlH}}_4$). This reaction is typically carried out in a solvent like diethyl ether, which helps stabilize the reactive intermediates.

This reaction perfectly yields diborane along with lithium and aluminium fluorides. Thus, statement two is also chemically sound.

Electron-Rich vs. Electron-Precise

Now, let's look at statement three, which focuses on the classification of covalent hydrides based on their electron count.

Phosphine (${\text{PH}}_3$) is formed by phosphorus, a group 15 element. Phosphorus has five valence electrons. It uses three of these to form covalent bonds with three hydrogen atoms, leaving one non-bonding pair of electrons (a lone pair). This extra pair makes phosphine an electron-rich hydride, capable of acting as a Lewis base.

Methane (${\text{CH}}_4$), on the other hand, is formed by carbon, a group 14 element. Carbon uses all four of its valence electrons to form four single bonds with hydrogen. It has exactly the required number of electrons to complete its stable octet, leaving absolutely no lone pairs. This makes methane an electron-precise hydride. Statement three is spot on.

Finally, we examine statement four. Are hydrogen fluoride ($\text{HF}$) and methane (${\text{CH}}_4$) considered molecular hydrides?

Yes! Elements of the p-block (groups 13 to 17) form covalent compounds with hydrogen. Unlike the extended ionic lattices of saline hydrides, these compounds exist as discrete, individual molecules held together by relatively weak intermolecular forces, such as van der Waals forces or hydrogen bonds. Because of their discrete molecular nature, they are aptly named molecular or covalent hydrides. So, statement four is correct as well.

Since all four statements are chemically accurate and conceptually flawless, the correct option must encompass all of them. This makes option (a) our final answer. This problem serves as a fantastic, all-encompassing review of the diverse world of hydrides!