Question

The correct statements among (a) to (b) are:
(a) Saline hydrides produce H₂​ gas when reacted with H₂​O.
(b) Reaction of LiAlH₄​ with BF₃​ leads to B₂​H₆​.
(c) PH₃​ and CH₄​ are electron - rich and electron precise hydrides, respectively.
(d) HF and CH₄​ are called molecular hydrides.
A. (c) and (d) only
B. (a), (b) and (c) only
C. (a), (b), (c) and (d)
D. (a), (c) and (d) only

Answer 1

Hint: Saline hydrides are compounds that form when hydrogen reacts with alkali metals or alkaline earth metals. Electron-rich hydrides are those that contain excess electrons as lone pairs on their central atom. These hydrides are formed by elements of groups 15, 16 and 17. Electron precise hydrides are those containing the exact number of valence electrons to form normal covalent bonds. They are usually formed by group 14 elements.

Complete Step by Step Solution:
To answer this question, we must go through each statement individually and check its correctness.
Statement (a)
Saline hydrides are those that form when hydrogen reacts with s-block elements which are found in groups 1 and 2 of the periodic table. The group 1 and group 2 elements are less electronegative than hydrogen. Examples include sodium hydride (\[NaH\]) and lithium hydride (\[LiH\]).
A characteristic property of saline hydrides is that they react vigorously with water to produce pure hydrogen gas. For example, \[NaH(s) + {H_2}O(l) \to NaOH(aq) + {H_2}(g)\]. Thus, statement (a) is correct.

Statement (b)
Lithium aluminium hydride (\[LiAl{H_4}\]) does react with boron trifluoride (\[B{F_3}\]) to form diborane (\[{B_2}{H_6}\]).
\[3LiAl{H_4} + 4B{F_3} \to 2{B_2}{H_6} + 3LiF + 3Al{F_3}\]
Thus, statement (b) is correct.

Statement (c)
Electron precise hydrides contain the exact number of electrons to form a normal covalent bond. It means that the number of valence electrons is equal to the number of covalent bonds that the hydride can form. Electron precise hydrides can be formed by group 14 elements. In methane (\[C{H_4}\]), carbon’s valence shell electronic configuration is \[2{s^2}2{p^2}\]i.e., it has four valence electrons, and it forms four covalent bonds with hydrogens in methane. Therefore, methane is an electron-precise hydride.

Electron-rich hydrides possess a greater number of valence electrons than the number of covalent bonds they can form. These extra valence electrons are present on their central atoms as lone pairs. Electron-rich hydrides are formed by elements of groups 15, 16 and 17. Phosphane’s (\[P{H_3}\]) central atom is phosphorus which belongs to group 15 and its valence shell electronic configuration is \[3{s^2}3{p^3}\] i.e., it has 5 valence electrons. But, in phosphine, phosphorus forms only 3 covalent bonds with hydrogens. The 2 extra electrons are present as lone pairs of the phosphorus atom as shown below.

Image: Structure of Phosphine
Therefore, phosphane is an electron-rich hydride. Thus, statement (c) is correct.

Statement (d)
When hydrogen reacts with elements of similar electronegativity, the hydrides formed are called molecular hydrides. Molecular hydrides are also called covalent hydrides because the central atom and hydrogen are bonded by covalent bonds. Molecular hydrides are generally formed by non-metals (p-block elements). Hydrogen fluoride (\[HF\]) and methane (\[C{H_4}\]) are both examples of molecular hydrides. Thus, statement (d) is correct.
Thus, option C is the correct answer to this question.

Note: Ketone does not give a silver mirror test but alpha hydroxy ketone gives a silver mirror test. An alpha hydroxy ketone is a molecule having adjacent ketone and alcohol groups. Example: Glucose, tartaric acid.