Question

Elements of which of the following group(s) of periodic tables do not form hydrides?
(A) Groups 7, 8, 9
(B) Group 13
(C) Groups 15, 16, 17
(D) Group 14

Answer 1

The term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. The general formula being $M{{H}_{X}}$ , where M represents element and X the number of hydrogen atoms. This general formula was applicable for all elements except noble gases. Hydride is a negative ion of hydrogen, ${{H}^{-}}$ , also called hydride ion. This negative charge of a hydride ion, the reason for reducing properties of hydrides.

Complete step by step answer:
The first period elements of group 7, 8, 9, 13, 14, 15, 16, and 17 are Mn, Fe, Co, B, C, N, O, and F, from the periodic table.
Let us consider, group-13, Boran forms more hydrides with the general formula ${{B}_{n}}{{H}_{n+4}}$ , for example, diborane (${{B}_{2}}{{H}_{6}}$ ), pentaborane (${{B}_{5}}{{H}_{9}}$ ), etc.
In group-14, carbon forms hydrides with hydrogen, for example, methane $C{{H}_{4}}$ , etc.
In group-15, ammonia is an important nitrogen hydride.
In group-16, oxygen forms hydrides are water,${{H}_{2}}O$ , hydrogen peroxide, ${{H}_{2}}{{O}_{2}}$ etc.
In group-17, fluorine forms hydrides as HF,
Group 7,8,9 does not form hydrides out of the periodic table. The property does not form hydrides of this 7, 8, 9 groups are known as the hydride gap.
So, the correct answer is “Option A”.

Note: Depending upon the physical and chemical properties, these hydrides are divided into three following categories, ionic, metallic, and covalent hydrides. Due to the hydrogen bonding of these three hydrides, they have high boiling points and strong dipole-dipole attractions because of the high ionic character of the compounds.