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Which of the following metals directly combine with hydrogen gas to give a hydride? (a) Au (b) \(\mathrm{Ni} \quad\) (c) \(\mathrm{Ca}\) (d) \(\mathrm{Cu}\)

Short Answer

Expert verified
Calcium (Ca) directly combines with hydrogen gas to give a hydride.

Step by step solution

01

Understanding Metal Hydrides

Metal hydrides form when metals combine with hydrogen. Generally, more reactive metals, particularly alkali metals (Li, Na, K) and alkaline earth metals (Be, Mg, Ca, etc.), readily form hydrides.
02

Assessing Reactivity with Hydrogen

Look at the given options and assess their reactivity with hydrogen. Au (gold) and Cu (copper) are not highly reactive and typically do not form hydrides under normal conditions. Ni (nickel) can adsorb hydrogen under certain conditions but does not easily form a stable hydride at standard temperature and pressure. Ca (calcium) is an alkaline earth metal and can form hydrides.
03

Determining the Correct Option

Among the options provided, calcium (Ca) is the most reactive with hydrogen gas and will combine directly to form a hydride, which is commonly known as calcium hydride (CaH2).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reactivity of Metals
The reactivity of metals is a key concept in understanding why certain metals can combine with hydrogen gas to form metal hydrides. Metals are elements that tend to lose electrons to form positive ions, a process called oxidation. The ease with which a metal can lose electrons and react with other elements, including hydrogen, is known as its reactivity.

Alkali metals and alkaline earth metals are at the top of the reactivity series, which means they are highly reactive and readily interact with hydrogen to form hydrides. For instance, when calcium, an alkaline earth metal, reacts with hydrogen, the metal donates electrons to hydrogen, forming calcium hydride. In contrast, less reactive metals like gold (Au) and copper (Cu) do not surrender electrons as easily and generally do not form hydrides under normal conditions.

The reactivity series helps predict which metals will react and how vigorously they will do so. It is a useful tool for chemists and students alike in anticipating the outcomes of reactions involving metals and other elements.
Hydrogen Gas
Hydrogen gas (H₂) plays a unique role in chemistry due to its simple diatomic molecule structure and high reactivity. In its pure form, hydrogen is a colorless, odorless, and highly flammable gas that exists as a diatomic molecule—meaning two hydrogen atoms are bonded together.

Hydrogen's reactivity is central to its ability to form compounds known as hydrides with a wide range of metals. During the formation of metal hydrides, hydrogen gas acts as a reducing agent by accepting electrons from the metal atoms. This is the key to the formation of ionic hydrides, especially with highly reactive metals that are more willing to donate electrons.

The behavior of hydrogen and its applications go beyond the formation of metal hydrides. It's also an important element in energy storage, as a clean fuel in the form of hydrogen fuel cells, and in the production of ammonia for fertilizers using the Haber process.
Alkaline Earth Metals
Alkaline earth metals, such as beryllium (Be), magnesium (Mg), calcium (Ca), and others, are in group 2 of the periodic table and are known for their reactivity, particularly in forming hydrides. These metals all have two electrons in their outer shell, which they are prone to lose to achieve a stable electronic configuration.

When alkaline earth metals react with hydrogen, they form ionic hydrides by donating their two valence electrons to hydrogen atoms. For example, calcium combines with hydrogen to form calcium hydride ({CaH2}). This reaction is indicative of a broader pattern: the heavier alkaline earth metals, those with a larger atomic number, tend to be more reactive with hydrogen.

Moreover, these metals have other significant applications and characteristics, such as their use in producing alloys, their contribution to the formation of bioactive structures in living organisms, like bones and shells, and in various industrial processes, reflecting their usefulness beyond hydride formation.

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Most popular questions from this chapter

Which of the following statements regarding hydrides is not correct? (a) lonic hydrides are crystalline, non-volatile and non-conducting in solid state. (b) Electron-deficient hydrides act as Lewis acids or electron acceptors. (c) Elements of group-13 form electron-deficient hydrides. (d) Elements of group \(15-17\) form electron-precise hydrides.

Which of the following represents the chemical equation involved in the preparation of \(\mathrm{H}_{2} \mathrm{O}_{2}\) from barium peroxide? (a) \(\mathrm{BaO}_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{BaSO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{H}_{2} \mathrm{O}_{2}\) (c) \(\mathrm{BaO}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{BaCO}_{3}+\mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 3 \mathrm{BaSO}_{4}+2 \mathrm{H}_{3} \mathrm{PO}_{4}\)

A metal \((M)\) produces a gas \((N)\) on reaction with alkalies like \(\mathrm{NaOH}\) and \(\mathrm{KOH}\). Same gas is produced when the metal reacts with dilute sulphuric acid Gas \((N)\) reacts with another toxic gas \((P)\) to form methanol at high temperature and pressure. ( \(N\) ) also reacts with metals like \((Q)\) to form electrovalent hydrides. \(M, N, P\) and \(Q\) respectively are (a) \(\mathrm{Zn}, \mathrm{H}_{2}, \mathrm{CO}, \mathrm{Na}\) (b) \(\mathrm{Na}, \mathrm{H}_{2}, \mathrm{Cl}_{2}, \mathrm{Ca}\) (c) \(\mathrm{Al}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{~B}\) (d) \(\mathrm{Mg}, \mathrm{H}_{2}, \mathrm{NO}_{2}, \mathrm{Al}\)

Peroxodisulphate, on hydrolysis yields (a) water (b) dihydrogen (c) hydrogen peroxide (d) deuterium.

Given below are two reactions of water with sodium and carbon dioxide. What is the nature of water in these reactions? (i) \(2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}\) (ii) \(6 \mathrm{CO}_{2}+12 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{O}_{2}\) (a) In (ii) water acts as an oxidising agent and in (i) it acts as a reducing agent. (b) In (i) water acts as an oxidising agent while in (ii) it acts as a reducing agent. (c) In both, (i) and (ii) hydrogen acts as a reducing agent. (d) In both, (i) and (ii) hydrogen acts as an oxidising agent.

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