Periodic Table and Elements

Several carbides are assumed to be salts of the acetylide anion C₂^2- (also called percarbide), which has a triple bond between the two carbon atoms. Alkali metals, alkaline earth metals, and lanthanoid metals form acetylides, e.g., sodium carbide Na₂C₂, calcium carbide CaC₂, and LaC₂ Lanthanoids also form acetylides with formula M₂C₃.

This option is correct because all the elements of 3d series are good reducing agents except copper and the highest oxidation state shown by any transition element is +8 (Ru in 4d series and Os in 5d series).

It is correct answer. Caesium has the largest atomic size in periodic table.

Low ionisation energy makes alkali metals lose their outermost single electron easily to form +ve charged ions or cations. Thus, they are highly reactive in nature.

The ionisation energy of Be is quite high. Hence, the energy of flame is not sufficient to excite Be electrons to higher energy level.

On moving down the group, ability to lose its valence electrons increases (with decrease in ionisation potential), hence, reducing ability of metals increases. Thus, the correct order of reducing ability of alkali metals is Li < Na < K < Rb < Cs. The correct order of melting points of alkali metals is Li (180°C) > Na (98°C) > K (64°C) > Rb (39°C) > Cs (29°C).

₅₈Ce: [Xe] 4f¹ 5d¹ 6s² Ce⁴⁺ ion is formed by losing 4 electrons from valence shell of Ce. Hence it has noble gas electronic configuration similar to Xe.

₅₆Ba: [Xe] 6s² Ba²⁺ ion is formed by losing 2 electrons from valence shell of Ba. Hence, it has noble gas electronic configuration similar to Xe.

Properties of sodium metal are as follows: (1) It is a soft metal and can be cut with a knife. (2) It reacts quickly with water and air. (3) It has a low melting point.

Metals readily lose electrons to form positive ions (cations) because they have few valence electrons and low ionization energies. This electron loss is the defining characteristic of metallic behavior in chemical reactions. Oxides, ores, and alloys are compounds or mixtures, not the elements themselves.

this answer is correct because Berrylium is having typically small size which leads to formation of covalent compounds and its anamolous propertis because of which it is not considered as an alkaline earth metal compound.

this option is correct because higher the hydration enthalpy more molecules of water will get attached to the molecule and hence higher will be the formation of hydrated compounds.

this option is correct because Be being small in size does not react directly to form hydride however its hydride can be obtained by the reduction of berrylum chloride with lithium aluminium hydride as per the following reaction:2BeCl₂ + LiAlH₄  ------------> 2BeH₂ + HCl + AlCl₃

this option is correct because actually solubility decreases as the atomic number of the metal ion increases and it becomes less stable with increasing atomic number of the metal ion involved.

this option is correct because in alkali metals when one elctron is detached from the outermost shell then it acquires noble gas configuration resuting in large second ionisation enthalpy and hence it does not show +2 oxidation state.