Classification of elements and Periodicity in properties MCQs for NEET — Chemistry Questions with Answers

(a) H^->H>H+

It is known that radius of a cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Whereas, the radius of anion is always greater than a cation due to decrease in effective nuclear charge.

(b) Na⁺>F^->O^2-

The given species are isoelectronic as they contain same number of electrons. For isoelectronic species,

      ionic radii $\propto$ 1/atomic number

                   Ion: Na⁺ F^- O^2-

Atomic number:    11  9   8 

Hence, the correct order of ionic radii is O^2->F^->Na⁺

(c) Similarly, the correct option is O^2->F^->Na⁺

(d) Ion            :        Al³⁺  Mg²⁺     N^3-

Atomic number:         13     12        7

Hence, the correct order is N^3->Mg²⁺>Al³⁺

Isoelectronic species contain same number of electrons Be²⁺ contains 2 electrons. Among the given options, only Li⁺ contains 2 electrons and therefore, it is isoelectronic with Be²⁺.

H⁺ $\to$no electron; Na⁺ $\to$ 10e^-

Li⁺ $\to$ 2e^-;  Mg2⁺ $\to$ 10e^-

Hence, Be²⁺ is isoelectronic with Li⁺

(d) Ionic compounds are more soluble in water or in an aqueous medium

Ionic character $\propto$size of cation (if anion is same)

The order of size of cation is

             Na⁺>Zn²⁺>Cu²⁺

.^.. The order of ionic character and hence of solubility in water is as

                    Na₂S>ZnS>CuS

Ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins decreases as the size of alkali metal ions increases.

Since, the order of size of alkali metal ions

           Li⁺<Na⁺<K⁺<Rb⁺

Thus, the ease of adsorption follows the order

         Rb⁺<K⁺<Na⁺<Li⁺

(a) Atomic radius of the elements decreases across a period from left to right due to increase in effective nuclear charge. On moving down a group, since, number of shells increases, so atomic radius increases. Amongst isoelectronic species, ionic radius increases with increase in negative charge or decrease in positive charge

In general, the atomic and ionic radii increases on moving down a group. But the elements of second transition series (eg, Zr, Nb, Mo etc) have the almost same radii as the elements of third transition series (eg, Hf, Ta, W etc). This is because of lanthanide contraction i.e., imperfect shielding of one 4f- electron by another.

(c)Key Idea Ionic radii $\propto$ charge on anion
$\propto \frac{1}{\mathrm{charge} \mathrm{on} \mathrm{cation}}$
During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a great force of attraction by the nucleus, i.e., attracted more towards the nucleus. In other words, nucleus hold the remaining electrons more tightly and this results in decreased radii.

However, in case of anion formation, the addition of electron(s) takes place in the same outer shell, thus the hold of nucleus on the electrons of outer shell decreases and this results in increased ionic radii.

Thus, the correct order of ionic radii is S^2- > Cl^- > K⁺ > Ca²⁺

 Key Idea Electron gain enthalpy, generally, increases in a period from left to right and decreases in a group on moving downwards. However, members of III period have somewhat higher electron gain enthalpy as compared to the corresponding members of second period, because of their small size.

O and S belong to VI A (16) group and CL and F belong to VII A (17) group. Thus, the electron gain enthalpy of Cl and F is higher as compared to O and S.

             Cl and F > O and S

Between Cl and F, Cl has higher electron gain enthalpy as in F, the incoming electron experiences a greater force of repulsion because of small size of F atom. Similar is true in case of O and S ie, the electron gain enthalpy of S is higher as compared to O due to its small size. Thus, the correct order of electron gain enthalpy of given elements is

    O<S<F<Cl

Key Idea In peroxides, the oxidation state of O is -1 and they give H₂O₂, with dilute acids, and have peroxide linkage.

In K0₂,

           +1 + (x $\times$ 2)=0

           x = -1/2 (thus, it is a superoxide, not a peroxide.)

 In BaO₂,

           +2 + (x $\times$2) = 0

           x = -1

Thus, it is a peroxide. Only it gives H₂O₂ when reacts with dilute acids and has peroxide linkage as

                    Ba²⁺ [O - O]^2-

                 (peroxide linkage)

 In ${\mathrm{MnO}}_2$ and ${\mathrm{NO}}_2$, Mn and N exhibit variable oxidation states, thus, the oxidation state of O in these is - 2. Hence, these are not peroxides. Thus, it is clear, that among the given molecules only ${\mathrm{BaO}}_2$ is a peroxide.

(a) Key Idea : The amount of energy required to remove an electron from a unipositive ion is called second IP which generally increases in a period from left to right.

The second ionisation energies of the first transition series increase almost regularly with increase in atomic number. However, the value for Cr is sufficiently higher than those of its neighbour, i.e., (Mn). This is due to stable configuration of Cr⁺ (3d⁵ exactly half filled).

Note: The half-filled and completely filled configurations are more stable.