NEET Practice Questions (MCQs) with Answers & Solutions

(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.

Key Idea: The smaller the size of ion, the greater the degree of hydration and

The mobility of ion in aqueous solution $\propto$ 1/size of ion in aqueous solution

Generally in a group size or radii of ions increases, so Li has smallest size thus bind maximum number of water molecules with itself and becomes largest in size (in aqueous solution). Hence, the order of ionic radii (size) of alkali metal ions in aqueous solution is

     Li⁺>Na⁺>K⁺>Rb⁺>Cs⁺

Thus, the order of mobility

      Cs⁺>Rb⁺>K⁺>Na⁺>Li⁺

(c) Key Idea : Generally bond dissociation energies decreases in a group. Bond dissociation energy also decreases with repulsion.
X-X Bond                F-F         Cl-Cl         Br-Br           I-I
Bond length ($\stackrel{\circ }{\mathrm{A}}$)     1.42        1.99          2.28           2.67
Bond dissociation    38            57           45.5           35.6
energy (kcal/mol)

In general the bond dissociation energy decreases as the bond length increases, but the
bond dissociation energy of F₂ is less than that of Cl₂ is due to greater interelectronic repulsions between the lone pair of electrons on the two bonded fluorine atoms. Hence, the order of bond dissociation energy is not as: F₂ > Cl₂ > Br₂ > I₂ 

(b) Thermal stability of carbonates increases in a group as we move from top to bottom and decreases in a period as we move from left to right. So, the correct order of thermal stability of given carbonates is:

(b) BeCO₃ < MgCO₃ < CaCO₃ < K₂CO₃

Be, Mg and Ca present in second group and K present in first group.

it is because of polarizability and energetics..generally larger cation form stable compound with larger anion ,or their stability increases with increase in ionic character. 

Hydration energy of sulphate decreases from top to bottom in a IInd group. Mg²⁺ is smaller than other ions of that group so Mg²⁺ is readily hydrated. MgSO₄ has higher hydration energy than lattice energy.