NEET Practice Questions (MCQs) with Answers & Solutions

(A)

     Basic strength of the oxides increases in the order ${\mathrm{Li}}_2\mathrm{O}<{\mathrm{Na}}_2\mathrm{O}<{\mathrm{K}}_2\mathrm{O}<{\mathrm{Rb}}_2\mathrm{O}<{\mathrm{Cs}}_2\mathrm{O}.$ The increase

     in basic strength is due to the decrease in I.E. and increase in electropositive character. The melting

     points of the halides decrease in order $\mathrm{NaF}>\mathrm{NaCl}>\mathrm{NaBr}>\mathrm{Nal}$, as the size of the halide ion

     increases. The decrease in melting point is due to increase the covalent character with increase in the

     size of anion according to Fazan's rule.

(C)

    $$\begin{gathered} {\mathrm{NH}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\stackrel{ }{\to }{\mathrm{NH}}_4{\mathrm{HCO}}_3 \\ \mathrm{NaCl}+{\mathrm{NH}}_4{\mathrm{HCO}}_3\stackrel{ }{\to }{\mathrm{NaHCO}}_3+{\mathrm{NH}}_4\mathrm{Cl} \\ 2{\mathrm{NaHCO}}_3\stackrel{ 150°\mathrm{C} }{\to }{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \\ 2{\mathrm{NH}}_4\mathrm{Cl}+\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\stackrel{ }{\to }2{\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CaCl}}_2 \left(\mathrm{by} \mathrm{product}\right) \end{gathered}$$

(B)

     (2) ${\mathrm{KO}}_2+2{\mathrm{H}}_2\mathrm{O}\stackrel{ }{\to }\mathrm{KOH}+{\mathrm{H}}_2{\mathrm{O}}_2+\frac{1}{2}{\mathrm{O}}_2$

     (3) $4{\mathrm{KO}}_2+2{\mathrm{CO}}_2\stackrel{ }{\to }2{\mathrm{K}}_2{\mathrm{CO}}_3+3{\mathrm{O}}_2$

(D)

     ${\mathrm{NaNO}}_2$ (sodium nitrite) acts both as an oxidizing agent and reducing agent because the nitrogen

     atom in it is in +3 oxidation state (+3 intermediate oxidation state for nitrogen)

     Oxidizing property : $2{\mathrm{NaNO}}_2+2\mathrm{Kl}+2{\mathrm{H}}_2{\mathrm{SO}}_4\stackrel{}{\to }{\mathrm{NaSO}}_4+{\mathrm{K}}_2{\mathrm{SO}}_4+2\mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{l}}_2$

     Reducing property : ${\mathrm{NaNO}}_2+{\mathrm{H}}_2{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{NaNO}}_3+{\mathrm{H}}_2\mathrm{O}.$

(C)    $2\mathrm{M}{\left({\mathrm{NO}}_3\right)}_2\stackrel{ ∆ }{\to }2\mathrm{MO}+4{\mathrm{NO}}_2+{\mathrm{O}}_2\left(\mathrm{M}=\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\right)$

(C)

     (1) The solubility increases down the group because the change in lattice energy is more as compared

           to hydration energy. Thermal stability and the basic character both increase down the group as

           metallic character increases.

      (2) It is a correct statement.

      (3) Beryllium and aluminium are diagonally related. Chlorides of both are covalent in nature and

           thus are soluble in organic solvents. Chlorides of both are electron deficient and thus act as

           strong Lewis acids.

Assertion and reason are correct statements and reason is the correct explanation of assertion.

(A) Beryllium and aluminium have diagonal relationship.

(B) $M+\left(x+y\right)N{H}_3\to {\left[M{\left(N{H}_3\right)}_x\right]}^{+}+{\left[e{\left(N{H}_3\right)}_y\right]}^{-}$

    Above 3M concentration, the solutions are no longer blue but are bronze/copper-bronze
            coloured with a metallic luster due to formation of metal ion clusters.

(B) (i) Bigger cation stabilises bigger anions and vice-versa.

(ii) Alkali metal carbonates and their bicarbonates stabilities depend on their metallic character.
Stability of carbonates and bicarbonates $\propto$ Metallic character of their i.e. electropositive character.