NEET Practice Questions (MCQs) with Answers & Solutions

(D). Melting points and boiling points decrease on moving down the group from Li to Cs. The cesium metal exists as liquid at normal temperature of about 30$°$C. The m.p. of Cs is 29$°$C.

(C). The electronic configuration of calcium is 2, 8, 8, 2. As such due to empty d-orbital calcium forms strong metallic bond. Hence its boiling point is high.

(A). Covalent halides are formed by Be and Al among the metals given. The halide of Al, ie, $A{l}_{2}C{l}_6$ becomes ionic in polar solvent. Hence Y is Al. Halide of Be is a polymer in solid state. Thus X is Be.

(D). The chemical reactions involved in Solvay process are as below :

$N{H}_3+C{O}_2+H_{2}O\to N{H}_{4}HC{O}_3$

$N{H}_{4}HC{O}_3+NaCl\stackrel{30°C}{\to }NaHC{O}_3\downarrow +N{H}_{4}Cl$

$\underset{\left(used again\right)}{2NaHC{O}_3}\stackrel{250°C}{\to }N{a}_{2}C{O}_3+H_{2}O+C{O}_2$;

$\underset{\left(used again\right)}{2N{H}_{4}Cl}+Ca{\left(OH\right)}_2\to \underset{slaked lime}{CaC{l}_2}+2H_{2}O+2N{H}_3$

(D). The blue coloured solution of an alkali metal in ammonia is explained on the basis of formation of ammoniated (solvated) metal cations and ammoniated (solvated) electrons in the metal-ammonia solution in the following way :

$M\to M^{+}+e^{-}$

$M^{+}+xN{H}_3\to {\left\{M{\left(N{H}_3\right)}_x\right\}}^{+}$

$e^{-}+yN{H}_3\to {\left[e{\left(N{H}_3\right)}_y\right]}^{-}$

$\underset{Solvated metal cation}{M+\left(x+y\right)N{H}_3}\to \underset{Solvated electron}{{\left[M{\left(N{H}_3\right)}_x\right]}^{+}+{\left[e{\left(N{H}_3\right)}_y\right]}^{-}}$

The blue colour of the solution is due to excitation of free electrons to higher energy levels. The absorption of photons takes place in the red region of the spectrum and hence, the solution appears blue in the transmitted light. As the concentration of the alkali metal increases, the metal ion cluster formation takes place and at very high concentration the solution becomes coloured like that of metallic copper.

(A). $\underset{\left(X\right)}{CaC{O}_3}\stackrel{∆}{\to }CaO+C{O}_2, CaO+H_{2}O\to \underset{\left(Y\right)}{Ca{\left(OH\right)}_2}$

 $\underset{\left(Y\right)}{Ca{\left(OH\right)}_2}+\underset{\left(excess\right)}{C{O}_2}\to \underset{\left(Z\right)}{Ca{\left(HC{O}_3\right)}_2}$

$\underset{\left(Z\right)}{Ca{\left(HC{O}_3\right)}_2}\stackrel{∆}{\to }\underset{\left(X\right)}{CaC{O}_3}+H_{2}O+C{O}_2$

(D). $2Mg+O_2\to 2MgO$

$2Mg{\left(N{O}_3\right)}_2\stackrel{∆}{\to }2MgO+4N{O}_2+O_2$

$2MgS{O}_4\stackrel{∆}{\to }2MgO+S{O}_3+S{O}_2+O_2$

$Mg+H_{2}O\left(steam\right)\to MgO+H_2$

(C). $Ca{C}_2{O}_4\leftrightharpoons C{a}^{2+}+C_2{O}_4^{2-}$

Since acetic acid is a weak electrolyte it does not impart sufficient $H^{+}$ so that it can combine with oxalate ion to form feebly ionised oxalic acid. Therefore, the equilibrium will not shift to the right.

(B). Low solubility of LiF in water is due to its high lattice energy. LiCl has some covalent character due to the high polarising power of $L{i}^{+}$. That is why LiCl is soluble in water as well as in acetone.

(A). The s-block elements are highly electropositive in nature. So, these are very reactive. That is why these elements do not occur free in nature. Usually, the s-group elements occur in nature as halides, carbonates and sulphates. These metals are obtained from their molten halides by electrolytic reduction.