S - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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51.
$${K_2}C{O_3}$$ cannot be prepared by Solvay's process because
A
$$KHC{O_3}$$ is less soluble than $$NaHC{O_3}$$
B
$$KHC{O_3}$$ is more soluble to be precipitated by $$KCl$$ and $$N{H_4}HC{O_3}$$
C
$${K_2}C{O_3}$$ is more soluble to be precipitated by $$KCl$$
D
$${K_2}C{O_3}$$ is less soluble than $$N{a_2}C{O_3}$$
Answer :
$$KHC{O_3}$$ is more soluble to be precipitated by $$KCl$$ and $$N{H_4}HC{O_3}$$
$$KHC{O_3}$$ is more soluble to be precipitated by the addition of $$N{H_4}HC{O_3}$$ to the saturated solution of $$KCl.$$
52.
Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?
A
$$MgC{O_3}$$
B
$$CaC{O_3}$$
C
$$SrC{O_3}$$
D
$$BaC{O_3}$$
Answer :
$$BaC{O_3}$$
As the electropositive character increases down the group, the stability of the carbonates increases.
53.
Complete the following equations :
$$\eqalign{
& \left( {\text{i}} \right)N{a_2}{O_2} + 2{H_2}O \to \underline {\left( W \right)} + {H_2}{O_2} \cr
& \left( {{\text{ii}}} \right)2K{O_2} + 2{H_2}O \to \underline {\left( X \right)} + \underline {\left( Y \right)} + {O_2} \cr
& \left( {{\text{iii}}} \right)N{a_2}O + C{O_2} \to \underline {\left( Z \right)} \cr} $$
54.
In the case of alkali metals, the covalent character decreases in the order :
A
$$MF > MCl > MBr > MI$$
B
$$MF > MCl > MI > MBr$$
C
$$MI > MBr > MCl > MF$$
D
$$MCl > MI > MBr > MF$$
Answer :
$$MI > MBr > MCl > MF$$
As the size of halide ion decreases, $${I^ - } > B{r^ - } > C{l^ - } > {F^ - },$$ tendency to get polarised by cation $${M^ + }$$ decreases and hence covalent character decreases.
55.
A substance which gives brick red flame and breaks down on heating to give oxygen and a brown gas is
A
magnesium nitrate
B
calcium nitrate
C
barium nitrate
D
strontium nitrate
Answer :
calcium nitrate
Calcium imparts brick red colour to the flame and calcium nitrate evolves $${O_2}$$ and a brown gas, $$N{O_2}$$ upon heating.
\[2Ca{{\left( N{{O}_{3}} \right)}_{2}}\xrightarrow{\Delta }2CaO+{{O}_{2}}\] \[+\underset{\left( \text{Brown gas} \right)}{\mathop{4N{{O}_{2}}\uparrow }}\,\]
56.
$$HI$$ can be prepared by all the following methods except
57.
A chemical $$'A'$$ is used for the preparation of washing soda to recover ammonia. When $$C{O_2}$$ is bubbled through an aqueous solution of $$'A',$$ the solution turns milky. It is used in white washing due to disinfectant nature. What is the chemical formula of $$'A'?$$
A
$$Ca{\left( {HC{O_3}} \right)_2}$$
B
$$CaO$$
C
$$Ca{\left( {OH} \right)_2}$$
D
$$CaC{O_3}$$
Answer :
$$Ca{\left( {OH} \right)_2}$$
$$Ca{\left( {OH} \right)_2}$$ is used in Solvay process i.e., preparation of washing soda $$\left( {N{a_2}C{O_3} \cdot 10{H_2}O} \right)$$ to recover $$N{H_3}.$$
$$2N{H_4}Cl + \mathop {Ca{{\left( {OH} \right)}_2}}\limits_{\left( A \right)} \to $$ $$2N{H_3} + CaC{l_2} + {H_2}O$$
$$\mathop {Ca{{\left( {OH} \right)}_2}}\limits_{\left( A \right)} + C{O_2} \to $$ $$\mathop {CaC{O_3} \downarrow }\limits_{{\text{Milkiness}}} + {H_2}O$$
$$Ca{\left( {OH} \right)_2}$$ is used in white washing due to its disinfectant nature.
58.
Molecular formula of Glauber’s saltis:
A
$$MgS{O_4}.7{H_2}O$$
B
$$CuS{O_4}.5{H_2}O$$
C
$$HFeS{O_4}.7{H_2}O$$
D
$$N{a_2}S{O_4}.10{H_2}O$$
Answer :
$$N{a_2}S{O_4}.10{H_2}O$$
Glauber’s salt is $$N{a_2}S{O_4}.10{H_2}O.$$
59.
The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders ?
A
$$CsH > RbH > KH > NaH > LiH$$
B
$$KH > NaH > LiH > CsH > RbH$$
C
$$NaH > LiH > KH > RbH > CsH$$
D
$$LiH > NaH > KH > RbH > CsH$$
Answer :
$$LiH > NaH > KH > RbH > CsH$$
The stability of alkali metal hydrides decreases from $$Li$$ to $$Cs.$$ It is due to the fact that $$M–H$$ bonds become weaker with increase in size of alkali metals as we move down the group from $$Li$$ to $$Cs.$$ Thus the order of stability of hydrides is $$LiH > NaH > KH > RbH > CsH$$
60.
The reducing power of a metal depends on various factors. Suggest the factor which makes
$$Li$$ metal the strongest reducing agent in aqueous solution :
A
Sublimation enthalpy
B
Ionisation enthalpy
C
Hydration enthalpy
D
Electron-gain enthalpy
Answer :
Hydration enthalpy
Due to high hydration energy of $$L{i^ + }$$ cation, the standard reduction potential of $$L{i^ + }$$ is more negative among all alkali metal cations hence $$Li$$ act as strong reducing agent in water.