P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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601.
Which of the following oxidation states are the most characteristics for lead and tin respectively?
A
+ 4, + 2
B
+ 2, + 4
C
+ 4, + 4
D
+ 2, + 2
Answer :
+ 2, + 4
The tendency to form + 2 ionic state increase on moving down the group due to inert pair effect.
Most characteristic oxidation state for lead and tin are + 2, + 4 respectively.
602.
The shape of gaseous $$SnC{l_2}$$ is
A
tetrahedral
B
linear
C
angular
D
$$T$$ - shaped
Answer :
angular
Shape of $$SnC{l_2}$$ is angular due to $$s{p^2}$$ hybridisation and having the following structure
603.
The correct order of acidic strength is
A
$${K_2}O > CaO > MgO$$
B
$$C{O_2} > {N_2}{O_5} > S{O_3}$$
C
$$N{a_2}O > MgO > A{l_2}{O_3}$$
D
$$C{l_2}{O_7} > S{O_2} > {P_4}{O_{10}}$$
Answer :
$$C{l_2}{O_7} > S{O_2} > {P_4}{O_{10}}$$
The species $$C{l_2}{O_7},S{O_2}$$ and $${P_4}{O_{10}}$$ are the anhydrides of $$HCl{O_4},{H_2}S{O_3}$$ and $${H_3}P{O_4}$$ respectively. The acid strength of these acids follows the order $$HCl{O_4} > {H_2}S{O_3} > {H_3}P{O_4}.$$ The corresponding anhydrides also followthe same order.
604.
Which of the following is not correctly matched?
605.
In the preparation of $$HN{O_3},$$ we get $$NO$$ gas by catalytic oxidation of ammonia. The moles of $$NO$$ produced by the oxidation of two moles of $$N{H_3}$$ will be ________.
A
2
B
3
C
4
D
6
Answer :
2
\[\underset{\left( \text{from air} \right)}{\mathop{4N{{H}_{3\left( g \right)}}+5{{O}_{2\left( g \right)}}}}\,\xrightarrow[500\text{ }K,\,9\text{ }bar]{Pt/Rh\text{ gauge catalyst}}\] \[4N{{O}_{\left( g \right)}}+6{{H}_{2}}{{O}_{\left( g \right)}}\]
Thus, 2 moles of $$NO$$ will be produced by the oxidation of 2 moles of $$N{H_3}.$$
606.
The gas evolved on heating $$Ca{F_2}$$ and $$Si{O_2}$$ with concentrated $${H_2}S{O_4},$$ on hydrolysis gives a white gelatinous precipitate. The precipitate is :
607.
Which among the following factors is the most important in making fluorine the strongest oxidizing halogen?
A
Hydration enthalpy
B
Ionization enthalpy
C
Electron affinity
D
Bond dissociation energy
Answer :
Bond dissociation energy
The fluorine has low dissociation energy of $$F{\text{ - }}F$$ bond and reaction of atomic fluorine is exothermic in nature
608.
The formation of $$O_2^ + {\left[ {Pt{F_6}} \right]^ - }$$ is the basis for the formation of xenon fluorides. This is because
A
$${O_2}$$ and $$Xe$$ have comparable sizes.
B
both $${O_2}$$ and $$Xe$$ are gases.
C
$${O_2}$$ and $$Xe$$ have comparable ionisation energies.
D
Both (A) and (C).
Answer :
Both (A) and (C).
(i) The first ionization energy of xenon ( 1, 170 $$kJ\,\,mo{l^ - }$$ ) is quite close to that of dioxygen $$\left( {1,180\,\,kJ\,\,mo{l^{ - 1}}} \right).$$
(ii) The molecular diameters of xenon and dioxygen are almost identical.
Based on the above similarities Barlett ( who prepared $$O_2^ + {\left[ {Pt{F_6}} \right]^ - }$$ compound ) suggested that since oxygen combines with $$Pt{F_6},$$ so xenon should also form similar compound with $$Pt{F_6}.$$
609.
The substance used as a smoke screen in
warfare is
A
$$SiC{l_4}$$
B
$$P{H_3}$$
C
$$PC{l_5}$$
D
$${\text{acetylene}}$$
Answer :
$$SiC{l_4}$$
Silicon chloride is easily hydrolysed to give white fumes, so it is used as a smoke screen in warfare.
$$SiC{l_4} + 4{H_2}O \to Si{\left( {OH} \right)_4} + 4HCl$$
610.
Which of the following is not correctly matched ?
A
$$S{F_4} - {\text{gas}}$$
B
$$Se{F_4} - {\text{liquid}}$$
C
$$Te{F_4} - {\text{solid}}$$
D
$$S{F_6} - {\text{solid}}$$
Answer :
$$S{F_6} - {\text{solid}}$$
All hexafluorides of group 16 elements are gaseous in nature.