P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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511.
Which is the correct arrangement of the compounds based on their bond strength?
A
$$HF > HCl > HBr > HI$$
B
$$HI > HBr > HCl > HF$$
C
$$HCl > HF > HBr > HI$$
D
$$HF > HBr > HCl > HI$$
Answer :
$$HF > HCl > HBr > HI$$
No explanation is given for this question. Let's discuss the answer together.
512.
Fill in the blanks.
The high reactivity of fluorine is due to its ________ dissociation energy. Its shows only ________ oxidation state. It has ________ electron affinity than chlorine. Among all hydrogen halides boiling point is highest for ________ .
A
low, - 1, lower, $$HF$$
B
high, + 1, higher, $$HF$$
C
low, + 1, lower, $$HCl$$
D
high, - 1, higher, $$HF$$
Answer :
low, - 1, lower, $$HF$$
No explanation is given for this question. Let's discuss the answer together.
513.
Anhydrous aluminium chloride $$\left( {A{l_2}C{l_6}} \right)$$ is covalent compound and soluble in water giving :
A
$$A{l^{3 + }}\,{\text{and}}\,C{l^ - }\,ions$$
B
$${\left[ {Al{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}\,{\text{and}}\,C{l^ - }\,ions$$
514.
In graphite, the layers of carbon atoms are held by
A
covalent bonds
B
coordinate bonds
C
van der Waals' forces
D
ionic bonds
Answer :
van der Waals' forces
No explanation is given for this question. Let's discuss the answer together.
515.
Which of the following is the best description for the behaviour of bromine in the reaction given below ?
$${H_2}O + B{r_2} \to HOBr + HBr$$
A
Proton acceptor only
B
Both oxidized and reduced
C
Oxidized only
D
Reduced only
Answer :
Both oxidized and reduced
$${H_2}O + \mathop B\limits^0 {r_2} \to HO\mathop {Br}\limits^{ + 1} + H\mathop {Br}\limits^{ - 1} $$
Thus here oxidation number of $$Br$$ increases from 0 to +1 and also decreases from 0 to –1. Thus it is oxidised as well as reduced.
516.
In the following sets of reactants which two sets best exhibit the amphoteric characters of $$A{l_2}{O_3}.\,x{H_2}O?$$
$$\eqalign{
& {\text{Set}}\,1{\text{:}}\,A{l_2}{O_3}.\,x{H_2}O\left( s \right)\,{\text{and}}\,O{H^ - }\left( {aq} \right) \cr
& {\text{Set}}\,{\text{2:}}\,A{l_2}{O_3}.\,x{H_2}O\left( s \right){\text{and}}\,{{\text{H}}_2}O\left( l \right) \cr
& {\text{Set}}\,{\text{3:}}\,A{l_2}{O_3}.\,x{H_2}O\left( s \right)\,{\text{and}}\,{H^ + }\left( {aq} \right) \cr
& {\text{Set}}\,{\text{4:}}\,A{l_2}{O_3}.\,x{H_2}O\left( s \right)\,{\text{and}}\,N{H_3}\left( {aq} \right) \cr} $$
A
1 and 2
B
1 and 3
C
2 and 4
D
3 and 4
Answer :
1 and 3
Aluminium oxide is amphoteric oxide because it shows the properties of the both acidic and basic oxides. It reacts with both acids and bases to form salt and water.
$$\eqalign{
& A{l_2}{O_3}.\,x{H_2}O + 2NaOH \to NaAl{O_2} + {H_2}O \cr
& A{l_2}{O_3}.\,x{H_2}O + HCl \to AlC{l_3} + {H_2}O \cr} $$
517.
Cane sugar on reaction with nitric acid gives
A
$$C{O_2}\,\,{\text{and}}\,\,S{O_2}$$
B
$$2HCOOH$$
C
$${\left( {COOH} \right)_2}$$
D
$${\text{no reaction}}$$
Answer :
$${\left( {COOH} \right)_2}$$
When nitric acid reacts with cane sugar, it forms oxalic acid.
518.
$${P_2}{H_4}$$ can be removed from phosphine containing traces of it
A
by passing impure $$P{H_3}$$ gas through a freezing mixture.
B
by passing the impure $$P{H_3}$$ gas through $$HI$$ and then its treatment with $$KOH\left( {aq} \right).$$
C
by both (A) and (B).
D
by none of these.
Answer :
by both (A) and (B).
Both the methods (A) and (B) can be used. When passed through freezing mixture the $${P_2}{H_4}$$ present condenses and pure $$P{H_3}$$ is obtained.
When passed through $$HI,P{H_3}$$ is absorbed forming $$P{H_4}I.P{H_4}I$$ when treated with $$KOH\left( {aq} \right)$$ yields pure phosphine.
$$\eqalign{
& P{H_3} + HI \to P{H_4}I \cr
& P{H_4}I + KOH\left( {aq} \right) \to KI + {H_2}O + P{H_3} \uparrow \cr} $$
519.
Basicity of orthophosphoric acid is
A
2
B
3
C
4
D
5
Answer :
3
Orthophosphoric acid $$\left( {{H_3}P{O_4}} \right)$$ have the following structure
It is clear from the structure that it contains three replaceable hydrogen atoms, so it gives three $${H^ + }$$ $$ions$$ on dissolution in water. So, the basicity of $${{H_3}P{O_4}}$$ is three.
$${H_3}P{O_4} \to 3{H^ + } + PO_4^{3 - }$$
520.
Identify the incorrect statement.
A
Graphite is thermodynamically most stable allotrope of carbon.
B
Other forms of elemental carbon like coke, carbon black, charcoal are impure forms of graphite.
C
All allotropes of carbon have thermodynamically different stability.
D
Charcoal and coke are obtained by heating wood in absence of air.
Answer :
Charcoal and coke are obtained by heating wood in absence of air.
Charcoal is obtained from wood while coke is obtained from coal.