P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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351.
Oxyacids of phosphorous and the starting materials for their preparation are given below.
Oxyacid
Materials for preparation
a.
$$HP{O_3}$$
1.
$${P_2}{O_5} + {H_2}{O_2} + {H_2}O$$
b.
$${H_3}P{O_3}$$
2.
$${P_4}{O_{10}} + {H_2}O$$
c.
$${H_3}P{O_4}$$
3.
$${P_2}{O_3} + {H_2}O$$
d.
$${H_3}P{O_5}$$
4.
$${P_4}{O_{10}} + {H_2}S{O_4}$$
Choose the correct answer from the codes given below :
A
a - 4, b - 3, c - 2, d - 1
B
a - 1, b - 3, c - 2, d - 4
C
a - 4, b - 3, c - 1, d - 2
D
a - 2, b - 3, c - 1, d - 4
Answer :
a - 4, b - 3, c - 2, d - 1
No explanation is given for this question. Let's discuss the answer together.
352.
Identify the incorrect statement among the following
A
$$B{r_2}$$ reacts with hot and strong $$NaOH$$ solution to give $$NaBr$$ and $${H_2}O.$$
B
Ozone reacts with $$S{O_2}$$ to give $$S{O_3}.$$
C
Silicon reacts with $$NaOH(aq)$$ in the presence of air to give $$N{a_2}Si{O_3}$$ and $${H_2}O.$$
D
$$C{l_2}$$ reacts with excess of $$N{H_3}$$ to give $${N_2}$$ and $$HCl.$$
Answer :
$$C{l_2}$$ reacts with excess of $$N{H_3}$$ to give $${N_2}$$ and $$HCl.$$
Chlorine reacts with excess of ammonia to produce ammonium chloride and nitrogen. $$3C{l_2} + 8N{H_3}\left( {{\text{excess}}} \right) \to 6N{H_4}Cl + {N_2}$$
353.
Which of the following elements can be involved in \[p\pi -d\pi \] bonding?
A
Carbon
B
Nitrogen
C
Phosphorus
D
Boron
Answer :
Phosphorus
No explanation is given for this question. Let's discuss the answer together.
354.
Fill in the blanks by choosing the appropriate option. The noble gases can form compounds with $$\underline {\left( {\text{i}} \right)} $$ and $$\underline {\left( {{\text{ii}}} \right)} .$$ The mixture of $$\underline {\left( {{\text{iii}}} \right)} $$ and $$\underline {\left( {{\text{iv}}} \right)} $$ is used for respiration by divers.
(i)
(ii)
(iii)
(iv)
(a)
iodine
oxygen
oxygen
argon
(b)
fluorine
oxygen
helium
oxygen
(c)
xenon
platinum
argon
krypton
(d)
helium
oxygen
xenon
argon
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(b)
No explanation is given for this question. Let's discuss the answer together.
355.
Hypo is used in photography to
A
reduce $$AgBr$$ grains to metallic silver
B
convert metallic silver to silver salt
C
remove undecomposed silver bromide as a soluble complex
D
remove reduced silver
Answer :
remove undecomposed silver bromide as a soluble complex
After the developing, the sensitive emulsion of silver bromide is still present on the plate in the parts unaffected by light. Therefore, it is necessary to remove it in order to get the permanent image. This process is called fixing of image. The negative plate after washing is dipped in a fixing solution of sodium thiosulphate (hypo). It dissolves unaffected silver bromide but leaves metallic silver unaffected.
$$2N{a_2}{S_2}{O_3} + AgBr \to $$ $$N{a_3}\left[ {Ag{{\left( {{S_2}{O_3}} \right)}_2}} \right] + NaBr$$
356.
Group 13 elements show + 1 and + 3 oxidation states. Relative stability of + 3 oxidation state may be given as
Stability of + 3 oxidation state decreases from $$Al$$ to $$Tl.$$ $$B$$ always shows + 3 oxidation state in all of its compounds.
357.
Which one has the highest boiling point?
A
$$Kr$$
B
$$Xe$$
C
$$He$$
D
$$Ne$$
Answer :
$$Xe$$
$$Xe\,$$ . As we move down the group, the melting and boiling points show a regular increase due to corresponding increase in the magnitude of their van der waal forces of attraction as the size of the atom increases.
358.
In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
360.
For $${H_3}P{O_3}$$ and $${H_3}P{O_4},$$ the correct choice is
A
$${H_3}P{O_3}$$ is dibasic and reducing
B
$${H_3}P{O_4}$$ is diabasic and non-reducing
C
$${H_3}P{O_4}$$ is tribasic and reducing
D
$${H_3}P{O_4}$$ is tribasic and non reducing
Answer :
$${H_3}P{O_3}$$ is dibasic and reducing
The structure of $${H_3}P{O_3}$$ is as given below.
There are only two $$ - OH$$ groups and hence it is dibasic. The oxidation number of $$P$$ in this acid is +3, whereas $$P$$ may have +5 oxidation state also. Therefore, $${H_3}P{O_3}$$ can be oxidised which means $${H_3}P{O_3}$$ is a reducing agent.