The p-Block Elements Test

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The p-Block Elements Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

________ bacteria are microbes that work in the absence of oxygen.

A
Anaerobic

B
Aerobic

C
Gram-positive

D
Gram-negative

Solution

Anaerobic bacteria works in the absence of oxygen.
Example is yeast.
Question 2
1 / -0

Hydride of nitrogen is called :

A
Hydrogen sulphide

B
Water

C
Ammonia

D
Nitrous oxide

Solution

Hydride of nitrogen is the binary compound formed by the combination of hydrogen and nitrogen .
The hydride formed with nitrogen is ammonia ($$NH_3$$).
Hence option C is correct.
Question 3
1 / -0

Nitrogen reacts with a metal $$M$$ to produce a compound $$X$$. This compound $$X$$, when dissolved in water forms pungent gas $$Y$$ which can be identified easily by the formation of white fumes in the presence of $$HCl$$.

Identify $$Y$$.

A
Nitrogen dioxide

B
Ammonia

C
Nitrous oxide

D
Nitric oxide

Solution

Calcium metal reacts with nitrogen to form calcium nitride.
Metal $$M$$ is calcium.

$$3\underset{M}{Ca}+N_2 \rightarrow Ca_3N_2.$$

The calcium nitride reacts with water to form $$NH_3$$, which will react with HCl to give white fumes.

Thus, $$Y$$ is $$NH_3$$.
Question 4
1 / -0

$$BrF_{5}$$ is a:

A
Interhalogen compound

B
Pseudohalogen compound

C
Both the above

D
None of the above

Solution

Inter halogen compounds are the compounds in which one halogen combines with other halogen.
$$BrF_5$$ is an inter halogen compound. In this compound, bromine combines with fluorine.
Pseudohalogens on the other hand are covalent dimers and contains two or more electronegative atoms out of which at least one is nitrogen.
Question 5
1 / -0

Calcium imide on hydrolysis will give gas (B) which on oxidation by bleaching powder gives gas (C). Gas (C) on reaction with magnesium gives compound (D). (D) on hydrolysis gives again gas (B). (B), (C) and (D) respectively are :

A
$${NH_3}$$, $${N_2}$$, $${Mg_3N_2}$$

B
$${N_2}$$, $${NH_3}$$, $$Mg_3N_2$$

C
$${N_2}$$, $${N_2O_5}$$, $${Mg(NO_3)_2}$$

D
$${NH_3}$$, $${NO_2}$$, $${Mg(NO_2)_2}$$

Solution

Reactions :
$$CaNH+H_2O\rightarrow CaO+NH_3$$
$$3CaOCl_2 + 2NH_3\rightarrow 3CaCl_2 + 3H_2O + N_2$$
$$3Mg + N_2 \rightarrow Mg_3N_2$$
$$Mg_3N_2 + 6 H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$$
Question 6
1 / -0

Which one of the following anions can undergo both oxidation and reduction?

A
$$Cr_2{O_7}^{2-}$$

B
$$N{O_3}^{-}$$

C
$$OCl^-$$

D
$${S}^{2-}$$

Solution

$$(A)$$ In $$Cr_2{O_7}^{2-}$$, the oxidation state of $$Cr$$ is $$+6$$, which is maximum oxidation of $$Cr$$. It can only lower down its oxidation state. So, it acts only as an oxidising agent.
$$(B)$$ In $$N{O_3}^{-}$$, the oxidation state of $$N$$ is $$+5$$, which is the maximum oxidation of $$N$$. It can only lower its oxidation state and so, it acts only as an oxidising agent.
$$(C)$$ In $$OCl^-$$, the oxidation state of $$Cl$$ is $$+1$$ and $$Cl$$ shows oxidation state from $$-1$$ to $$+7$$ and so, it can undergo both oxidation and reduction.
$$(D)$$ In $${S}^{2-}$$, the oxidation state of $$S$$ is $$-2$$, which is the minimum oxidation state of $$S$$ and so, it can only increase its oxidation state and hence, it acts only as a reducing agent.
Therefore, $$OCl^-$$ can undergo both oxidation as well as reduction.
Question 7
1 / -0

Which one of the following compounds on strong heating evolves ammonia gas?

A
$$(NH_{4})_{2}SO_{4}$$

B
$$NH_4NO_{3}$$

C
$$(NH_{4})_{2}Cr_{2}O_{7}$$

D
$$NH_{4}NO_{2}$$

Solution

$$(NH_4)_2Cr_2O_7 \to N_2$$

$$NH_4NO_3 \to N_2O$$;

$$NH_4NO_2 \to N_2$$

$$(NH_4)_2SO_4 \to NH_3$$

Ammonia is obtained by heating ammonium sulphate.

$$(NH_4)_2SO_4 \rightarrow NH_3 + NH_4HSO_4$$
Question 8
1 / -0

Consider the following transformations:

(I)     $$XeF_{6} + NaF \,\, \rightarrow \,\, Na^{+}[XeF_{7}]^{-}$$

(II)    $$2PCl_{5}(s) \,\, \rightarrow \,\, [PCl_{4}]^{+}\,[PCl_{6}]^{-}$$

(III)   $$[Al(H_{2}O)_{6}]^{3+} \, + \, H_{2}O \,\, \rightarrow \,\, [Al(H_{2}O)_{5}OH]^{2+} \, + \, H_{3}O^{+}$$

Correct transformations are :

A
I,II,III

B
I,III

C
I,II

D
II,III

Solution

All three are correct as they are properly balanced.

$$[Al(H_{2}O)_{6}]^{3+} \, + \, H_{2}O \,\, \rightarrow \,\, [Al(H_{2}O)_{5}OH]^{2+} \, + \, H_{3}O^{+}$$

$$XeF_{6} + NaF \,\, \rightarrow \,\, Na^{+}[XeF_{7}]^{-}$$

$$2PCl_{5}(s) \,\, \rightarrow \,\, [PCl_{4}]^{+}\,[PCl_{6}]^{-}$$
Question 9
1 / -0

Among the following the correct statement is :

A
Phosphates have no biological significance in humans

B
between nitrates and phosphates,phosphates are less abundant in earth's crust

C
between nitrates and phosphates,nitrates are less abundant in earth's crust

D
oxidation of nitrates is possible in soil

Solution

Nitrates are less abundant than phosphates in the earth's crust because they are soluble in water. Moreover they can also be reduced by numerous microbes present in earth's crust.
Question 10
1 / -0

The weakest $$Cl-O$$ bond is found in :

A
$$ClO_{4}^{-}$$

B
$$ClO_{2}^{-}$$

C
$$ClO^{-}$$

D
$$ClO_{3}^{-}$$

Solution

The anions given are conjugate bases of acids. According to Lowry Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base. These anions are stabilised to greater extent, it has lesser attraction for proton and therefore, will behave as weaker base (lesser tendency for the reaction to go in backward direction). Consequently, the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjgate base has weak acid and vice versa. The charge stabilization increases in the order : $$ClO^- < ClO_2^- < ClO_3^- < ClO_4^-$$. This means that $$ClO^-$$ will have minimum stability and therefore, will have maximum attraction for $$H^+$$. In other words $$ClO^-$$ will be strongest base and so its conjugate acid HClO will be the weakest acid.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 31

Result

The p-Block Elements Test - 31

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SHARING IS CARING

Question 1

Anaerobic

Aerobic

Gram-positive

Gram-negative

Solution

Question 2

Hydrogen sulphide

Water

Ammonia

Nitrous oxide

Solution

Question 3

Nitrogen dioxide

Ammonia

Nitrous oxide

Nitric oxide

Solution

Question 4

Interhalogen compound

Pseudohalogen compound

Both the above

None of the above

Solution

Question 5

$${NH_3}$$, $${N_2}$$, $${Mg_3N_2}$$

$${N_2}$$, $${NH_3}$$, $$Mg_3N_2$$

$${N_2}$$, $${N_2O_5}$$, $${Mg(NO_3)_2}$$

$${NH_3}$$, $${NO_2}$$, $${Mg(NO_2)_2}$$

Solution

Question 6

$$Cr_2{O_7}^{2-}$$

$$N{O_3}^{-}$$

$$OCl^-$$

$${S}^{2-}$$

Solution

Question 7

$$(NH_{4})_{2}SO_{4}$$

$$NH_4NO_{3}$$

$$(NH_{4})_{2}Cr_{2}O_{7}$$

$$NH_{4}NO_{2}$$

Solution

Question 8

I,II,III

I,III

I,II

II,III

Solution

Question 9

Phosphates have no biological significance in humans

between nitrates and phosphates,phosphates are less abundant in earth's crust

between nitrates and phosphates,nitrates are less abundant in earth's crust

oxidation of nitrates is possible in soil

Solution

Question 10

$$ClO_{4}^{-}$$

$$ClO_{2}^{-}$$

$$ClO^{-}$$

$$ClO_{3}^{-}$$

Solution

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