The p-Block Elements Test

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The p-Block Elements Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Calcium cyanamide on treatment with steam produces :

A
$$NH_3+CaO$$

B
$$NH_3+CaHCO$$

C
$$NH_3+CaCO_3$$

D
$$NH_3+Ca(OH)_2$$

Solution

Reaction:
$$CaCN_2+3H_2O\rightarrow CaCO_3+NH_3$$
Question 2
1 / -0

$$NH_4CI(s)$$ is heated in a test tube, vapours are brought in contact with red litmus paper, which changes to blue and then to red. it is because of :

A
formation of $$NH_4OH$$ and $$HCI$$

B
formation of $$NH_3$$ and $$HCI$$

C
greater diffusion of $$NH_3$$ than $$HCI$$

D
grater diffusion of $$HCI$$ than $$NH_3$$

Solution

$$NH_4Cl\xrightarrow{\Delta} NH_3+HCl$$
First HCl reacts with red litmus and converts it into blue but as diffusion of ammonia is high so after that ammonia reacts with blue litmus and converts back to red.
Question 3
1 / -0

$$Fe^{2+}$$ reduces $$NH_2OH$$ to :

A
$$NH_3$$

B
$$N_3H$$

C
$$N_2H_4$$

D
$$N_2$$

Solution

$$ \begin{array}{l} \text { when } \mathrm{Fe}^{+2} \text {reacts with } \mathrm{NH}_{2} \mathrm{OH} \text { , } \mathrm{Fe}^{+2} \text {gets oxidised to } \mathrm{Fe}^{+3} \text {while } \mathrm{N} \mathrm{H}_{2} \mathrm{OH} \text { is reduced to } \mathrm{NH}_{3} \text { . } \\ \text { Thus, } \mathrm{Fe}^{+2} \text {reduces } \mathrm{NH}_{2} \mathrm{OH} \text { to } \mathrm{NH}_{3} \text { . } \end{array} $$
Question 4
1 / -0

For inert gases, $$\displaystyle \frac{C_p}{C_v}= $$ _____ .

A
1.33

B
1.40

C
1.66

D
can't be predicted

Solution

Inert gases are monoatomic so their $$C_v = \dfrac{3}{2}$$ and $$C_p={5}{2}$$ and

$$\displaystyle \frac{C_p}{C_v}= \dfrac{5\times 2}{3\times 2} =\dfrac{5}{3} = 1.66$$
Question 5
1 / -0

The gas not absorbed by coconut charcoal is :

A
He

B
Ne

C
Ar

D
Kr

Solution

Activated charcoal is used to treat poisonings, reduce intestinal gas (flatulence) and absorb gases apart from $$He$$. $$He$$ is not absorbs as its size is very small.
Question 6
1 / -0

Hydrates of helium and neon have not been prepared because of :

A
low polarisability

B
small size

C
low boiling point

D
all of the above

Solution

When water is allowed to freeze in the presence of Ar, Kr or Xe under pressure, atoms of noble gas get trapped in the crystal lattice of ice giving clathrates corresponding to the composition, 8G.46H2O (G = noble gas). These clathrates are also known as the noble gas hydrates.

But due to small size, low boiling point and low polarisability, He and Ne are unable to trap so they do not form.
Question 7
1 / -0

Which of the following are coloured and paramagnetic both ?
$$NO_2, Cu^+, O_2, O_3, Hg^{2+}_2, Cd^{2+}, Al, C(graphite)$$

A
$$ NO_2$$

B
$$ NO_2$$ , $$ O_3$$

C
$$ Hg^{2+}_2 $$

D
$$ NO_2$$, $$Cu^+$$

Solution

$$NO_2$$ contains 1 unpaired electron. Hence, it is colored and para magnetic. N atom has 5 valence electrons. Out of which it uses 2 valence electrons to form a sigma bond and a pi bond with O atom. N also uses 2 valence electrons to form a coordinate bond with another O atom. Thus 4 out of 5 valence electrons of N are involved in the bond formation. There is one unpaired electron left on nitrogen. Due to this nitrogen dioxide is para magnetic and brown colored gas.
Question 8
1 / -0

Consider the following properties of the noble gases:
$$I$$: they readily from compounds which are colourless.
$$II$$ : they generally do not form ionic compounds.
$$III$$ : they have variable oxidation states in their compounds.
$$IV$$ : they generally do not form covalent compounds.

Select the the correct properties.

A
$$I, II$$ and $$III$$ only

B
$$ II$$ and $$III$$ only

C
$$I$$ and $$III$$ only

D
$$I$$ only

Solution

In noble gases, as their octet is complete, so they don't have tendency to form compounds. Only $$Xe$$ forms few compounds with F, O and other noble gases and these compounds have covalent character.

$$Xe$$ shows variable oxidation in these compounds like $$XeOF$$ and $$XeF_2$$.
Question 9
1 / -0

High reactivity of white phosphorus is due to :

A
unusual bonding that produces considerable strain

B
high solubility in water

C
oxide layer made on it

D
all of the above

Solution

From figure it can be seen that,
The high reactivity of white phosphorus is due to increased steric (strain) factor.
Question 10
1 / -0

In $$P_4$$ (tetrahedral) :

A
each P is joined to four P

B
each P is joined to three P

C
each P is joined to two P

D
$$P_4$$ does not exist

Solution