The p-Block Elements Test

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The p-Block Elements Test - 62

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Weekly Quiz Competition

Question 1
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Directions For Questions

Species having $$X - O - H$$ linkage (X = non-metal with positive oxidation state) are called oxy acids and parent acid of a non-metal may exist in two form (a) -ic form of parent oxy acid (b) -us form of parent oxy acid.

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$$X - O - X$$ bond (where X = central atom) is not present in species.

A
$$Cl_{2}O_{7}$$

B
$$H_{2}N_{2}O_{2}$$

C
$$N_{2}O_{5}$$

D
$$H_{2}S_{2}O_{7}$$

Solution

$$H_2N_2O_2$$ is a compound which have not present in species X-O-X bond
Question 2
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In the synthesis of sodium carbonate by solvay (or ammonia-soda) process, the recovery of ammonia is done by treating $$NH_4Cl$$ with $$Ca(OH)_2$$. The by-product obtained in this process is:

A
$$CaCl_2$$

B
$$NaCl$$

C
$$NaOH$$

D
$$NaHCO_3$$

Solution

$$2NH_4Cl+Ca(OH)_2 \rightarrow 2NH_3+H_2O+CaCl_2$$
$$CaCl_2$$ is obtained as by-product in this process.
Question 3
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The inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total $$p-$$ and $$s-$$electrons, is:

A
$$He$$

B
$$Ne$$

C
$$Ar$$

D
$$Kr$$

Solution

$$He, Ne$$ and $$Ar$$ do not have any $$d$$ electrons, while the difference of their $$s$$ and $$p$$ electrons is not zero.

The electronic configuration of krypton (atomic number is 36) is: $$1s^2,2s^2,2p^6,3s^2,3p^6,3d^{10},4s^2,4p^6$$.

Total $$s-$$electrons are 8 and total $$p-$$electrons are 18.

Thus, the difference is 10 and total d-electrons are 10 which are equal.

So, option $$D$$ is correct answer.
Question 4
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The following acids have been arranged in order of decreasing acid strength. Identify the correct order.
$$ClOH (I),\ BrOH (II),\ IOH (III)$$

A
$$I>II>III$$

B
$$II>I>III$$

C
$$III>II>I$$

D
$$I>III>II$$

Solution

The acidic character of hypohalous acid depends on the electronegativity of halogen. As electronegativity value of halogen increases the acidic strength of hyophalous acid.
Question 5
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Which of the following molecule doesn't have $$d\pi -p\pi$$ bonding?

A
$$SO_2$$

B
$$P_4O_{10}$$

C
$$H_3PO_4$$

D
$$B_3N_3H_6$$

Solution

The structural study of $$B_3N_3H_6$$ shows that it has hexagonal ring with alternating B and N atom along with H atom. Each B and N are $$sp^2$$ hybridized and form three sigma bond one to terminal H atom and two to the adjacent N or B atom. There is a p$$\pi$$-p$$\pi$$ donation of filled nitrogen p$$_z$$ atomic orbital to adjacent Boron p$$_z$$ atomic orbitals. So, borazine contain 3 pi and 12 sigma bonds. Hence, there is no involvement of any d-orbital in hybridization of $$B_3N_3H_6$$.
Question 6
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Which of the following reactions does not produce ammonia as one of the products?

A
$$Mg_3N_2 + H_2O\rightarrow$$ Products

B
$$(NH_4)_2SO_4 + CaO \rightarrow$$ Products

C
$$NaNO_2 + Zn\xrightarrow{NaOH}$$ Products

D
$$NH_4NO_3 \xrightarrow[ ]{\Delta }$$ Products
Question 7
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In spite of having less electron affinity than chlorine, fluorine is the strongest oxidizing agent among the halogens. The most important reason for this phenomenon is :

A
lower bond dissociation energy of $$F_{2}$$ than $$Cl_{2}$$

B
much lower heat of hydration of $$F^-$$ than $$Cl^-$$

C
much greater heat of hydration of $$F^-$$ than $$Cl^-$$

D
greater electronegativity of $$F$$ than $$Cl$$

Solution

In spite of having less electron affinity than chlorine, fluorine is the strongest oxidising agent among the halogens. The most important reason for this phenomenon is much greater heat of hydration of $$F^-$$ than $$Cl^-$$ which compensates for the lower electron affinity of $$F$$ than of $$Cl$$. Hence, fluorine is the strongest oxidizing agent.
Question 8
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Which of the following arrangement of bond dissociation enthalpy of halogens is correct?

A
$$F-F>Cl-Cl>Br-Br$$

B
$$F-F< Cl-Cl < Br-Br$$

C
$$F-F > Cl-C l< Br-Br$$

D
$$F-F < Cl-Cl > Br-Br$$

Solution

Bond energy is inversely dependent on bond length. The order of bond length is $$Br_2 > Cl_2 > F_2.$$.

Hence, the order of bond energy should be $$Br_2 < Cl_2 < F_2.$$. But, bond dissociation enthalpy of $$F_{2}$$ is less than $$Cl_{2}$$ because of small size of $$F_{2}$$ and large inter-electronic repulsions.

Thus, order of bond energy is $$Cl_{2} > F_{2}$$ and $$ Br_2 < Cl_2$$.
Question 9
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Ammonia can be dried by :

A
conc. $$H_2SO_4$$

B
$$P_4O_{10}$$

C
$$CaO$$

D
$$CaCl_2$$

Solution

$$NH_3$$ is basic in nature and cannot be dried by dehydrating agents of acidic nature such as conc. $$ H_2SO_4, \ P_4O_{10}$$ and $$CaCl_2$$.

$$2NH_3+H_2SO_4 \rightarrow (NH_4)_2SO_4$$

$$12NH_3 + P_4O_{10} + 6H_2O \rightarrow 4(NH_4)_3PO_4$$

$$CaCl_2+8{NH}_3\rightarrow CaCl_2.8{NH}_3$$.

$$CaO$$ is used in drying $$NH_3$$ because $$CaO$$ being basic in nature does not react chemically with $$NH_3$$.

Hence, option C is correct.
Question 10
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Which of the following set of elements have the strongest tendency to form anions?

A
N, O and F

B
P, S and Cl

C
N, P and Cl

D
N, P and S

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The p-Block Elements Test - 62

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The p-Block Elements Test - 62

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Question 1

$$Cl_{2}O_{7}$$

$$H_{2}N_{2}O_{2}$$

$$N_{2}O_{5}$$

$$H_{2}S_{2}O_{7}$$

Solution

Question 2

$$CaCl_2$$

$$NaCl$$

$$NaOH$$

$$NaHCO_3$$

Solution

Question 3

$$He$$

$$Ne$$

$$Ar$$

$$Kr$$

Solution

Question 4

$$I>II>III$$

$$II>I>III$$

$$III>II>I$$

$$I>III>II$$

Solution

Question 5

$$SO_2$$

$$P_4O_{10}$$

$$H_3PO_4$$

$$B_3N_3H_6$$

Solution

Question 6

$$Mg_3N_2 + H_2O\rightarrow$$ Products

$$(NH_4)_2SO_4 + CaO \rightarrow$$ Products

$$NaNO_2 + Zn\xrightarrow{NaOH}$$ Products

$$NH_4NO_3 \xrightarrow[ ]{\Delta }$$ Products

Question 7

lower bond dissociation energy of $$F_{2}$$ than $$Cl_{2}$$

much lower heat of hydration of $$F^-$$ than $$Cl^-$$

much greater heat of hydration of $$F^-$$ than $$Cl^-$$

greater electronegativity of $$F$$ than $$Cl$$

Solution

Question 8

$$F-F>Cl-Cl>Br-Br$$

$$F-F< Cl-Cl < Br-Br$$

$$F-F > Cl-C l< Br-Br$$

$$F-F < Cl-Cl > Br-Br$$

Solution

Question 9

conc. $$H_2SO_4$$

$$P_4O_{10}$$

$$CaO$$

$$CaCl_2$$

Solution

Question 10

N, O and F

P, S and Cl

N, P and Cl

N, P and S

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