The p-Block Elements Test

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The p-Block Elements Test - 54

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Weekly Quiz Competition

Question 1
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Directions For Questions

The heavier elements of group $$13, 14$$ and $$15$$ besides their group oxidation state exhibit another oxidation state which is two units lower than the group oxidation state. The stability of lower oxidation state increases down the group.
The display of lower oxidation state is due to inert pair effect.

...view full instructions

Which of the following statements is incorrect?

A
Boron shows only $$+3$$ oxidation state.

B
In $$Ga+3$$ oxidation state is more stable than $$+1$$ oxidation state.

C
In $$Sn+2$$ oxidation state is more stable than $$+4$$ oxidation state.

D
In $$Tl+1$$ oxidation state is more stable than $$+3$$ oxidation state.

Solution

In the group 13, 14, 15 the stability of lower oxidation state increases down in the group.

But, $$Sn^{+4}$$ is more stable than $$Sn^{+2}$$ due to its configuration.

The electronic configuration of Sn is $$5s^{2} 5p^{4}$$.

In $$Sn^{4}$$, it lose four electrons and achieve full filled electronic configuration.

Due to its stability , $$Sn^{+4}$$ is more stable than $$Sn^{2}$$.

Among the all statements mentioned, $$\textbf{(C) is incorrect}$$.
Hence, option will be $$(C)$$.
Question 2
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Directions For Questions

The heavier elements of group $$13, 14$$ and $$15$$ besides their group oxidation state exhibit another oxidation state which is two units lower than the group oxidation state. The stability of lower oxidation state increases down the group.
The display of lower oxidation state is due to inert pair effect.

...view full instructions

Inert pair effect is not exhibited by:

A
$$Pb$$

B
$$Bi$$

C
$$Tl$$

D
$$B$$

Solution

$$\textbf{Inert pair effect}$$ is the tendency of the two electrons in the outermost atomic s-orbital to remain unshared in compounds of post-transition metals.

Inert pair effect is not exihibited by $$\textbf{Boron}$$ due to some relativistic effects caused by the enormous speed of electrons on lower energy-levels in heavier atoms.It shows only single state i.e. +3.

$$\textit{Hence,}$$
$$\textbf{correct option is (D) B}$$ .
Question 3
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Correct match is:

A
Ordinary form of borax $$: Na_2B_4O_7 \cdot 5H_2O$$

B
Colemanite $$: Ca_2B_6O_{11} \cdot 5H_2O$$

C
Boronatrocalcite $$: 2Mg_3B_8O_{15} \cdot MgCl_2$$

D
Octahedral form of borax $$: Na_2B_4O_7 \cdot 10H_2O$$

Solution
Question 4
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Boron does not form $${B}^{3+}$$ cation easily. It is due to

A
energy required to form $${B}^{3+}$$ ion is very high

B
Boron is non-metal

C
Boron is semi-metal

D
none of the above

Solution

The correct option is (A)
Explanation:
As the Boron atom is small in size a large amount of energy is needed to remove 3 electrons from the boron atom. So Boron does not form B3+ ion.
The atomic number of Boron is 5. Its electronic configuration is 1s [2]2s[2]2p[1].
When one electron is removed from the p orbital a He-like fulfilled s orbital is left. This is highly stable. So the second ionization enthalpy is quite high.
Again when one electron is removed a half filled orbital is left. So the third ionization enthalpy is also quite high.
Since the total energy needed to make B3+ is the total of all the ionization enthalpies an enormous amount of energy is required to form it.
Question 5
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Aqueous solutions of borax acts as a buffer because:

A
it contains weak acid and its salt with strong base

B
it contains tribasic acid and strong base

C
it contains number of neutral water molecules

D
none of the above

Solution
Question 6
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Which of the following is most stable?

A
$$Pb^{2+}$$

B
$$Ge^{2+}$$

C
$$Si^{2+}$$

D
$$Sn^{2+}$$

Solution

The inert pair effect is the tendency of the two electrons in the outermost atomic s-orbital to remain unshared in compounds of post-transition metals. As a result, the inert pair of ns electrons remains more tightly held by the nucleus and hence participates less in bond formation.

Hence, the tendency to show lower oxidation state increases down the group due to the inert pair effect.
Since, all the given ions are of same group, the most stable ion with +2 state is Pb²⁺.

Thus, option A is correct.
Question 7
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Which of the following halides is most acidic?

A
$$BCl_3$$

B
$$SbCl_3$$

C
$$BiCl_3$$

D
$$CCl_4$$

Solution

A species which is more electron deficient will attract electron easily and hence will be more acidic. Boron atom in $$BCl_3$$, has six electrons in the outermost orbit and thus it can accept a pair of electrons form a donor molecule to complete its octet. The effective overlap between orbital's decreases due to large size of p orbital of halides. Hence $$BCl_3$$ is most acidic than other options.
Question 8
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The stable compound in $$+1$$ oxidation state are formed by:

A
$$B$$

B
$$Al$$

C
$$Tl$$

D
$$Ga$$

Solution
Question 9
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$$Tl I_3$$ is an ionic compound which furnishes the following ions:

A
$$Tl^{3+}$$ and $$I_3^-$$

B
$$Tl^{3+}$$ and $$I^-$$

C
$$Tl^{+}$$ and $$I^-$$

D
$$Tl^{+}$$ and $$I_3^-$$

Solution

$$Tl I_3$$ is an ionic compound which furnishes the following ions:

$$TlI_{3}\rightleftharpoons Tl^{+}+I_{3}^{-}$$

$$Tl$$ is in $$+1$$ oxidation state because of inert pair effect.
Question 10
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Directions For Questions

In the modern periodic table. the elements are placed in order of increasing atomic number. The long form of periodic table shows all the elements in numerical order. The start of a new period always corresponds to the introduction if the first electron into the $$s-$$ orbital of a new principal quantum number. The number of elements in each period corresponds to the number of electrons required to fill the orbitals according to Aufbau principle.

$$1s, 2s 2p, 3s 3p, 4s 3d 4p, 5s 4d 5p, 6s 4f 5d 6p, 7s 5f 6s 7p$$

The main groups correspond to elements in which $$s$$ and $$p$$ orbitals are being filled. The transition elements are those in which $$d-$$ orbitals are being filled. Each group contains elements of similar electronic configuration. According to the recommendation of $$IUPAC$$, the group of elements are numbered from $$1$$ to $$18$$. The elements corresponding to filling of $$4f-$$ orbitals are called lanthanoids and those corresponding to the filling of $$5f-$$ orbitals are called actinoids. These elements are accommodated into two separate horizontal rows below the table.

There are only $$81$$ table elements known. Form these elements one or more isotopes do not undergo spontaneous radioactive decay. No stable isotope exists for any elements above bismuth. Two elements, uranium and thorium for which only radioactive isotopes exist, are found quite abundantly on earth because the half-lives of some of their isotopes are almost as great as the age of the earth.

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Outermost configuration of two $$p-$$ block elements $$A$$ and $$B$$ are $$ns^{2}np^{1}$$ and $$ns^{2}np^{6}$$ respectively. The elements are if $$n=4$$;

A
$$A=Ga, B=Xe$$

B
$$A=In, B=Xe$$

C
$$A=Ga, B=Kr$$

D
$$A=Zn, B=Kr$$

Solution

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The p-Block Elements Test - 54

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Question 1

Boron shows only $$+3$$ oxidation state.

In $$Ga+3$$ oxidation state is more stable than $$+1$$ oxidation state.

In $$Sn+2$$ oxidation state is more stable than $$+4$$ oxidation state.

In $$Tl+1$$ oxidation state is more stable than $$+3$$ oxidation state.

Solution

Question 2

$$Pb$$

$$Bi$$

$$Tl$$

$$B$$

Solution

Question 3

Ordinary form of borax $$: Na_2B_4O_7 \cdot 5H_2O$$

Colemanite $$: Ca_2B_6O_{11} \cdot 5H_2O$$

Boronatrocalcite $$: 2Mg_3B_8O_{15} \cdot MgCl_2$$

Octahedral form of borax $$: Na_2B_4O_7 \cdot 10H_2O$$

Solution

Question 4

energy required to form $${B}^{3+}$$ ion is very high

Boron is non-metal

Boron is semi-metal

none of the above

Solution

Question 5

it contains weak acid and its salt with strong base

it contains tribasic acid and strong base

it contains number of neutral water molecules

none of the above

Solution

Question 6

$$Pb^{2+}$$

$$Ge^{2+}$$

$$Si^{2+}$$

$$Sn^{2+}$$

Solution

Question 7

$$BCl_3$$

$$SbCl_3$$

$$BiCl_3$$

$$CCl_4$$

Solution

Question 8

$$B$$

$$Al$$

$$Tl$$

$$Ga$$

Solution

Question 9

$$Tl^{3+}$$ and $$I_3^-$$

$$Tl^{3+}$$ and $$I^-$$

$$Tl^{+}$$ and $$I^-$$

$$Tl^{+}$$ and $$I_3^-$$

Solution

Question 10

$$A=Ga, B=Xe$$

$$A=In, B=Xe$$

$$A=Ga, B=Kr$$

$$A=Zn, B=Kr$$

Solution

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