The p-Block Elements Test

Result

The p-Block Elements Test - 33

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A Si and a Ge diode has identical physical dimensions. The band gap in Si is larger than that in Ge. An identical reverse bias is applied across the diodes

A
The reverse current in Ge is larger than that in Si

B
The reverse current in Si is larger than that in Ge

C
The reverse current is identical in the two diodes

D
The relative magnitude of the reverse currents cannot be determined from the given data only

Solution

The reverse current is identical in two diodes if the identical reverse bias is applied across the diodes.
Question 2
1 / -0

$$\displaystyle Al{ H }_{ 3 }$$ is an example for :

A
interstitial hydride

B
electron rich hydride

C
ionic hydride

D
electron deficient hydride

Solution

According to the Bohr's octet rule, each element tends to completely fill its outermost orbit with 8 electrons in it. Electronic configuration of aluminium is 2,8,3 that means it still needs 5 more electrons to complete its octet. That's why aluminium is electron deficient.

Hence, option $$D$$ is correct.
Question 3
1 / -0

Which of the following metals has the largest abundance in the earth's crust?

A
Aluminium

B
Calcium

C
Magnesium

D
Sodium

Solution

Aluminium is the most abundant metal and third most abundant element in earth's crust.
Question 4
1 / -0

Why carbon can form stable $$C = C$$ bond where as silicon can't form $$Si = Si$$ ?

A
Poor overlap of $$p\pi - d\pi$$ in silicon

B
Poor overlap of $$p-s$$ in silicon

C
Poor overlap of $$p\pi-s$$ in silicon

D
Silicon do not have vacant orbital

Solution

Silicon can't form $$Si=Si$$ due to poor overlapping of $$p\pi-d\pi$$ bond. Si has no tendency to form double bonds with itself due to its large size. The bond formation is difficult du to the large size.
Question 5
1 / -0

Boiling point increases in order $$CH_4$$ < $$SH_4$$ < $$GeH_4$$ because :

A
atomic number of central metal atom increases

B
molecular mass increases

C
dipole - dipole interaction increases

D
bond length increases

Solution

Boiling point increases in order $$CH_4 < SH_4 < GeH_4$$
molecular mass increases. The atomic mass increases in the order $$C < S < Ge$$. Hence, the molecular mass increases in the same order.
Question 6
1 / -0

Amphoteric oxide is:

A
$$Al_2O_3$$

B
$$SiO_2$$

C
$$P_4O_{10}$$

D
$$ClO_7$$
Solution
Explanation:
- The chemicals which react with both acids and bases are called amphoteric oxides.
- $$Al_2O_3$$ reacts with both acids as well as bases. So it is an amphoteric oxide.
- $$SiO_2$$, $$P_4O_{10}$$ and $$ClO_7$$ are acidic oxides, hence none of these three are amphoteric.
- Therefore, $$SiO_2$$, $$P_4O_{10}$$ and $$Cl_2O_7$$ are not amphoteric oxides.
Hence, the correct answer is option $$A$$.
Question 7
1 / -0

A type of ion found in aluminum oxide is :

A
$$X^+$$

B
$$X^{2+}$$

C
$$X^{3+}$$

D
$$XO_3^{2-}$$

E
$$XO_4^{2-}$$

Solution

A type of ion found in aluminum oxide is $$\displaystyle X^{3+}$$.
Aluminum oxide is $$\displaystyle Al_2O_3$$. It contains $$\displaystyle Al^{3+} $$ ion and $$\displaystyle O^{2-}$$ ion.
Question 8
1 / -0

Heavier elements of p-block do show higher order oxidation state because of :

A
lanthanide contraction

B
inert pair effect

C
screening effect

D
absence of vacant d-orbital

Solution

Heavier element of p-block does show higher oxidation state because of the inert pair effect.

The inert pair effect is the tendency of the electrons in the outermost atomic s orbital to remain unionized or unshared in compounds of post-transition metals.

The term inert pair effect is often used in relation to the increasing stability of oxidation states that are two less than the group valency for the heavier elements of groups 13, 14, 15 and 16.

Hence the correct option is B.
Question 9
1 / -0

Which one of the following is paramagnetic in nature?

A
$$NO$$

B
$$N_2O$$

C
$$N_2O_5$$

D
All of the above

Solution

Whenever two electrons are paired together in an orbital, or their total spin is 0, they are diamagnetic electrons. Atoms with all diamagnetic electrons are called diamagnetic atoms.
A paramagnetic electron is an unpaired electron. $$NO$$ has one unpaired electron. Hence, it is paramagnetic in nature.
Question 10
1 / -0

Out of $$HClO, HBrO$$ and $$HIO$$ which one is most acidic :

A
$$HBrO$$

B
$$HClO$$

C
$$HIO$$

D
All are equally acidic

Solution

As $$Cl$$ is most electronegative out of $$Cl, Br,I$$, it will make the $$Cl-O$$ bond more polar and weaker compared to $$Br$$ and $$I$$.

Therefore, $$HClO$$ will be most acidic.

Hence the correct option is B.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 33

Result

The p-Block Elements Test - 33

Score

-

Rank

-

SHARING IS CARING

Question 1

The reverse current in Ge is larger than that in Si

The reverse current in Si is larger than that in Ge

The reverse current is identical in the two diodes

The relative magnitude of the reverse currents cannot be determined from the given data only

Solution

Question 2

interstitial hydride

electron rich hydride

ionic hydride

electron deficient hydride

Solution

Question 3

Aluminium

Calcium

Magnesium

Sodium

Solution

Question 4

Poor overlap of $$p\pi - d\pi$$ in silicon

Poor overlap of $$p-s$$ in silicon

Poor overlap of $$p\pi-s$$ in silicon

Silicon do not have vacant orbital

Solution

Question 5

atomic number of central metal atom increases

molecular mass increases

dipole - dipole interaction increases

bond length increases

Solution

Question 6

$$Al_2O_3$$

$$SiO_2$$

$$P_4O_{10}$$

$$ClO_7$$

Solution

Question 7

$$X^+$$

$$X^{2+}$$

$$X^{3+}$$

$$XO_3^{2-}$$

$$XO_4^{2-}$$

Solution

Question 8

lanthanide contraction

inert pair effect

screening effect

absence of vacant d-orbital

Solution

Question 9

$$NO$$

$$N_2O$$

$$N_2O_5$$

All of the above

Solution

Question 10

$$HBrO$$

$$HClO$$

$$HIO$$

All are equally acidic

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.