The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 27

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

Ammonium dichromate is used in fire works. The green colored powder blow in the air is:

A
$$Cr{O}_{3}$$

B
$${Cr}_{2}{O}_{3}$$

C
$$Cr$$

D
$$CrO ({O}_{2})$$

Solution

$$(\mathrm{NH}_4)_2\mathrm{Cr}_2\mathrm{O}_7\xrightarrow{\Delta} \mathrm{Cr}_2\mathrm{O}_3+\mathrm{N}_2+4\mathrm{H}_2\mathrm{O}$$

The green-coloured $$ Cr_2O_3$$ powder is blown in the air by the large volume of $$ N_2$$ and water vapour produced and settles like dust from the volcano.

Hence, the correct option is $$\text{B}$$
Question 2
1 / -0

More number of oxidation states are exhibited by the actinoids than by lanthanoids. The main reason for this is:

A
greater metallic character of the lanthanoids than that of the corresponding actinoids.

B
more active nature of the actinoids.

C
more energy difference $$5f$$ and $$6d$$-orbitals than that between $$4f$$ and $$5d$$-orbitals.

D
lesser energy difference between $$5f$$ and $$6d$$-orbitals than that between $$4f$$ and $$5d$$-orbitals.

Solution

Actinoids display more oxidation states because of very small energy gap between 5f, 6d and 7s sub shells. Thus, the outermost electrons get easily excited to the higher energy level giving variable oxidation state.

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

The colorless species is:

A
$$V{Cl}_{3}$$

B
$$VO{SO}_{4}$$

C
$${Na}_{3}{VO}_{4}$$

D
$$\left[ V{ \left( { H }_{ 2 }O \right) }_{ 6 }{ SO }_{ 4 } \right] .{ H }_{ 2 }O$$

Solution

$$ VCl_3$$ is green.

$$ VOSO_4$$ is blue.

$$ Na_3VO_4$$ is colorless vanadium is present in +5 oxidation state .it does not have any unpaired electron.

$$\left[ V{ \left( { H }_{ 2 }O \right) }_{ 6 }{ SO }_{ 4 } \right]$$ .$${ H }_{ 2 }O$$ is purple.
Question 4
1 / -0

Titanium shows magnetic moment of $$1.73BM$$; then find out its oxidation state:

A
$$+1$$

B
$$+4$$

C
$$+3$$

D
$$+2$$

Solution

$$1.73 BM$$ $$ = \sqrt {n\times(n+2)} $$.

Solving this we get n $$=$$ 1.

So Titanium has a electron configuration of [Ar] $$ 4s^03d^1$$. So its oxidation state is +3.
Question 5
1 / -0

In $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ every $$Cr$$ is linked to:

A
two $$O$$-atoms.

B
three $$O$$-atoms.

C
four $$O$$-atoms.

D
five $$O$$-atoms.

Solution

Every $$Cr$$ is linked to four other $$O$$ atoms in $$Cr_2O_7^{2-}$$.
Question 6
1 / -0

Which pair of compounds is expected to show similar color in aqueous medium?

A
$$Fe{Cl}_{3}$$ and $$Cu{Cl}_{2}$$

B
$$VO{Cl}_{2}$$ and $$Cu{Cl}_{2}$$

C
$$VO{Cl}_{2}$$ and $$Fe{Cl}_{2}$$

D
$$Fe{Cl}_{2}$$ and $$Mn{Cl}_{2}$$

Solution

The color of a transition metal ion is related to its number of unpaired d electrons. Hence, $$ CuCl_2$$ and $$ VOCl_2$$ are expected to have same color.
Question 7
1 / -0

The compound of vanadium has magnetic moment of $$1.73B$$. The vanadium chloride has the formula:

A
$$V{Cl}_{2}$$

B
$$V{Cl}_{3}$$

C
$$V{Cl}_{4}$$

D
$$V{Cl}_{5}$$

Solution

$$ \sqrt {n(n+2)} = 1.73$$ Solving this we get the number of unpaired electrons as 1. $$ V^{4+}$$ will result in 1 unpaired electron. Hence the solution is $$ VCl_4$$
Question 8
1 / -0

The "Spin-only" magnetic moment [in units of Bohr magneton,$$({\mu}_{B})]$$ of $${Ni}^{2+}$$ in aqueous solution would be (At. no. $$Ni=28$$):

A
$$0$$

B
$$1.73$$

C
$$2.84$$

D
$$4.90$$

Solution

The electronic configuration of $$Ni^{+2}$$ is $$1s^22s^22p^63s^23p^63d^8$$

Number of unpaired electrons in $$Ni^{+2}$$ is 2 formula used for the spin only magnetic moment

$$\mu = \sqrt{n(n+2)}$$

$$\mu = \sqrt{2(2+2)}$$

$$\mu = \sqrt{8}$$

$$\mu = 2.84BM$$

Hence, the correct option is $$\text{C}$$
Question 9
1 / -0

Which of the following elements involves gradual filling of $$5f$$-level?

A
Lanthanoids

B
Actinoids

C
Transition metals

D
Coinage metals

Solution

Actinoids involves gradual filling of $$5f$$-level.
Hence option B is correct.
Question 10
1 / -0

Among the following metals, interatomic forces are probably the weakest in :

A
$$Cu$$

B
$$Ag$$

C
$$Zn$$

D
$$Hg$$

Solution

Among the metals, interatomic forces are probably the weakest in $$Hg.$$

This can be understand from the fact that $$Hg$$ is liquid whereas $$Cu, Ag$$ and $$Zn$$ are solids at room temperature.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 27

Result

The d- and f- Block Elements Test - 27

Score

-

Rank

-

SHARING IS CARING

Question 1

$$Cr{O}_{3}$$

$${Cr}_{2}{O}_{3}$$

$$Cr$$

$$CrO ({O}_{2})$$

Solution

Question 2

greater metallic character of the lanthanoids than that of the corresponding actinoids.

more active nature of the actinoids.

more energy difference $$5f$$ and $$6d$$-orbitals than that between $$4f$$ and $$5d$$-orbitals.

lesser energy difference between $$5f$$ and $$6d$$-orbitals than that between $$4f$$ and $$5d$$-orbitals.

Solution

Question 3

$$V{Cl}_{3}$$

$$VO{SO}_{4}$$

$${Na}_{3}{VO}_{4}$$

$$\left[ V{ \left( { H }_{ 2 }O \right) }_{ 6 }{ SO }_{ 4 } \right] .{ H }_{ 2 }O$$

Solution

Question 4

$$+1$$

$$+4$$

$$+3$$

$$+2$$

Solution

Question 5

two $$O$$-atoms.

three $$O$$-atoms.

four $$O$$-atoms.

five $$O$$-atoms.

Solution

Question 6

$$Fe{Cl}_{3}$$ and $$Cu{Cl}_{2}$$

$$VO{Cl}_{2}$$ and $$Cu{Cl}_{2}$$

$$VO{Cl}_{2}$$ and $$Fe{Cl}_{2}$$

$$Fe{Cl}_{2}$$ and $$Mn{Cl}_{2}$$

Solution

Question 7

$$V{Cl}_{2}$$

$$V{Cl}_{3}$$

$$V{Cl}_{4}$$

$$V{Cl}_{5}$$

Solution

Question 8

$$0$$

$$1.73$$

$$2.84$$

$$4.90$$

Solution

Question 9

Lanthanoids

Actinoids

Transition metals

Coinage metals

Solution

Question 10

$$Cu$$

$$Ag$$

$$Zn$$

$$Hg$$

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.