The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Which has the highest ionic radii among given ions?
$$Ce^{3+}, La^{3+}, Pm^{3+}$$ and $$Yb^{3+}$$

A
$$Yb^{3+}$$

B
$$Ce^{3+}$$

C
$$La^{3+}$$

D
$$Pm^{3+}$$

Solution

Electronic configuration of ions are:
1. $$Yb^{ 3+ }:Xe4f^{ 13 }$$
2.$$Ce^{ 3+ }:Xe4f^{ 1 }$$
3.$$La^{ 3+ }:Xe4f^0$$
4.$$Pm^{ 3+ }:Xe4f^4$$
Since, the least shielding tendency of f-orbital electron, effective nuclear charge increases and therefore size decreases along the period. Therefore $$La^{ 3+ }$$ has highest ionic radii.
This phenomenon is called lanthanoid contraction.
Question 2
1 / -0

The magnetic moment of an ion in its +3 oxidation state is 3.85 BM. The number of unpaired electrons present in the ion is _____________.

A
1

B
4

C
3

D
5

Solution

Given that magnetic moment is 3.85 BM

Magnetic moment = $$\sqrt{n(n+2)}$$
where n is no. of unpaired electron.

$$\therefore$$ 3.85 = $$\sqrt{n(n+2)}$$

No. of unpaired electron present in ion is 3.

Hence, the correct option is $$C$$
Question 3
1 / -0

IP values of Sc, Y and La are in the order:

A
Sc > Y > La

B
Sc < Y < La

C
Sc = Y = La

D
La > Sc > Y

Solution

IP values of Sc, Y and La are in the order $$\displaystyle Sc > Y > La $$

On moving down the group, the IP values decreases as the effective nuclear charge decreases due to poor shielding by inner electrons.
Question 4
1 / -0

Number of $$Cr-O$$ bonds in dichromate ion$$(Cr_{2}O_{7}^{2-})$$ is:

A
6

B
7

C
8

D
4

Solution

From structure, it is clear that it has $$8\ Cr-O$$ bond, out of which $$6\ Cr-O$$ bonds are equal.
Question 5
1 / -0

In dichromate ion ________________________.

A
4 Cr-O bonds are equivalent

B
6 Cr-O bonds are equivalent

C
8 Cr-O bonds are equivalent

D
all Cr-O bonds are equivalent

Solution

In dichromate dianion 6 Cr-O bonds are equivalent.
Question 6
1 / -0

Platinum, palladium and iridium are called noble metals because:

A
Alfred Nobel discovered them

B
They are inert towards many common reagents

C
They are shining, lustrous and pleasing to look at

D
They are found in active state

Solution

Due to their electronic configuration, they are inert to many common reagents so they are called noble metals.
Question 7
1 / -0

Arrange $$Ce^{3+}, La^{3+}, Pm^{3+}$$ and $$Yb^{3+}$$ in increasing order of their ionic radii.

A
$$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$$

B
$$Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}$$

C
$$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$$

D
$$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$$

Solution

As we move from left to right in a period, radii decrease due to an increase in nuclear charge and lanthanide contraction.

So the order is $$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}.$$
Question 8
1 / -0

Transition elements show positive oxidation state, generally, due to:

A
large atomic size

B
low ionization energy

C
low electronegativity

D
high electronegativity
Question 9
1 / -0

Many lanthanoid elements are used to prepare:

A
ceramic materials

B
water softener

C
superconducting materials

D
enzyme catalysts

Solution
Question 10
1 / -0

Select the correct order with respect to paramagnetic property of the ions.

A
$$Mn^{2{+}} > Cr^{2+} > V^{2{+}} > Ti^{2+}$$

B
$$Ti^{2{+}} > V^{2{+}} > Mn^{2{+}} > Cr^{{2+}}$$

C
$$V^{2{+}} > Ti^{2{+}} > Mn^{2{+}} > Cr^{2{+}}$$

D
$$Ti^{3{+}} > V^{2{+}} > Mn^{2{+}} > Cr^{2{+}}$$

Solution

The expression for magnetic moment is $$\sqrt{n(n+2)}$$, where $$n$$ is number of unpaired electrons.

So, greater is the number of unpaired electrons, greater will be the magnetic moment of an ion.

$$Mn^{2+}$$ has $$5$$ unpaired electrons.

$$Cr^{2+}$$ has $$4$$ unpaired electrons.

$$V^{2+}$$ has $$3$$ unpaired electrons.

$$Ti^{2+}$$ has $$2$$ unpaired electrons.

Hence, the correct order is as follows:

$$Mn^{2+}>Cr^{2+}>V^{2+}>Ti^{2+}$$

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 20

Result

The d- and f- Block Elements Test - 20

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SHARING IS CARING

Question 1

$$Yb^{3+}$$

$$Ce^{3+}$$

$$La^{3+}$$

$$Pm^{3+}$$

Solution

Question 2

1

4

3

5

Solution

Question 3

Sc > Y > La

Sc < Y < La

Sc = Y = La

La > Sc > Y

Solution

Question 4

6

7

8

4

Solution

Question 5

4 Cr-O bonds are equivalent

6 Cr-O bonds are equivalent

8 Cr-O bonds are equivalent

all Cr-O bonds are equivalent

Solution

Question 6

Alfred Nobel discovered them

They are inert towards many common reagents

They are shining, lustrous and pleasing to look at

They are found in active state

Solution

Question 7

$$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$$

$$Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}$$

$$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$$

$$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$$

Solution

Question 8

large atomic size

low ionization energy

low electronegativity

high electronegativity

Question 9

ceramic materials

water softener

superconducting materials

enzyme catalysts

Solution

Question 10

$$Mn^{2{+}} > Cr^{2+} > V^{2{+}} > Ti^{2+}$$

$$Ti^{2{+}} > V^{2{+}} > Mn^{2{+}} > Cr^{{2+}}$$

$$V^{2{+}} > Ti^{2{+}} > Mn^{2{+}} > Cr^{2{+}}$$

$$Ti^{3{+}} > V^{2{+}} > Mn^{2{+}} > Cr^{2{+}}$$

Solution

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