The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

A transition metal $$X$$ has a configuration $$\left[Ar\right]3d^{5}$$ in its $$+3$$ oxidation state. Its atomic number is:-

A
$$22$$

B
$$26$$

C
$$28$$

D
$$19$$
Question 2
1 / -0

Magnetic moments of $$V(Z = 23), Cr(Z = 24)$$ and $$Cu(Z = 29)$$ are x,y and z respectively, then which of the following is correct?

A
$$x = y = z$$

B
$$y > x > z$$

C
$$x > y > z$$

D
$$z > y > x$$

Solution

$$V\left( z=23 \right) \quad \left[ Ar \right] { 3d }^{ 3 }\rightarrow n=3$$

$$Cr\left( z=24 \right) \quad \left[ Ar \right] { 4s }^{ 1 }{ 3d }^{ 5 }\rightarrow n=6$$

$$Cu\left( z=29 \right) \quad \left[ Ar \right] { 4s }^{ 1 }{ 3d }^{ 10 }\rightarrow n=1$$

where n= no. of unpaired electrons

So, $$y>x>z$$

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

Which of the following ion will not form coloured aqueous solution?

A
$$Ni^{2+}$$

B
$$Fe^{3+}$$

C
$$Ti^{4+}$$

D
$$Cu^{2+}$$

Solution

$$ \displaystyle Ni^{2+}$$ ion with $$ \displaystyle 3d^8$$ outer electronic configuration has two unpaired electrons. It will form a green coloured aqueous solution.

$$ \displaystyle Fe^{3+}$$ ion with $$ \displaystyle 3d^5$$ outer electronic configuration has five unpaired electrons. It will form yellow coloured aqueous solution.

$$ \displaystyle Ti^{4+}$$ ion will not form coloured aqueous solutionaas it does not contain unpaired electrons.
$$ \displaystyle Ti^{4+}$$ ion with $$ \displaystyle 3d^0$$ outer electronic configuration has zero unpaired electrons.

$$ \displaystyle Cu^{2+}$$ ion with $$ \displaystyle 3d^9$$ outer electronic configuration has one unpaired electron. It will form blue coloured aqueous solution.

Hence, option $$C$$ is correct.
Question 4
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(i) $$\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+AcOH\rightarrow \underset{ppt}{M}$$
(ii) $$\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+HCN\rightarrow \underset{ppt.}{N}$$
Which is the correct option regarding the given reactions?

A
M and N are the same and characterized as $$BaCrO_4$$

B
M is $$BaCrO_4$$ and N stands for no precipitate

C
M is $$BaCrO_4$$ and N is $$BaCO_3$$

D
M stands for no precipitate and N is $$BaCrO_4$$

Solution

(i) $$\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+AcOH\rightarrow \underset{ppt} BaCrO_4+Na_2CO_3$$

(ii) $$\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+HCN\rightarrow \underset{ppt.} BaCrO_4+Na_2CO_3$$
So M and N are $$BaCrO_4$$.
Hence option A is correct.
Question 5
1 / -0

Which one of the following shows maximum paramagnetic character?

A
$${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 3- }$$

B
$${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 4- }$$

C
$${ \left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$$

D
$${ \left[ Cu{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ }$$

Solution

(i) $$[Fe(CN)_6]^{3-}$$

Here, oxidation state of $$Fe = +3$$

Electronic configuration of $$Fe$$ (ground state) $$= [Ar]3d^64s^2$$

Electronic configuration of $$Fe$$ (excited state) $$= [Ar]3d^54s^0$$

Number of unpaired electrons $$=5$$

(ii) $$[Fe(CN)_6]^{4-}$$

Here, oxidation state of $$Fe = +2$$

Electronic configuration of $$Fe$$ (ground state) $$= [Ar]3d^64s^2$$

Electronic configuration of $$Fe$$ (excited state) $$= [Ar]3d^64s^0$$

Number of unpaired electrons $$=4$$

(iii) $$[Cr(H_2O)_6]^{3+}$$

Here, oxidation state of $$Cr = +3$$

Electronic configuration of $$Cr^{3+}= [Ar]3d^34s^0$$

Number of unpaired electrons $$=3$$

(iv) $$[Cu(H_2O)_6]^{2+}$$

Here, oxidation state of $$Cu = +2$$

Electronic configuration of $$Cu^{2+} = [Ar] 3d^94s^0$$

Number of unpaired electrons $$=1$$

A number of unpaired electrons present is directly proportional to a paramagnetic character.

So, $$[Fe(CN)_6]^{3-}$$ has maximum paramagnetic character.
Question 6
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Catalytic activity of transition metals depends on:

A
their ability to exist in different oxidation states

B
the size of the metal atoms

C
the number of empty atomic orbitals available

D
all of the above

Solution
Question 7
1 / -0

Find the spin magnetic moment of the $$Fe^{3+}$$.

A
$$5.91$$

B
$$4.89$$

C
$$3.87$$

D
none of these

Solution
Question 8
1 / -0

What would be the correct order if Zn,Fe,Al and Cu are arranged in increasing order of reactivity ?

A
Cu, Fe, Zn, Al

B
Al, Cu, Fe, Zn

C
Zn, Al, Cu, Fe

D
Fe, Zn, Al, Cu

Solution
Question 9
1 / -0

The incorrect statement is :

A
$$Cu^{2+}$$ salts give red colour borax bead test in reducing flame.

B
$$Cu^{2+}$$ and $$Ni^{2+}$$ ions give black colour precipitate with $$H_2S$$ in the presence of $$HCl$$ solution.

C
Ferric ion gives blood red colour with potassium thiocyanate.

D
$$Cu^{2+}$$ ion gives chocolate coloured precipitate with potassium ferrocyanide solution.
Question 10
1 / -0

Which has the maximum magnetic moment?

A
$$Mn^{2+}$$

B
$$Fe^{2+}$$

C
$$Ti^{2+}$$

D
$$Cr^{2+}$$

Solution

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 62

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The d- and f- Block Elements Test - 62

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Question 1

$$22$$

$$26$$

$$28$$

$$19$$

Question 2

$$x = y = z$$

$$y > x > z$$

$$x > y > z$$

$$z > y > x$$

Solution

Question 3

$$Ni^{2+}$$

$$Fe^{3+}$$

$$Ti^{4+}$$

$$Cu^{2+}$$

Solution

Question 4

M and N are the same and characterized as $$BaCrO_4$$

M is $$BaCrO_4$$ and N stands for no precipitate

M is $$BaCrO_4$$ and N is $$BaCO_3$$

M stands for no precipitate and N is $$BaCrO_4$$

Solution

Question 5

$${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 3- }$$

$${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 4- }$$

$${ \left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$$

$${ \left[ Cu{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ }$$

Solution

Question 6

their ability to exist in different oxidation states

the size of the metal atoms

the number of empty atomic orbitals available

all of the above

Solution

Question 7

$$5.91$$

$$4.89$$

$$3.87$$

none of these

Solution

Question 8

Cu, Fe, Zn, Al

Al, Cu, Fe, Zn

Zn, Al, Cu, Fe

Fe, Zn, Al, Cu

Solution

Question 9

$$Cu^{2+}$$ salts give red colour borax bead test in reducing flame.

$$Cu^{2+}$$ and $$Ni^{2+}$$ ions give black colour precipitate with $$H_2S$$ in the presence of $$HCl$$ solution.

Ferric ion gives blood red colour with potassium thiocyanate.

$$Cu^{2+}$$ ion gives chocolate coloured precipitate with potassium ferrocyanide solution.

Question 10

$$Mn^{2+}$$

$$Fe^{2+}$$

$$Ti^{2+}$$

$$Cr^{2+}$$

Solution

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