The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 61

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Weekly Quiz Competition

Question 1
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A student made the following observations in the laboratory.

I. Clean copper metal did not react with $$1$$ molar $$Pb(NO_{3})_{2}$$ solution
II. Clean lead metal dissolved in a $$1$$ molar $$AgNO_{3}$$ solution and crystals of $$Ag$$ metal appeared
III. Clean silver metal did not react with $$1$$ metal $$Cu(NO_{3})_{2}$$ solution

The order of decrease in reducing character of three metals is:

A
$$Pb, Cu, Ag$$

B
$$Pb, Ag, Cu$$

C
$$Cu, Ag, Pb$$

D
$$Cu, Pb, Ag$$

Solution

The order of decrease in reducing character of three metals is $$Pb >Cu > Ag$$.

$$Pb$$ has maximum reducing power. Hence, $$Pb$$ can reduce both copper ions and silver ions.

$$Ag$$ has a minimum reducing power. Hence, it can neither reduce copper ions nor reduce lead ions.

The reducing power of $$Cu$$ is higher than that of $$Ag$$ but lower than that of $$Pb$$. Hence, $$Cu$$ can reduce silver ions but $$Cu$$ cannot reduce lead ions.
Question 2
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The compound pictured below undergoes decomposition when heated.
Which of the following CORRECTLY represents a possible set of products of this decomposition?

A

B

C

D

Solution

The given compound is (NH$$_4)_2 Cr_2O_7$$
Now,
(NH$$_4)_2 Cr_2O_7$$ (s) $$^\Delta\rightarrow$$ $$Cr_2O_3$$ (s) $$+N_2(g) +4H_2O$$ (g)
When ammonium dichromate is heated , it gives out orange sparks and green $$Cr_2O_3$$ crystals and thrown out into the alr producing volcanic eruption like effect.
Question 3
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The magnetic moment of $$\left[ Co{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] { Cl }_{ 3 }$$ is:

A
$$1.73$$

B
$$2.83$$

C
$$6.6$$

D
Zero

Solution

In $${ \left[ Co{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] }^{ 3+ }$$, the oxidation state of $$Co$$ is $$+3$$.

$${ \left[ Co{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] }^{ 3+ }$$ does not contain unpaired electron hence, its magnetic moment is zero.
Question 4
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Actinides exhibit a larger number of oxidation state than that of corresponding lanthanides. The reason behind this aspect is :

A
lesser energy difference between 5f and 6d-orbitals than between 4f and 5d-orbitals

B
larger atomic size of actinides than the lanthanides

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

D
greater reactive nature of the actinides than the lanthanides

Solution

Because of lesser energy difference between $$5f$$ and $$6d$$ orbitals than that between $$4f$$ and $$5d$$-orbitals actinides exhibit larger number of oxidation state than that of corresponding lanthanides. There is no large energy difference for pulling out the electrons in case of actinides.
Question 5
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In the oxidation of oxalic acid by potassium permanganate, in the presence of dil. $$H_2SO_4$$ one of the products _______ acts as an autocatalyst.

A
$$K_2SO_4$$

B
$$MnSO_4$$

C
$$MnO_2$$

D
$$Mn_2O_3$$

Solution

In the oxidation of oxalic acid by potassium permanganate, in the presence of dil. $$\displaystyle H_2SO_4$$ one of the products $$\displaystyle MnSO_4$$ acts as an autocatalyst.

$$\displaystyle 2KMnO_4 +5H_2C_2O_4+3H_2SO_4\rightarrow 2MnSO_4+K_2SO_4+10CO_2 + 8H_2O$$
Question 6
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The magnetic moment of a transition metal ion is $$\sqrt { 15 } BM$$. Therefore, the number of unpaired electrons present in it is:

A
$$4$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

The magnetic moment of a transition metal ion is $$\sqrt { 15 } BM$$. Therefore, the number of unpaired electrons present in it is $$3$$.
If n is the number of unpaired electrons, then the magnetic moment is given by the expression:

$$ \displaystyle \mu = \sqrt {n(n+2)}$$
$$ \displaystyle \sqrt { 15 } = \sqrt {n(n+2)}$$
$$ \displaystyle { 15 } = {n(n+2)}$$
$$ \displaystyle 15 = n^2 + 2n$$
$$ \displaystyle n^2+2n- { 15 } = 0$$

This is the quadratic equation with solution

$$ \displaystyle n = \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$$
$$ \displaystyle n = \dfrac {-2 \pm \sqrt {2^2-4(1)(- { 15 } )}}{2(1)}$$
$$ \displaystyle n = \dfrac {-2 \pm 8}{2(1)}$$
$$ \displaystyle n=3$$ or $$ \displaystyle n=-5$$.

The negative value is discarded as the number of unpaired electrons cannot be negative.

Hence, $$ \displaystyle n=3$$ .
Question 7
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Which one of the following complex ions has the highest magnetic moment?

A
$$[Cr(NH_{3})_{6}]^{3+}$$

B
$$[Fe(CN)_{6}]^{3-}$$

C
$$[Fe(CN)_{6}]^{4-}$$

D
$$[Zn(NH_{3})_{6}]^{2+}$$

Solution

Higher the number of unpaired electrons, the higher is the magnetic moment.

In $$[Cr(NH_{3})_{6}]^{3+}, Cr$$ is present as $$Cr^{3+}$$. Thus, $$Cr$$ in $$[Cr(NH_{3})_{6}]^{3+} = [Ar] 3d^{3} 4s^{0} 4p^{0}$$. So, it will have 3 unpaired electrons.

In $$[Fe(CN)_{6}]^{3-}$$, one unpaired electron is present and in $$[Fe(CN)_{6}]^{4-}$$ and $$[Zn(NH_{3})_{6}]^{2+}$$, both, unpaired electrons are absent. $$CN^-$$ is a strong field ligand, thus causes pairing.

Hence, $$[Cr(NH_{3})_{6}]^{3+}$$ has the highest magnetic moment among the given complex ions. $$[\mu = \sqrt {3(3 + 2)} = 3.87\ BM] $$
Question 8
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In the dichromate dianion, the nature of bonds are:

A
four equivalent $$Cr - O$$ bonds

B
six equivalent $$Cr - O$$ bonds and one $$O - O$$ bond

C
six equivalent $$Cr - O$$ bonds and one $$Cr - Cr$$ bond

D
six non-equivalent $$Cr - O$$ bonds

E
six equivalent $$Cr - O$$ bonds and one $$Cr - O - Cr$$ bond

Solution

In the dichromate dianion structure, the negative charge can be on any $$O$$ attached to $$Cr$$, therefore all $$Cr-O$$ are equivalent. Also have $$Cr-O-Cr$$ bond.
Thus, option D is correct.
Question 9
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Which of the following complex compounds will exhibit the highest paramagnetic behaviour?

$$[At. no. Ti = 22, Cr = 24, Co = 27, Zn = 30]$$

A
$$[Ti(NH_{3})_{6}]^{3+}$$

B
$$[Cr(NH_{3})_{6}]^{3+}$$

C
$$[Co(NH_{3})_{6}]^{3+}$$

D
$$[Zn(NH_{3})_{6}]^{2+}$$

Solution

(a) Electronic configuration of $$Ti^{3+}$$ in $$[Ti(NH_{3})_{6}]^{3+}$$
$$Ti^{3+} = 3d^{1}$$ [ref. image (a)]

(b) Electronic configuration of $$Cr^{3+}$$ in $$[Cr(NH_{3})_{6}]^{3+}$$
$$Cr^{3+} = 3d^{3}$$; [ref. image (b)]

(c) Electronic configuration of $$Co^{3+}$$ in $$[Co(NH_{3})_{6}]^{3+}$$;
$$Co^{3+} = 3d^{6}$$.
In the presence of strong field ligand $$NH_{3}$$, pairing of electrons takes place and hence, octahedral complex, $$[Co(NH_{3})_{6}]^{3+}$$ is diamagnetic.
$$[Co(NH_{3})_{6}]^{3+} \begin{bmatrix}\text {inner orbital or}\\ \text {low spin complex}
\end{bmatrix} (6NH_{3} molecles)$$   [ref. image (c)]

(d) Electronic configuration of $$Zn^{2+}$$ in $$[Zn(NH_{3})_{6}]^{2+}$$
$$\therefore [Zn(NH_{3})_{6}]^{2+}$$ is an outer orbital complex and is diamagnetic.   [ref. image (d)]

$$[Cr(NH_{3})_{6}]^{3+}$$ complex has the highest unpaired electron. Hence it will be highly paramagnetic.

Option B is correct.
Question 10
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The magnetic moment of a divalent ion in aqueous solution with atomic number 25 is:

A
5.47 B.M

B
2.9 B.M

C
6.9 B.M

D
6.7 B.M

Solution

electronic configuration of the element with atomic number 25(divalent ion) will be $$[Ar]3d^5$$,having $$5$$ unpaired electrons.

magnetic moment will be $$\sqrt{n(n+2)}$$

where n is unpaired electrons

n=5

$$\sqrt {5(5+2) }=\sqrt {35}=5.47$$