The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 46

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Question 1
1 / -0

$$AgCl$$ and $$NaCl$$ are colourless. $$NaBr$$ and $$NaI$$ are also colourless but $$AgBr$$ and $$AgI$$ are coloured. This is due to:

A
$$ Ag^+$$ polarizes $$Br^-$$ and $$I^- $$

B
$$ Ag^+ $$ has unpaired d - orbital

C
$$ Ag^+ $$ depolarizes $$ Br^-$$ and $$I^- $$

D
None is correct

Solution

According to Fajan's rule, smaller the size of cation and larger the size of anion, the greater is the covalent character of the ionic compound.

In $$AgBr$$ and $$AgI$$, $$Ag^+$$ being smaller in size it polarizes $$Br^-$$ and $$I^-$$ due to which they are colored.

Hence, the correct option is (A).
Question 2
1 / -0

$$ Ti^{2+}$$ is purple while $$ Ti^{4+} $$ is colorless,because:

A
There is no crystal field effect in $$ Ti^{4+} $$

B
$$ Ti^{4+}$$ has $$3d^2 $$ configuration

C
$$ Ti^{4+}$$ has $$3d^0 $$ configuration

D
$$ Ti^{4+} $$ is a very small cation when compared to $$ Ti^{2+} $$ and hence does not absorb any radiation

Solution

Electronic configuration of $$Ti=[Ar]3d^24s^2$$
So,
$$Ti^{2+}=[Ar]3d^2$$
$$Ti^{4+}=[Ar]3d^0$$

$$Ti^{2+}$$ ion has two unpaired electrons due to which d-d transition is possible by absorption of light in visible region. Because of which $$Ti^{2+}$$ ion is purple and $$Ti^{4+}$$ is colorless due to absence of unpaired electrons.

Hence, the correct option is (B).
Question 3
1 / -0

A green mass is formed in the charcoal cavity test when a colorless salt (X) is fused with cobalt nitrate (X) may contain

A
Aluminium

B
Copper

C
Barium

D
Zinc

Solution
Question 4
1 / -0

In the dichoromate dianion:-

A
Four Cr-O bonds are equivalent

B
Six Cr-O bonds are equivalent

C
All Cr-O bonds are equivalent

D
All Cr-O bonds are non-equivalent

Solution
Question 5
1 / -0

In the dichromate dianoin:

A
Four Cr-O bonds are equivalent

B
Six Cr-O bonds equivalent

C
All Cr-O bonds are equivalent

D
All Cr-O bonds are non-equivalent

Solution

Due to resonance, six Cr-O bonds are equivalent in dichromate ion.
Question 6
1 / -0

Increasing basic properties of $$ TiO_2$$, $$ZrO_2$$ and $$HfO_2 $$ are in order.

A
$$ TiO_2 < ZrO_2 < HfO_2 $$

B
$$ HfO_2 < ZrO_2 < TiO_2 $$

C
$$ HfO_2 < TiO_2 < ZrO_2 $$

D
$$ ZrO_2 < TiO_2 < HfO_2 $$

Solution
Question 7
1 / -0

In context of the lanthanoids , which of the following statements is not correct ?

A
There is gradual decrease in the raddi of the members with increasing atomic number in the series

B
All the members exhibit $$+3$$ oxidation state

C
Because of similar properties, the separation of lanthanoids is not easy

D
Availability of $$4f$$ electrons results in the formation of compounds in $$+4$$ state for all the members of the series.

Solution

All lanthanides belong to the same group so they exibit similar properties.
4f subshell of lanthanides is well shielded by 5d and 6s orbitals so electrons of 4f subshell don't participate in bonding.
Radius of lanthanides gradually decreases as atomic number increses due to poor shielding effect od f subshell.
Lanthanides usually show +3 oxidation state.
Question 8
1 / -0

One heating ammonium dichromate, the gas evolved is:

A
Oxygen

B
Ammonia

C
Nitrous oxide

D
Nitrogen

Solution

$$ (NH_4) Cr_2 O_7 \xrightarrow{\triangle} \ N_2 \uparrow +Cr_2O_3(s)+4H_2O (l)$$

One heating ammonium dichromate, the gas evolved is nitrogen gas.
Question 9
1 / -0

A solution on treatment with $$NH_3$$ turns blue, it contains:

A
$$Cu^{2+}$$

B
$$Ca^{2+}$$

C
$$Co^{2+}$$

D
$$Mn^{2+}$$

Solution
Question 10
1 / -0

Hg is found in nature as HgS because :

A
$$Hg^{2+}$$ is soft acid and $$S^{2-}$$ soft base

B
Lattice energy of HgS is high

C
Hg and S are close in periodic table

D
Ionisation energy of Hg is very high

Solution

Mercuric ion ($$Hg^{2+}$$) is a soft acid and sulfide ion ($$S^{2-}$$) is a soft base. Hence $$Hg^{2+}$$ reacts more strongly with $$S^{2-}$$ to form mercuric sulfide $$HgS$$.

Hence, mercury is available in the form of cinnabar, $$HgS$$, whereas carbonate, oxide and silicate ions are hard bases. Therefore, mercuric ion ($$Hg^{2+}$$) does not prefer to combine with these (carbonate, oxide and silicate) ions.