D and F Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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211.
In the equation
$$4M + 8C{N^ - } + 2{H_2}O + {O_2} \to 4{\left[ {M{{\left( {CN} \right)}_2}} \right]^ - } + 4O{H^ - }$$
Identify the metal $$M.$$ It is
A
copper
B
iron
C
gold
D
zinc
Answer :
gold
This is cyanide process for gold $$\left( {M = Au} \right).$$
212.
Arrange $$C{e^{ + 3}},L{a^{ + 3}},P{m^{ + 3}}\,{\text{and}}\,Y{b^{ + 3}}$$ in inreasing order of their ionic radii.
In lanthanides there is a regular decrease in the atomic radii as well as ionic radii of trivalent ions as the atomic number increases from $$Ce\,{\text{to}}\,Lu.$$ This decrease in size of atoms and ions is known as lanthanide contraction. Although the atomic radii do show some irregualrities but ionic radii decreases from $$La\,{\text{to}}\,Lu.$$ Thus the correct order is.
$$\mathop {Y{b^{ + 3}}}\limits_{86.8{\text{pm}}} < \mathop {P{m^{ + 3}}}\limits_{97{\text{pm}}} < \mathop {C{e^{ + 3}}}\limits_{102{\text{pm}}} < \mathop {L{a^{ + 3}}}\limits_{103{\text{pm}}} $$
213.
Which of the following statements about the interstitial compounds is incorrect?
A
They retain metallic conductivity
B
They are chemically reactive
C
They are much harder than the pure metal
D
They have higher melting points than the pure metals
Answer :
They are chemically reactive
Interstitial compounds are obtained when small atoms like $$H, B, C, N,$$ etc., fit into the interstitial space of lattice metals. These retain metallic conductivity. These resemble the parent metal in chemical properties (reactivity) but differ in physical properties like hardness, melting point, etc.
214.
Due to lanthanoid contraction which of the following properties is not expected to be similar in the same vertical columns of second and third row transition elements?
A
Atomic radii
B
Ionisation energies
C
Magnetic moments
D
Lattice energies
Answer :
Magnetic moments
Magnetic moments depend upon the number of unpaired electrons.
215.
Match the column I with column II and mark the appropriate choice.
Column I
Column II
a.
$$FeS{O_4} \cdot 7{H_2}O$$
1.
Green
b.
$$NiC{l_2} \cdot 6{H_2}O$$
2.
Light pink
c.
$$MnC{l_2} \cdot 4{H_2}O$$
3.
Pale green
d.
$$CoC{l_2} \cdot 6{H_2}O$$
4.
Pink
e.
$$C{u_2}C{l_2}$$
5.
Colourless
A
a - 3, b - 4, c - 1, d - 2, e - 5
B
a - 2, b - 3, c - 4, d - 1, e - 5
C
a - 5, b - 2, c - 3, d - 4, e - 1
D
a - 3, b - 1, c - 2, d - 4, e - 5
Answer :
a - 3, b - 1, c - 2, d - 4, e - 5
No explanation is given for this question. Let's discuss the answer together.
216.
The basic character of the transition metal monoxides follows the order $$\left( {{\text{Atomic no's}}{\text{.}}\,\,Ti = 22,V = 23,Cr = 24,Fe = 26} \right)$$
A
$$VO > CrO > TiO > FeO$$
B
$$CrO > VO > FeO > TiO$$
C
$$TiO > FeO > VO > CrO$$
D
$$TiO > VO > CrO > FeO$$
Answer :
$$TiO > VO > CrO > FeO$$
Oxides of transition metals in low oxidation states + 2 and + 3 $$\left( {MO,{M_3}{O_4}\,{\text{and}}\,{M_2}{O_3}} \right)$$ are generally basic except $$C{r_2}{O_3}$$ which is amphoteric in character.
Basic character generally decreases with increase in atomic number.
217.
When copper is heated with $$conc.$$ $$HN{O_3}$$ it produces
A
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,NO$$
B
$$Cu{\left( {N{O_3}} \right)_2},\,NO\,\,{\text{and}}\,\,N{O_2}$$
C
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,\,{N_2}O$$
D
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,\,N{O_2}$$
Nitric acid acts as an oxidising agent while reacting with copper. When $$dil.$$ $$HN{O_3}$$ reacts, reaction proceeds as:
$$3Cu + 8HN{O_3}\left( {dil.} \right) \to $$ $$3Cu{\left( {N{O_3}} \right)_2} + 2NO + 4{H_2}O$$
and when conc. $$HN{O_3}$$ is used, reaction proceeds as
$$Cu + 4HN{O_3}\left( {conc.} \right) \to $$ $$Cu{\left( {N{O_3}} \right)_2} + 2N{O_2} + 2{H_2}O$$
218.
$$KMn{O_4}$$ acts as an oxidising agent in acidic medium. The number of moles of $$KMn{O_4}$$ that will be needed to react with one mole of sulphide ions in acidic solution is
219.
$$E_{\frac{{M{n^{3 + }}}}{{M{n^{2 + }}}}}^ \circ $$ is highly positive than that of $$E_{\frac{{C{r^{3 + }}}}{{C{r^{2 + }}}}}^ \circ $$ or $$E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^ \circ $$ because
A
$$M{n^{2 + }}\left( {{d^5}} \right)$$ can be easily oxidised to $$M{n^{3 + }}\left( {{d^4}} \right)$$ due to low ionisation enthalpy
B
third ionisation enthalpy of $$Mn$$ is much larger due to stable half filled $${{d^5}}$$ electronic configuration of $$M{n^{2 + }}$$
C
$$M{n^{3 + }}$$ is more stable than $$M{n^{2 + }}$$ due to higher oxidation state
D
second ionisation enthalpy of $$Mn$$ is higher than third ionisation enthalpy.
Answer :
third ionisation enthalpy of $$Mn$$ is much larger due to stable half filled $${{d^5}}$$ electronic configuration of $$M{n^{2 + }}$$
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