Case study (4 Marks)

MCQ 14 Marks

Read the passage given below and answer the following questions: Transition elements are elements that have partially filled $d-$orbitals. The configuration of these elements corresponds to$ (n - 1)d^{1-10} \ ns^{1-2}.$ It is important to note that the elements mercury, cadmium and zinc are not considered transition elements because of their electronic configurations, which corresponds to $(n - 1)d^{10} \ ns^2.$ Some general properties of transition elements are: These elements can fonn coloured compounds and ions due to $d-d$ transition; These elements exhibit many oxidation states; A large variety of ligands can bind themselves to these elements, due to this, a wide variety of stable complexes formed by these ions. The boiling and melting point of these elements are high. These elements have a large ratio of charge to the radius. In these questions $(Q.$ No. $i-iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion: Tungsten has very high melting point.

Reason: Tungsten is a covalent compound.

Assertion: $Zn, Cd$ and $Hg$ are normally not considered transition metals.

Reason: $d-$Orbitals in $Zn, Cd$ and $Hg$ elements are completely filled, hence these metals do not show the general characteristics properties of the transition elements.

Assertion: Copper metal gets readily corroded in acidic aqueous solution such as $\ce{HCl}$ and dil. $\ce{H_2SO_4}.$

Reason: Free energy change for this process is positive.

Assertion: Tailing of mercury occurs on passing ozone through it.

Reason: Due to oxidation of mercury.

Assertion: Transition metals are poor reducing agents.

Reason: Transition metals fonn numerous alloys with other metals.

A
Assertion and reason both are correct statements and reason is correct explanation for assertion.
B
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
C
Assertion is correct statement but reason is wrong statement.
D
Assertion is wrong statement but reason is correct statement.

Answer

Tungsten is a transition element and is very hard due to high metallic bonding.

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.
$(d)$ Assertion is wrong statement but reason is correct statement.

Explanation:
Non$-$oxidising acids $\ce{(HCl}$ and dil. $\ce{H_2SO_4)}$ do not have any effect on copper. However they dissolve the metal in presence of air. As it is a non$-$spontaneous process so, $\Delta\text{G}$ cannot $be-ve.$

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:
When mercury is exposed to ozone it gets superficially oxidised and loses its meniscus and sticks to the glass.

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:
In actual practice transition metals react with acid very slowly and act as poor reducing agents. This is due to the protection of metal as a result of formation of thin oxide protective film. Further, their poor tendency as reducing agent is due to high ionisation energy, high heat of vapourisation and low heat of hydration.

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Question 24 Marks

Read the passage given below and answer the following questions:
The $f-$block elements are those in which the differentiating electron enters the $(n -2)f$ orbital. There are two series of $F-$block elements corresponding to filling of $4f$ and $5f-$orbitals. The series of $4f-$orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of $f^0, f^7$ and $F^{14}$ configurations, though the most conunon oxidation states is $+3.$ There is a regular decrease in size oflanthanides ions with increase in atomic number which is known as lanthanide contraction.
The following questions are multiple choice questions. Choose the most appropriate answer:

The atomic numbers of three lanthanide elements $X, Y$ and $Z$ are $65, 68$ and $70$ respectively, their $Ln^{3+}$ electronic configuration is:

$\ce{4f^8, 4f^{11}, 4f^{13}}$
$\ce{4f^{11}, 4f^8, 4f^{13}}$
$\ce{4f^0, 4f^2, 4f^{11}}$
$\ce{4f^3, 4f^7, 4f^9}$

Lanthanide contraction is observed in:

$Gd$
$At$
$Xe$
$Te$

Which of the following is not the configuration oflanthanoid?

$\ce{[Xe]4f^{10}6s^2}$
$\ce{[Xe]4f^15d^16s^2}$
$\ce{[Xe]4d^{14}5d^{10}6s^2}$
$\ce{[Xe]4f^75d^16s^2}$

Name a member of the lanthanoid series which is well known to exhibit $+4$ oxidation state.

Cerium $(Z = 58)$
Europium $(Z = 63)$
Lanthanum $(Z = 57)$
Gadolinium $(Z = 64)$

Identify the incorrect statement among the following.

Lanthanoid contraction is the accumulation of successive shrinkages.
The different radii of $Zr$ and $Hf$ due to consequence of the lanthanoid contraction.
Shielding power of $4f$ electrons is quite weak.
There is a decrease in the radii of the atoms or ions as one proceeds from $La$ to $Lu.$

Answer

$(a) \ce{4f^8, 4f^{11}, 4f^{13}}$

Terbium $\ce(65), {4f^8;}$ Dysprosium $\ce{(Dy), 4f^9;}$ Ytterbium $\ce{(Yb), 4f^{13}.}$

$(a) \ce{Gd}$
$(c) \ce{[Xe]4d^{14}5d^{10}6s^2}$
$(a)$ Cerium $(Z = 58)$
$(b)$ The different radii of $Zr$ and $Hf$ due to consequence of the lanthanoid contraction.

The almost identical radii of $\ce{Zr (160\ pm)}$ and $\ce{Hf (159\ pm),}$ a consequence of lanthanoid contraction.

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Question 34 Marks

Read the passage given below and answer the following questions :
The transition elements have incompletely filled $d-$subshells in their ground state or in any of their oxidation states. The transition elements occupy position in betweens- and $p-$blocks in groups $3-12$ of the Periodic table. Starting from fourth period, transition elements consists of four complete series : $Sc$ to $Zn, Y$ to $Cd$ and $La, Hf$ to $Hg$ and $Ac, Rf$ to $Cn$. In general, the electronic configuration of outer orbitals of these elements is $(n - 1)d^{1-10} n^{1-2}$. The electronic configurations of outer orbitals of $Zn, Cd, Hg$ and $Cn$ are represented by the general formula $(n - 1)d^{10}ns^2$. All the transition elements have typical metallic properties such as high tensile strength, ductility, malleability. Except mercury, which is liquid at room temperature, other transition elements have typical metallic structures. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. Transition metal also forms alloys. An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
The following questions are multiple choice questions. Choose the most appropriate answer:

Which of the following characteristics of transition metals is associated with higher catalytic activity?

High enthalpy of atomisation.
Variable oxidation states.
Paramagnetic behaviour.
Colour of hydrated ions.

Transition elements form alloys easily because they have.

Same atomic number.
Same electronic configuration.
Nearly same atomic size.
Same oxidation states.

The electronic configuration of tantalum $(Ta)$ is:

$[Xe]4f^05d^16s^2$
$[Xe]4f^{14}5d^26s^2$
$[Xe]4f^{14}5d^36s^2$
$[Xe]4f^{14}5d^46s^2$

Which one of the following outer orbital configurations may exhibit the largest number of oxidation states?

$3d^54s^1$
$3d^54s^2$
$3d^24s^2$
$3d^34s^2$

The correct statement$(s)$ among the following is/ are :

All $d$ and $f-$block elements are metals.
All $d$ and $f-$block elements form coloured ions.
All $d$ and $f-$block elements are paramagnetic.

$(I)$ only
$(I)$ and $(II)$ only
$(II)$ and $(III)$ only
$(I), (II)$ and $(III)$

Answer

$(b)$ Variable oxidation states.

The transition metals and their compounds are known for their catalytic activity.
This activity is ascribed to their ability to adopt multiple oxidation states to form complexes.

$(c)$ Nearly same atomic size.

Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals.

$(c)\ [Xe]4f^{14}5d^36s^2$
$(b) \ d^54s^2$

Greater the number of valence electrons, more will be the number of oxidation states exhibited by the element.

$(a)\ (i)$ only

All the $d-$block elements are metals, they exhibit most properties of metals like lustre, malleability, ductility, high density, high melting and boiling point, hardness, conduction of heat and electricity, etc.
All the $f-$block elements are also metals but they are not good conductors of heat and electricity.

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Question 44 Marks

Read the passage given below and answer the following questions:
The unique behaviour of $Cu,$ having a positive $E^\circ$ accounts for its inability to liberate $H_2$ from acids. Only oxidising acids $($nitric and hot concentrated sulphuric acid$)$ react with $Cu,$ the acids being reduced. The stability of the half$-$filled $(d^5)$ subshell in $Mn^{2+}$ and the completely filled $(d^{10})$ configuration in $Zn^{2+}$ are related to their $\text{E}^\circ\frac{\text{M}^{3+}}{\text{M}^{2+}}$ values. The low value for $Sc$ reflects the stability of $Sc^{3+}$ which has a noble gas configuration. The comparatively high value for $Mn$ shows that $Mn^{2+}(d^5)$ is particularly stable, whereas a comparatively low value for $Fe$ shows the extra stability of $Fe^{3+}(d^5)$. The comparatively low value for $V/s$ related to the stability of $v^{2+} ($half$-$filled $t_{2g}$ level$)$.
The following questions are multiple choice questions. Choose the most appropriate answer :

Standard reduction electrode potential of $\frac{\text{Zn}^{2+}}{\text{Zn}}$ is $0.76V$. This means

$ZnO$ cannot be reduced to $Zn$ by $H_2$ under standard conditions.
$Zn$ cannot liberate $H_2$ with concentrated acids.
$Zn$ is generally the anode in an electrochemical cell.
$Zn$ is generally the cathode in an electrochemical cell.

$\text{E}^\circ$ values for the couples $\frac{\text{Cr}^{3+}}{\text{Cr}^{2+}}$ and $\frac{\text{Mn}^{3+}}{\text{Mn}^{2+}}$ are $-0.41$ and $+1.51$ volts respectively. These values suggest that.

$Cr^{2+}$ acts as a reducing agent whereas $Mn^{3+}$ acts as an oxidizing agent.
$Cr^{2+}$ is more stable th an $Cr^{3+}$ state.
$Mn^{3+}$ is more stable than $Mn^{2+}$.
$Cr^{2+}$ acts as an oxidizing agent whereas $Mn^{3+}$ acts as a reducing agent..

The reduction potential values of $M, N$ and $O$ are $+2.46, -1.13$ and $-3.13V$ respectively. Which of the following order is correct regarding their reducing property?

$O > N > M$
$O > M > N$
$M > N > O$
$M > O > N$

Which of the following statements are true?

$Mn^{2+}$ compounds are more stable than $Fe^{2+}$ towards oxidation to $+3$ state.
Titanium and copper both in the first series of transition metals exhibits $+1$ oxidation state most frequently.
$Cu^+$ ion is stable in aqueous solutions.
The $\text{E}^\circ$ value for the $\frac{\text{Mn}^{3+}}{\text{Mn}^{2+}}$ couple is much more positive than that for $\frac{\text{Cr}^{3+}}{\text{Cr}^{2+}}$ or $\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}.$.

$(II)$ and $(III)$
$(I)$ and $(IV)$
$(I)$ and $(III)$
$(II)$ and $(IV)$

The stability of $\text{Cu}^{2+}_\text{(aq)}$ rather than $\text{Cu}^{+}_\text{(aq)}$ is due to.

More negative $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
Less negative $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
More positive $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
Less positive $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$

Answer

$(a)\ ZnO$ cannot be reduced to $Zn$ by $H_2$ under standard conditions.
$(a)\ Cr^{2+}$ acts as a reducing agent whereas $Mn^{3+}$ acts as an oxidizing agent.

Lesser and negative reduction potential indicates that $cr^{2+}$ is a reducing agent.
Higher and positive reduction potential indicates that $Mn^{3+}$ is a stronger oxidizing agent.

$(a)\ O > N > M$

The electrode which has more reduction potential is a good oxidizing agent and has least reducing power.

$(b)\ (i)$ and $(iv)$

Explanation:

It is because $Mn^{2+}$ has $3d^5$ electronic configuration which has extra stability.
Not titanium but copper, because with $+1$ oxidation state an extra stable configuration$, 3d^{10}$ results.
It is not stable as it undergoes disproportionation;

$2\text{Cu}^+_\text{(aq)}\rightarrow\text{Cu}^{2+}_\text{(aq)}+\text{Cu}_\text{(s)}.$ The $\text{E}^\circ$ value for this is favourable.

Much larger third ionisation energy of $Mn\ ($where the required change is $d^5$ to $d^4)$ is mainly responsible for this.

$(a)$ More negative $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$

The stability of $\text{Cu}^{2+}_\text{(aq)}$ rather than $\text{Cu}^{+}_\text{(aq)}$ is due to the much more negative $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}$ than $\text{Cu}^{+},$ which more than compensates for the second ionisation enthalpy of $Cu.$

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Question 54 Marks

Read the passage given below and answer the following questions : Transition metal oxides are compounds fanned by the reaction of metals with oxygen at high temperature. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic $V_2O_3$ to less basic $V_2O_4$ and to amphoteric $V_2O_5· V_2O_4$ dissolves in acids to give $VO^{2+}$ salts. Transition metal oxides are commonly utilized for their catalytic activity and semi conductive properties. Transition metal oxides are also frequently used as pigments in paints and plastic. Most notably titatnium dioxide. One of the earliest application of transition metal oxides to chemical industry involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid. Since then, many other applications have emerged, which include benzene oxidation to maleic anhydride on vandium oxides; cyclohexane oxidation to adipic acid on cobalt oxides. An important property of the catalyst material used in these processes is the ability of transition metals to change their oxidation state under a given chemical potential of reductants and oxidants. The following questions are multiple choice questions. Choose the most appropriate answer:

Which oxide of vanadium is most likely to be basic and ionic?

$VO$
$V_2O_3$
$VO_2$
$V_2O_5$

Vanadyl ion is:

$\text{VO}^{2+}$
$\text{VO}^{+}_2$
$\text{V}_{2}\text{O}^+$
$\text{VO}^{3-}_4$

Which of the following statements is false?

With fluorine vanadium can form $VF_5.$
With chlorine vanadium can form $VCl^5.$
Vanadium exhibits highest oxidation state in oxohalides $VOCl_3, VOBr_3$ and fluoride $VF_5.$
With iodine vanadium cannot form $Vl_5$ due to oxidising power of $V^{5+}$ and reducing nature of $I^-.$

The oxidation state of vanadium in $V_2O_5$ is:

$\frac{+5}{2}$
$+7$
$+5$
$+6$

Identify the oxidising agent in the following reaction.

$V_2O_{5 }+ 5Ca \rightarrow 2V + 5CaO$

$V_2O_5$
$Ca$
$V$
None of these.

Answer

$(a)\ VO$

Oxide of $V$ in lowest oxidation state, i.e.$, VO$ is basic and ionic in character.

$(a)\ \text{VO}^{2+}$

Vanadyl ion is $vo^{2+}$ where $V$ is in $+4$ oxidation state.

$(b)$ With chlorine vanadium can form $VCl^5$.
$(c)\ +5$
$(a)\ V_2O_5$

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Chemistry Case study (4 Marks)

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