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Chemistry 2 Marks Questions

The d & f Block Elements · MP Board (MPBSE) STD 12 Science (Madhya Pradesh - English Medium). 36 questions.

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Question 12 Marks
How would you account for the following:
The $d^1$ configuration is very unstable in ions.
Answer
Some species with $d^1$ configuration are reducing and tends to loose one electron to acquire $d^4$ stable configuration. Some other species with $d^1$ configuration like $Cr(V)$ and $Mn(VI)$ undergo disproportionation.
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Question 22 Marks
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
Ionisation enthalpies.
Answer
In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of $4f$ electrons in the third transition series. Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the $2^{nd}$ transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the $1^{st}$ transition series.
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Question 32 Marks
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Oxidation state
Answer
Atomic and ionic sizes: Both lanthanoids and actinoids show decrease in size of their atoms or ions in +3 oxidation state as we go from left to right. In lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, is called actinoid contraction. The contratibn is greater from element tot i element in actinodes due to poorer shielding by 5f electrons.
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Question 42 Marks
Explain giving reason:
Transition metals and many of their compounds show paramagnetic behaviour.
Answer
Magnetic properties: Transition elements and many of their compounds are paramagnetic, i.e., they are weakly attracted by a magnetic field. This is due to the presence of unpaired electrons in atorns, ions or molecules. The paramagnetic character increases as the number of. unpaired electrons increases. The paramagnetic character is measured in terms of magnetic moment and is given by $\mu=\sqrt{\text{n}(\text{n}+2}$ where n - number of unpaired electrons.
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Question 52 Marks
What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Answer
A reaction in which the same species is simultaneously oxidised as well as reduced is called a disproportionation reaction.
Thus we can say taht disproportionation reaction is a special type of redox reaction. In which an element in one oxidation state simutaneously undergoes both oxidation and reduction. For example, Cr(V) and Mn(VI) species undergo disproportionation reaction in acidic medium as follows:
$3\text{CrO}^{3-}_4+8\text{H}^+\rightarrow2\text{CrO}^{2-}_4+\text{Cr}^{3+}+4\text{H}_2\text{O}$
(Cr in +5 o. s) (Cr in +6 o. s)
$3\text{MnO}^{3-}_4+4\text{H}^+\rightarrow2\text{MnO}^{2-}_4+\text{MnO}_2+2\text{H}_2\text{O}$
Mn in +6 o.s +7 o.s 4 o. s
Here, we can say that Cr in + 5 oxidation state undergo disproportionation into its +6 and +3 states. Similarly, Mn in +6 oxidation state undergo disproportionation into +7 and +4 oxidation states.
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Question 62 Marks
Give examples and suggest reasons for the following features of the transition metal chemistry:The highest oxidation state is exhibited in oxoanions of a metal.
Answer
The highest oxidation state of transition element is seen in oxides and these oxides are acidic in nature. These acidic oxides dissolve in base and form oxoanions of transition metal. Thus oxoanions of transition element show highest oxidation state of transition metal. Examples are:
The highest oxidation state of $Cr$ is $+6$ and it is observed that in its oxo-anion, ${CrO_4}^{2–}($chromate ion$)$ and ${Cr_2O_7}^{2–} ($dichromate ion$).$
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Question 72 Marks
Write the electronic configurations of the elements with the atomic numbers $61, 91, 101,$ and $109.$
Answer
Atomic number Electronic configuration
$61$ $[Xe]^{54}\ 4f^5\ 5d^06s^2$
$91$ $[Rn]^{85}\ 5f^{13}\ 6d^17s^2$
$101$ $[Rn]^{86}\ 5f^{13}\ 5d^07s^2$
$109$ $[Rn]^{86}\ 5f^{14}\ 6d^17s^2$
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Question 82 Marks
Name the members of the lanthanoid series which exhibit $+4$ oxidation states and those which exhibit $+2$ oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
Answer
The lanthanides that exhibit $+2$ and $+4$ states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.
$+2$ $+4$
$Nd\ (60)$ $Ce\ (58)$
$Sm\ (62)$ $Pr\ (59)$
$Eu\ (63)$ $Nd\ (60)$
$Tm\ (69)$ $Tb\ (65)$
$Yb\ (70)$ $Dy\ (66)$
Ce after forming $\mathrm{Ce}^{4+}$ attains a stable electronic configuration of $[\mathrm{Xe}]$.
Tb after forming $\mathrm{Tb}^{4+}$ attains a stable electronic configuration of $[\mathrm{Xe}] 4 \mathrm{f}^7$.
Eu after forming $\mathrm{{Eu}^{2+}}$ attains a stable electronic configuration of $[\mathrm{Xe}] 4 \mathrm{f}^7$.
Yb after forming $\mathrm{Yb}^{2+}$ attains a stable electronic configuration of $[\mathrm{Xe}] 4 \mathrm{f}^{14}$.
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Question 92 Marks
Why are $\mathrm{Mn}^{2+}$ compounds more stable than $\mathrm{Fe}^{2+}$ towards oxidation to their $+3$ state?
Answer
Electronic configuration of $\mathrm{Mn}^{2+}$ is $[\mathrm{Ar}]^{18} 3 \mathrm{d}^5$
Electronic configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}]^{18} 3 \mathrm{d}^6$
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in $(+2)$ state has a stable $\mathrm{d}^5$ configuration. This is the reason $\mathrm{Mn}^{2+}$ shows resistance to oxidation to $\mathrm{Mn}^{3+}$ Also, $\mathrm{Fe}^{2+}$ has $3 \mathrm{d}^6$ configuration and by losing one electron, its configuration changes to a more stable $3 \mathrm{~d}^5$ configuration. Therefore, $\mathrm{Fe}^{2+}$ easily gets oxidized to $\mathrm{Fe}^{+3}$ oxidation state.
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Question 102 Marks
Which metal in the first series of transition metals exhibits $+1$ oxidation state most frequently and why$?$
Answer
Copper metal $\left(\mathrm{Cu}\right.$, at. no. 29) shows $+1$ oxidation state i.e., it exists as $\mathrm{Cu}^{+}$in large number of copper compounds e.g., cuprous oxide $\left(\mathrm{Cu}_2 \mathrm{O}\right)$, cuprous sulphide $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$; cuprous chloride $\left(\mathrm{Cu}_2 \mathrm{Cl}_2\right)$ etc. The electronic configuration of $\mathrm{Cu}^{+}$is $[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^0$. This configuration is very stable as all five $3$d orbitals are fully filled.
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Question 112 Marks
Predict which of the following will be coloured in aqueous solution? $\mathrm{Ti}^{3+}, \mathrm{V3}^{+}, \mathrm{Cu}^{+}, \mathrm{Sc}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{CO}^{2+}$. Give reasons for each.
Answer
Among the above mentioned ions, $\mathrm{Ti}^{3+}, \mathrm{v}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{CO}^{2+}$ are coloured. These ions are coloured due to presence of unpaired electrons, they can undergo $d-d$ transitions.
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Question 122 Marks
How would you account for the following:
Of the $\mathrm{d}^4$ species, $\mathrm{Cr}^{2+}$ is strongly reducing while manganese$(III)$ is strongly oxidising.
Answer
Of $\mathrm{d}^4$ species, $\mathrm{Cr}^{2+}$ has $3 d^4$ configuration and tends to loose one electron to acquire $\mathrm{d}^3$ configuration as it is highly stable and best metallic specie available for complex formation. $\mathrm{Cr}^{3+}$ can accommodate six lone pair of electrons from ligands due to $\mathrm{sp}^3 \mathrm{d}^2$ hybridisation e.g. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} \mathrm{Mn}^{3+}$ although have $\mathrm{d}^4$ configuration but tends to become $\mathrm{Mn}^{2+}$ stable specie by acquiring one electron to attain $\mathrm{d}^5$ configuration. It becomes exactly half filled on one hand and more energy is released in gain of electron due to higher nuclear charge.
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Question 132 Marks
What are interstitial compounds? Why are such compounds well known for transition metals?
Answer
Transition metals form large number of interstitial compounds. They are able to entrap small atoms of elements like H, G, N, B, etc., in their crystal lattice and even can make weak bonds with them. Due to formation of interstitial compounds, their malleability and ductility decreases and tensile. strength increases. Steel and cast iron are hard in comparison to wrought iron due to the presence of trapped carbon atoms in interstitial spaces.
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Question 142 Marks
Explain giving reason:Transition metals and their many compounds act as good catalyst.
Answer
Catalytic properties: Many of transition metals and their compounds act as catalyst in variety of reactions, e.g., finely divided iron in manufacture of $NH^3$ by Haber's process, $V^2O^5$ or Pt in manufacture of $H^2SO^4$ by Contact process, etc.). The catalytic activity is due to following two reasons.
  1. The ability of transition metal ion to pass easily from one oxidation state to another and thus providing a new path to reaction with lower activation energy.
  2. The surface of transition metal acts as very good adsorbent and thus provides increased concentration of reactants on their surface causing the reaction to occur.
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Question 152 Marks
Give examples and suggest reasons for the following features of the transition metal chemistry
The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
Answer
Transition metal oxides in lowest oxidation state are ionic in nature due to strongly metallic character of transition elements. In higher oxidation state transition element shows non-metallic character and as a result their oxides become covalent and acidic in nature. This is illustrated by following example of various oxides of manganese Oxide: $MnO\ Mn_2O_3\ Mn_3O_4\ MnO_2\ Mn_2O_7$
Ox. side of Mn:
$\underbrace{+2\ +3}\ \underbrace{+8/3\ +4} \ \underbrace{ \ \ \ \ \ \ \ +7}\\\text{basic} \ \ \ \ \ \text{amphoteric } \ \ \ \text{acidic}\\\text{oxide} \ \ \ \ \ \ \ \ \ \text{oxide} \ \ \ \ \ \ \ \ \ \ \text{oxide} \ \ \ \ $
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Question 162 Marks
Give examples and suggest reasons for the following features of the transition metal chemistry:
A transition metal exhibits highest oxidation state in oxides and fluorides.
Answer
The oxidation state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Oxidation results in an increase in the oxidation state. Reduction results in a decrease in the oxidation state. If an atom is reduced, it has a higher number of valence shell electrons, and therefore a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants.Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
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Question 172 Marks
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.
Answer
Inner transition metals are those elements in which the last electron enters the f-orbital also transition elements are the elements which have partly filled f-orbitals. These are also called f-block elements. There are two series (each of 14 elements) of inner transition elements.
  1. Lanthanoids (also called 4f series). These are from atomic numbers 58-71.
  2. Actinoids (also called 5f series ). These are from atomic numbers 90-103.
Among the given atomic numbers, only 59, 95 and 102 are the atomic numbers of inner transition elements. The other three atomic numbers represent the transition elements.
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Question 182 Marks
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Answer
Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as $d^0,\ d^5,\ d^{10}$. Since these states are exceptionally stable, their ionization enthalpies are very high.
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Question 192 Marks
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
Atomic sizes.
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Question 202 Marks
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer
The oxidation states displayed by the first half of the first row of transition metals are given in the table below.
  $Sc$ $Ti$ $V$ $Cr$ $Mn$
Oxidation state   $+2$ $+2$ $+2$ $+2$
$+3$ $+3$ $+3$ $+3$ $+3$
  $+4$ $+4$ $+4$ $+4$
    $+5$ $+5$ $+6$
      $+6$ $+7$
It can be easily observed that except Sc, all others metals display $+2$ oxidation state. Also, on moving from Sc to $Mn,$ the atomic number increases from $21$ to $25.$ This means the number of
electrons in the $3d-$orbital also increases from $1$ to $5.$
$ \mathrm{Sc}(+2)=d^1 $
$ \mathrm{Ti}(+2)=d^2 $
$ \mathrm{~V}(+2)=d^3 $
$ \mathrm{Cr}(+2)=d^4 $
$ \mathrm{Mn}(+2)=d^5$
$+2$ oxidation state is attained by the loss of the two $4$ selectrons by these metals. Since the number of delectrons in $(+2)$ state also increases from $Ti(+2)$ to $Mn(+ 2),$ the stability of $+2$ state increases (as d-orbital is becoming more and more half-filled). $Mn(+2)$ has $d5$ electrons (that is half-filled dshell, which is highly stable).
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Question 212 Marks
Explain giving reason:
The transition metals generally form coloured compounds.
Answer
Formation of coloured compounds (both in solid state as well as in aqueous solution) is another very common characteristics of transition metals. This is due to absorption of some radiation from visible light to cause $d-d$ transition of electrons in transition metal atom. The d-orbitals do not have same energy and under the influence of ligands, the d-orbitals split into two sets of orbitals having different energies; transition of electrons can take place from one set of d orbitals to another set within the same sub shell. Such transitions are called $d-d$ transitions. The energy difference for these d-d transitions fall in the visible region. When white light is incident on compounds of transition metals, they absorb a particular frequency and rernaining colours are emitted imparting a characteristic colour to the complex. $Zn^{2+}$ and $Ti^{4+}$ salts are white because they do not absorb any radiation in visible region.
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Question 222 Marks
Compare the chemistry of the actinoid with that of lanthanoids with reference to:
Chemical reactivity.
Answer
Chemical reactivityIn the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).
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Question 232 Marks
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
Oxidation states
Answer
In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.
However, $+2$ and $+3$ oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the $+2$ and $+3$ oxidation states. The stability of the $+2$ and $+3$ oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.
For example $\Big[\stackrel{{\text{II}}}{{\text{Fe}}}(\text{Cn})_6\Big]^{4-}, \ \Big[\stackrel{{\text{III}}}{{\text{Co}}}(\text{NH}_3)_6\Big]^{3+},\ \Big[\text{Ti}(\text{H}_2\text{O})_6\Big]^{3+}$ are stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example: $WCl_6,\ ReF_7,\ RuO_4$ etc.
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Question 242 Marks
Compare the chemistry of the actinoid with that of lanthanoids with reference to:
Oxidation states.
Answer
Oxidation states
The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7 slevels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.
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Question 252 Marks
What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Answer
Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements. Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.
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Question 262 Marks
In what way is the electronic configuration of the transition elements different from that of the non transition elements?
Answer
Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is $(n -1)d^{1-10}ns^{0-2}.$
The non-transition elements either do not have a d-orbital or have a fully filled d-orbital. Therefore, the electronic configuration of non-transition elements is $ns^{1-2}$ or $ns^2np^{1-6}.$
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Question 272 Marks
Explain giving reason:
The enthalpies of atomisation of the transition metals are high.
Answer
Because of large number of unpaired electrons in d-orbitals of their atoms they have stronger interatomic intefactions and hence stronger metallic bonding between atoms resulting in higher enthalpies of atomisation.
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Question 282 Marks
How would you account for the following:
Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
Answer
$Co^{2+}$ is stable in aqueous solution because it get surrounded and weakly bonded to water molecules. In presence of strong ligands and air it gets oxidised to $Co(III)$ as strong ligands get coordinated more strongly with $Co(III)$. The electronic configuration of $Co(II)$ and $Co(III)$ are:
$Co(II) = [Ar]_{18} 4s^03d^7$ and
$Co(III) = [Ar]_{18}4s^03d^6$​​​​​​​
In $Co(III)$ specie, 6 lone pairs of electrons from ligands are accommodated by $sp^3d^2$ hybridisation which is not possible in $Co(II).$
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Question 292 Marks
How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
Answer
The variability of oxidation states, a characteristic of transition elements, arises due to incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Cr}^{3+}$. This is in contrast with the variability of oxidation states of non-transition elements where oxidation states normally differ by a unit of two. i.e., $\mathrm{Sn}^{2+}, \mathrm{Sn}^{4+}, \mathrm{p}^{3+}$ and $\mathrm{p}^{\text {s+ }}$, etc. in the p -block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of J-block.For example, in group 6, Mo $(VI)$ and $W\ (VI)$ are found to be more stable than $\mathrm{Cr}(\mathrm{VI})$. Thus $\mathrm{Cr}(\mathrm{VI})$ in the form of dichromate in acidic medium is a strong oxidising agent, whereas $\mathrm{MOO}_3$ and $\mathrm{WO}_3$ are not.
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Question 302 Marks
Compare the chemistry of the actinoid with that of lanthanoids with reference to:Electronic configuration:
Answer
Electronic configuration:
The general electronic configuration for lanthanoids is $[Xe]^{54}4f^{0-14} 5d^{0-16}S^2$ and that for actinoids is $[Rn]^{86}5f^{1-14}6d^{0-17}s^2.$ Unlike $4f$ orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.
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Question 312 Marks
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: $3d^3,$
$3d^5, 3d^8$ and $3d^4?$
Answer
  Electronic configuration in
ground state		 Stable oxidation states
(i) $3d^3 ($Vanadium$)$ $+2, +3, +4$ and $+5$
(ii) $3d^5 ($Chromium$)$ $+3, +4, +6$
(iii) $3d^5 ($Manganese$)$ $+2, +4, +6, +7$
(iv) $3d^8 ($Cobalt$)$ $+2, +3$
(v) $3d^4$ There is no $3d^4$ configuration in
ground state.
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Question 322 Marks
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Atomic and ionic sizes
Answer
Oxidation state: Lanthanoids show lirnited oxidation states(+2, +3, +4) out of which +3 is most common whereas actinoids show +3, +4, +5, +6, +7 oxidation states.This is because of large energy gap between 4f 5d and 6s orbitals. However, actinoids show a large number of oxidation states because of small energy ap- between 5f 6d and Is orbitals.
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Question 332 Marks
For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems the values for some metals are as follows:
$Cr^{2+}/Cr$ $- 0.9 V$ $Cr^3/Cr^{2+}$ $- 0.4 V$
$Mn^{2+}/Mn$ $- 1.2 V$ $Mn^{3+}/Mn^{2+}$ $+ 1.5 V$
$Fe^{2+}/Fe$ $- 0.4 V$ $Fe^{3+}/Fe^{2+}$ $+ 0.8 V$
Use this data to comment upon:
  1. The stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}.$
  2. The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
Answer
  1. $Cr^{3+}/Cr^{2+}$ has negative reduction potential. Hence, $Cr^{3+}$ cannot be reduced to $Cr^{2+}.$ $Mn^{3+}/Mn^{2+}$ has a large positive reduction potential. Hence, $Mn^{3+}$ can be easily reduced to $Mn^{2+}.$ $Fe^{3+}/Fe^{2+}$ has small positive reduction potential. Hence, $Fe^{3+}$ is more stable than $Mn^{3+}$ but less stable than $Cr^{3+}$.
  2. From the $E^0$ values, the order of oxidation of the metal to the divalent cation is: $Mn > Cr > Fe.$
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Question 342 Marks
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer
  1. Vanadate, $\text{VO}_3^-$
Oxidation state of V is +5.
  1. Chromate, $\text{CrO}^{2-}_4$
Oxidation state of Cr is +6.
  1. Permanganate, $\text{MnO}^-_4$
Oxidation state of Mn is +7.
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Question 352 Marks
Compare the stability of $+2$ oxidation state for the elements of the first transition series.
Answer
In general, the stability of $+2$ oxidation state in first transition series decreases from left to right due to increase in the sum of first and second ionisation energies. However $Mn^{2+}$ is more stable due to half filled d-orbitals $(3d^5)$ and $Zn^{2+}$ is more stable due to completely filled d orbitals $(3d^{10}).$
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Question 362 Marks
Calculate the number of unpaired electrons in the following gaseous ions: $Mn^{3+}, Cr^{3+}, V^{3+}$ and $Ti^{3+}.$ Which one of these is the most stable in aqueous solution?
Answer
The number of unpaired electrons can be determined from their electronic configurations and are tabulated below:
  Specie Electronic Configuration No. of unpaired electron
1. $Mn^{3+}$ $[Ar] 3d^44s^0$ $4$
2. $Cr^{3+}$ $[Ar] 3d^34s^0$ $3$
3. $V^{3+}$ $[Ar] 3d^24s^0$ $2$
4. $Ti^{3+}$ $[Ar] 3d^14s^0$ $1$
Out of these, $Cr^{3+}$ is most stable in aqueous solution as its hydration energy is highest.
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