3 Marks Question

Question 513 Marks

How would you account for the following?
i. The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
ii. The $E ^{\circ}$ value for the $Mn ^{3+} / Mn ^{2+}$ couple is much more positive than that for $Cr ^{3+} / Cr ^{2+}$ couple or $Fe ^{3+} / Fe ^{2+}$ couple.
iii. The highest oxidation state of a metal is exhibited in its oxide or fluoride.

Answer

i. Due to Lanthanoid Contraction/or its meaning.
ii. Due to stable half - filled $3 d^5$ configuration of $Mn ^{2+} /$ high 3 rd ionisation enthalpy of Mn .
iii. Becuase Oxygen or Fluorine is highly electronegative and small size element.

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Question 523 Marks

How would you account for the following:
$\mathrm{Cr}^{2+}$ is reducing in nature while with the same d -orbital configuration ( $\mathrm{d}^4$ ) $\mathrm{Mn}^{3+}$ is an oxidising agent.

Answer

$\mathrm{Cr}^{2+}$ is reducing as its configuration changes from $\mathrm{d}^4$ to $\mathrm{d}^3$, a more stable half-filled $\mathrm{t}_{2 g}$ configuration while $\mathrm{Mn}^{3+}$ is oxidising as $\mathrm{Mn}^{3+}$ to $\mathrm{Mn}^{2+}$ change results in a more stable half-filled $\mathrm{d}^5$ configuration.

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Question 533 Marks

Assign reasons for the following:
The transition metals and many of their compounds act as good catalysts.

Answer

The catalytic activity of transition metals is attributed to the following reasons:

Because of their variable oxidation states, transition metals form unstable intermediate compounds and provide a new path with lower activation energy for the reaction.
In some cases, the transition metal provides a suitable large surface area with free valencies on which reactants are adsorbed.

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Question 543 Marks

On what basis can you say that scandium $(Z = 21)$ is a transition element but zinc $(Z = 30)$ is not?

Answer

On the basis of incompletely filled 3d-orbitals in case of scandium atom in its ground state $(3d^1)$, it is regarded as a transition element. On the other hand, zinc atom has completely filled d-orbitals $(3d^{10})$ in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.

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Question 553 Marks

$\text{E}^\ominus$ V of Cu is + $0.34V$ while that of Zn is – $0.76V$. Explain.

Answer

High ionisation enthalpy to change Cu(s) to $Cu^{2+}$ is not balanced by hydration enthalpy. Therefore, it exhibits a positive $E°$ value. However, Zn exhibits a lower value of ionization enthalpy because a stable $3d^{10}$ configuration is attained after losing two electrons. The hydration energy for $Zn^{2+}$ is comparable to that of $Cu^{2+}$. Therefore, $E°$ for Zn is negative.

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Question 563 Marks

Complete the following chemical equations:

$\text{MnO}_{4}^{-}+\text{C}_{2}\text{O}_{4}^{2-}+\text{H}+\rightarrow$
$\text{KMnO}_{4}\xrightarrow{\text{heated}}$
$\text{Cr}_{2}\text{O}_{4}^{2-}+\text{H}_{2}\text{S}+\text{H}^{+}+\rightarrow$

Answer

$\text{5C}_{2}\text{O}_{4}^{2-}+\text{2MnO}_{4}^{-}+\text{16H}^{4+}\xrightarrow{\ \ \ \ \ }\ \text{2Mn}^{2+}+\text{8H}_{2}\text{O}+\text{10CO}_{2}.$
$\text{KMnO}_{4}\xrightarrow{\text{heat}}\ \text{K}_{2}\text{MnO}_{2}+\text{MnO}_{2}+\text{O}_{2}$
$\text{Cr}_{2}\text{O}_{7}^{2-}+\text{3H}_{2}\text{S}+\text{8H}^{+}\xrightarrow{\ \ \ \ \ \ }\ \text{2Cr}^{3+}+\text{3S}+\text{7H}_{2}\text{O}$.

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Question 573 Marks

Give reasons:
a. $E ^{\circ}$ value for $Mn ^{3+} / Mn ^{2+}$ couple is much more positive than that for $Fe ^{3+} / Fe ^{2+}$.
b. Iron has higher enthalpy of atomization than that of copper.
c. $Sc ^{3+}$ is colourless in aqueous solution whereas $Ti ^{3+}$ is coloured.

Answer

a. $E ^{\circ} Mn ^{3+} / Mn ^{2+}=$ more positive
$E ^{\circ} Fe ^{3+} / Fe ^{2+}=$ less positive
$Mn ^{2+}=3 d^5 4 s^{\circ}$
$Mn ^{3+}=3 d^4$
$Mn ^{3+}$ will have a tendency to go in the +2 state because $Mn ^{2+}$ is more stable than $Mn ^{3+}$, whereas in the case of Fe , $Fe ^{3+}$ is more stable than $Fe ^{2+}$.
b. Iron has more number of unpaired electrons. Hence, it will have stronger interatomic interaction than copper.
c. $Sc ^3$ is colourless as it does not have unpaired electrons
$Ti ^{3+}$ has one unpaired electron. Hence, it is coloured.

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Question 583 Marks

Write the electronic configuration of the element with atomic number $102$.
What is lanthanoid contraction? What is its effect on the chemistry of the elements which follow the lanthanoids?

Answer

$[Rn] 5f^{14}7s^2$.
The steady decrease in atomic size of lanthanoids with increase in atomic number due to filling of electrons in inner orbitals.

Due to lanthanoid contraction the properties of 4d & 5d series elements become nearly same.

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Question 593 Marks

A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with $KOH$ in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. $H_2SO_4$ and $NaCl$, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

Answer

$A = KMnO_4, B = K_2MnO_4, C = MnO_2, D = MnCl_2$
Explanation:
A violet compound of manganese which is potassium permanganate $(KMnO_4)$, decomposes to liberate potassium manganate$(K_2MnO_4)$ and manganese dioxide$(MnO_2)$ along with oxygen. $KMnO_4(A) \rightarrow K_2MnO_4(B) +MnO_2(C) + O_2$ Manganese dioxide $(MnO_2)$reacts with KOH to give potassium manganate $(K_2MnO_4) MnO_2(C) + KOH + O_2\rightarrow 2K_2MnO_4(B) +2H_2O$ On heating Manganese dioxide $(MnO_2)$ with $NaCl$ and $H_2SO_4$, we get Manganese(II) chloride$(MnCl_2)$, chlorine gas and other products, $MnO_2(C) + 4NaCl + 4H_2SO_4\rightarrow MnCl_2(D) +4NaHSO_4+2H_2O +Cl_2$

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Question 603 Marks

The halides of transition elements become more covalent with increasing oxidation state of the metal.

Answer

As the oxidation state increases, size of the ion of transition element decreases. As per Fajan's rule, as the size of metal ion decreases, covalent character of the bond formed increases.

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Question 613 Marks

How would you account for the following situations?

The transition metals generally form coloured compounds.
With $3d^4$ configuration, $Cr^{2+}$ acts as a reducing agent but $Mn^{3+}$ acts as an oxidising agent. (Atomic Numbers, $Cr = 24, Mn = 25$).
The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.

Answer

Transition metal contain unpaired electrons and are excited to higher energy levels/d-d transition/absorption in visible region.
$Cr^{2+}$ is reducing as its configuration changes from $d^4$ to $d^3$, the latter having half filled $t_{2g}$ level whereas $Mn^{3+}$ to $Mn^{2+}$ results in half filled configuration.
Because $5f, 6d,$ and $7s$ are of comparable energy.

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Question 623 Marks

How do the oxides of transition elements in lower oxidation states differ from those in higher oxidation state in the nature of metal-oxygen bonding and why?

Answer

In the lower oxidation state the transition metal oxides are basic and they are acidic if the metal is in higher oxidation state. The oxides are amphoteric when the metal is in intermediate oxidation state. For example.

+3	+4	+7
$Mn_2O_3$	$MnO_2$	$Mn_2O_7$
Basic	Amphoteric	Acidic

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Chemistry 3 Marks Question