1 Marks Question

Question 11 Mark

Write down the electronic configuration of: $CO^{2+}$

Answer

$CO^{2+}: 1s^2 2s^2 2p^63s^23p^63d^7$
Or, $[Ar]^{18}d^7$

Question 21 Mark

Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Answer

$Mn (Z = 25) = 3d^54s^2$
Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from $+2$ to $+7.$

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Question 31 Mark

Silver atom has completely filled d orbitals $(4d^{10})$ in its ground state. How can you say that it is a transition element?

Answer

The outer electronic configuration of Ag $(Z = 47)$ is $4d^{10}5s^1.$
It shows $+1$ and $+2O.S$. $($in $AgO$ and $AgF_2)$. And in $+2O.S. ,$
the electronic configuration is $d^9$
i.e., $d -$ subshell is incompletely filled. Hence, it is a transition element.

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Question 41 Mark

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Electronic configuration

Answer

Electronic configuration:
The general electronic configuration of lanthanoids is $[Xe]^{54} 4f^{1-14} 5d^{0·1} 6s^2$ and that of actinoids is $[Rn]^{86} 5f^{0-14} 6d^{0-1} 7s^2,$
lanthanoids belong to 4f series whereas actinoids belong to 5f-series.

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Question 51 Mark

Write down the electronic configuration of:$Mn^{2+}$

Answer

$Mn^{2+}: 1s^2 2s^2 2p^63s^23p^63d^5$

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Question 61 Mark

Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer

This is due to poor shielding by 5 f -electrons in the actinoids than that by $4 \mathrm{f} \mathrm{e}^{-1} \mathrm{~s}$ in lanthanoids.

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Question 71 Mark

Write down the electronic configuration of:
$Cr^{3+}$

Answer

$Cr^{3+}: 1s^2 2s^2 2p^63s^23p^63d^3$
Or, $[Ar]^{18}3d^3$

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Question 81 Mark

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Answer

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 91 Mark

Which is a stronger reducing agent $Cr^{2+}$ or $Fe^{2+}$ and why?

Answer

$Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}.$
This is because $E^\circ(Cr^{3+}/Cr^{2+})$ is negative $(-0.41\ V)$
whereas $E^\circ(Fe^{3+}/Fe^{2+})$ is positive (+ 0.77 V).
Thus, $Cr^{2+}$ is easily oxidised to $Fe^{3+}$ but $Fe^{2+}$ cannot be easily oxidised to $Fe^{3+}.$

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Question 101 Mark

Write down the electronic configuration of:
$ \mathrm{Cu}^{+}$

Answer

$ \mathrm{Cu}^{+}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} $
$\text { Or, }[\mathrm{Ar}]^{18} 3 d^{10}$

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Question 111 Mark

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Answer

Lawrencium (Lr) is the last element in the series of antinoids.
Its electronic configuration is [Rn] $5f^{14}\ 6d^1\ 7s^2$. The possible oxidation state of Lawrencium is +3.

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Question 121 Mark

Calculate the ‘spin only’ magnetic moment of $M^{2+}_{(aq)}$ ion $(Z = 27).$

Answer

Atomic number $(27) = [Ar] 3d^74s^2$
$M^{2+}= [Ar] 3d^7$
Thus it has three unpaired electrons
therefore magnetic moment is
$\mu=\sqrt{\text{n}(\text{n}+2)}$
Where n = total number of unpaired electron
$\mu=\sqrt{3(3+2)}$
$\mu=\sqrt{15}$
$\mu=3.8\text{M}$

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Question 131 Mark

Explain why $Cu^+$ ion is not stable in aqueous solutions?

Answer

In an aqueous medium, $Cu^{2+}$ is more stable than $Cu^+.$ This is because although energy is required to remove one electron from $Cu^+$ to $Cu^{2+}$, high hydration energy of $Cu^{2+}$ compensates for it. Therefore, $Cu^+$ ion in an aqueous solution is unstable. It disproportionate to give $Cu^{2+}$ and $Cu.$
$2\text{Cu}^+_{(\text{aq})}\rightarrow\text{Cu}^+_{(\text{aq})}+\text{Cu}_{(\text{s})}$

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Question 141 Mark

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
$\mathrm{H}_2 \mathrm{S}$

Answer

$\mathrm{H}_2 \mathrm{S}: \mathrm{H}_2 \mathrm{S}$ is oxidised to sulphur.
$\text{Cr}_2\text{O}_7^{2-}+3\text{H}_2\text{S}+8\text{H}^+\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}+3\text{S}$

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Question 151 Mark

Write down the electronic configuration of:
$Th^{4+}$

Answer

$T h^{4+}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, 4 f^{14}, 5 s^2, 5 p^6, 5 d^{10}, 6 s^2, 6 p^6$

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Question 161 Mark

The $\text{E}^\ominus(\text{M}^{2+}/\text{M})$ value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high $\Delta_\text{a}\text{H}^\ominus$ and low $\Delta_\text{hyd}\text{H}^\ominus$).

Answer

$\text{E}^\ominus(\text{M}^{2+}/\text{M})$ for any metal is related to the sum of enthelpy changes taking place in following steps:
$\text{M(s)}+\Delta_\text{a}\text{H}\rightarrow\text{M(g)}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
Cu has a high enthelpy of atomisation $(\Delta_\text{a}\text{H})$ and a low enthelpy of hydration $(\Delta_\text{hyd}\text{H})$. The high energy required to transform Cu(s) to $Cu^{2+}$ (aq) is not balenced by its hydration enthelpy. Hence $\text{E}^\ominus(\text{Cu}^{2+}/\text{Cu})$ is positive.

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Question 171 Mark

What are the different oxidation states exhibited by the lanthanoids?

Answer

In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

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Question 181 Mark

Write down the electronic configuration of: $Lu^{2+}$

Answer

$ L u^{2+}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^6 4 f^{14} 5 d^1$
$ \text { Or, }[X e]^{54} 2 f^{14} 3 d^3$

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Question 191 Mark

Write down the electronic configuration of: $Pm^{3+}$

Answer

$ \mathrm{Pm}^{3+}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^6 4 f^4 $
$\text { Or, }[\mathrm{Xe}]^{54} 3 d^3$

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Question 201 Mark

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
Iodide

Answer

$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from +6 to +3. The oxidising action can be represented as follows:
$\text{Cr}_2\text{O}^{2-}_7+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$
Iodide: Iodide ion (J-) is oxidised to I by the acidfied solution of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$.
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{I}^-\rightarrow3\text{I}_2+6\text{e}^-\\ \overline{\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+6\text{I}^-\rightarrow3\text{I}_2+2\text{Cr}^{3+}+7\text{H}_2\text{O}}$

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Question 211 Mark

In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol$^{–1}.$ Why?

Answer

The extent of metallic bonding an element undergoes deideds the enthalpy of atomization the more extensive the metallic bonding of an element the more will be its enthalpy of atomization.
Sc & Zn belongs to $3^{rd}$ group of periodic table. In all transition metals (except Zn, electronic configuration: $3d^{10} 4s^2$), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.

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Question 221 Mark

Use Hund’s rule to derive the electronic configuration of $Ce^{3+}$ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Answer

$Ce: 1s^2 2s^23p^63s^23p^63d^{10} 4s^2 4p^6 4d^{10}5s^25p^6 4f^15d^16s^2$
Magnetic moment can be calculated as:
$\mu=\sqrt{\text{n}(\text{n}+2)}$
Where,
n = number of unpaired electrons
The electronic configuration of $Ce^{3+}: 1s^2 2s^2 2p^63s^23p^63d^{10} 4s^2 4p^6 4d^{10}5s^25p^6 4f^1$
In $Ce^{3+}, n = 1$
Therefore, $\mu=\sqrt{2(2+2)}$
$=\mu=\sqrt{2\times4}$
$=\mu=\sqrt{8}$
$=\mu=2\sqrt{2}$
= 2.828 BM

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Question 231 Mark

Write down the electronic configuration of: $Ce^{4+}$

Answer

$ \mathrm{Ce}^{4+}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^6 $
$ \mathrm{Or},[\mathrm{Xe}]^{54}$

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Question 241 Mark

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
iron(II) solution

Answer

Iron (II) solution:Ferrous salts $\left(\mathrm{Fe}^{2+}\right)$ are oxidised ferric $\left(\mathrm{Fe}^{3+}\right)$ salts when they are treated with acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$.
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^{+}+6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{Fe}^{2+}\rightarrow6\text{Fe}^{3+}+6\text{e}^-\\\overline{\text{Cr}_2\text{O}^{2-}_7+6\text{Fe}^2+14\text{H}^+\rightarrow2\text{Cr}^{3+}+6\text{Fe}^{3+}+7\text{H}_2\text{O}}$

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Question 251 Mark

Why do transition elements show similarities along the horizontal period?

Answer

All transition elements contain incompletely filled d-subshell whereas the outer shell electronic configuration remains the same.

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Question 261 Mark

Answer the following questions:
Which element of the first transition series has lowest enthalpy ofatomisation?

Answer

because of the completely filled 3d sub shell no unpaired electron is left for metallic bonding.

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Question 271 Mark

On the basis of Lanthanoid contraction, explain the following:
Stability of the complexes of lanthanoids.

Answer

Stability of complexes from La to Lu, increases as the size of the central atom decreases.

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Question 281 Mark

What is the common oxidation state of Cu, Ag and Au?

Answer

The common oxidation state of Cu, Ag and Au is +1.

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Question 291 Mark

Why are transition elements known as d-block elements?

Answer

The last electron enters (n - 1) d-orbital, i.e., d-orbital of the penultimate shell. Hence, these are known as d-block elements.

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MCQ 301 Mark

Electronic configuration of a transition element $X$ in $+3$ oxidation state is $[Ar]3d^5$. What is its atomic number?

A
$25$
✓
$26$
C
$27$
D
$24$

Answer

Correct option: B.

$26$

Electronic configuration of element $X = [Ar] 3d^{5+}$ oxidation state $= 18 + 5 + 3 = 26.$

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Question 311 Mark

Arrange the following in increasing order of acidic character: $\mathrm{CrO}_3, \mathrm{CrO}, \mathrm{Cr}_2 \mathrm{O}_3$.

Answer

$\mathrm{CrO} < \mathrm{Cr}_2 \mathrm{O}_3 < \mathrm{CrO}_3$. Higher the oxidation state, more will be acidic character.

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Question 321 Mark

How many elements are present in the d-block of the periodic table?

Answer

40 elements.

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Question 331 Mark

On the basis of Lanthanoid contraction, explain the following:
Trends in acidic character of lanthanoid oxides.

Answer

The acidic nature of lanthanoid oxides increases from La to Lu.

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Question 341 Mark

Give the general electronic configuration of actinoids.

Answer

$5 f^{1-14} 6 d^{0-1} 7 s^2$

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Question 351 Mark

Which divalent metal ion has maximum paramagnetic character among the first transition metals? Why?

Answer

$\mathrm{Mn}^{2+}$ has the maximum paramagnetic character because of the maximum number of unpaired electrons, viz., 5 .

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Question 361 Mark

Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number.

Answer

Formula of oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number (6) is $\text{Cr}_2\text{O}^2_{−7}.$
2Cr + (-2 ×7) = -2
2Cr - 14 = -2
2Cr = 12
Cr = +6
Oxidation of Cr in $\text{Cr}_2\text{O}^2_{−7}$ is +6 which is equal to its group number 6.

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Question 371 Mark

Write the outer electronic configuration of lanthanoids.

Answer

$4 \mathrm{f}^{1-14} 5 \mathrm{~d}^{0-1} 6 \mathrm{~s}^2$

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Question 381 Mark

Name one ore each of manganese and chromium.

Answer

$\mathrm{MnO}_2$ (pyrolusite) is an ore of Mn whereas $\mathrm{FeCr}_2 \mathrm{O}_4$ (chromite) is an ore of Cr .

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Question 391 Mark

Why are transition elements so named?

Answer

Transition elements are so named because their properties are in between those of s-block and p-block.

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Question 401 Mark

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

Answer

Cerium (Ce).

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Question 411 Mark

What is the general formula by which the electronic configuration of the transition elements is represented?

Answer

$(n-1) d^{1-10} n s^{1-2}$

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Question 421 Mark

Answer the following questions:
Which element of the first transition series has highest third ionisation enthalpy?

Answer

The trend in the third ionisation enthalpies is not complicated by the 4 s orbital factor and shows the greater difficulty of removing an electron from the $\mathrm{d} 5\left(\mathrm{Mn}^{2+}\right)$ and $\mathrm{d} 10\left(\mathrm{Zn}^{2+}\right)$ ions.

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Question 431 Mark

Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.

Answer

$\mathrm{MnO}_4^{-} / \mathrm{KMnO}_4$

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Question 441 Mark

Why does a transition series contain 10 elements?

Answer

There are five d-orbitals in an energy level and each orbital can contain two electrons. As we move from one element to the next, an electron is added and for complete filling of the five d-orbitals, 10 electrons are required.

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Question 451 Mark

A transition metal which does not exhibit variation in oxidation state in its compounds.

Answer

Scandium (Sc).

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Question 461 Mark

Name a transition element which does not exhibit variable oxidation state

Answer

Scandium (Z = 21) does not exhibit variable oxidation states.

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Question 471 Mark

What is meant by 'lanthanoid contraction'?

Answer

The regular decrease in the atomic and ionic radii/(having the same charge) of Lanthanoids with increasing atomic number is known as Lanthanoid contraction.

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Question 481 Mark

Write any one use of pyrophoric alloys.

Answer

Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for lighters.

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Question 491 Mark

Analysis shows that FeO has a non-stoichiometric composition with formula $Fe_{0.95}O$. Give reason.

Answer

FeO has a non-stoichiometric composition with the formula $Fe_{0.95}O$, because Fe is present in both +2 and +3 oxidation states.

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Question 501 Mark

How would you account for the increasing oxidising power in the series $\text{VO}^+_2<\text{Cr}2\text{O}^{2-}_7<\text{MnO}^-_4$

Answer

This is due to the increasing stability of the lower species to which they are reduced.

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Question 511 Mark

On the basis of lanthanoid contraction, explain the following:
Radii of 4d and 5d block elements.

Answer

Radii of 4d and 5d block elements will be almost same.

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Question 521 Mark

Account for the following:
Actinoids show wide range of oxidation states.

Answer

This is due to comparable energies of 5f, 6d and 7s orbitals.

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Question 531 Mark

Answer the following questions:
Which element of the first transition series has highest second ionisation enthalpy?

Answer

Exchange the second ionization enthalpy shows unusually high values for Cr and Cu first transition series where the $d^5$ and $d^{10}$ configuration of the $\mathrm{M}^{+}$ions are disrupted, with considerable loss of energy.

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Question 541 Mark

What are the two important oxidation states of Group 6 elements of the periodic table?

Answer

+3 and +6.

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Question 551 Mark

Fe has higher melting point than Cu. Why?

Answer

This is because $\mathrm{Fe}\left(3 d^6 4 s^1\right)$ has four unpaired electrons in 3 d -subshell. While $\mathrm{Cu}\left(3 \mathrm{d}^{10} 4 \mathrm{s}^1\right)$ has only one electron in the 4 s -subshell. Hence, metallic bonds in Fe are much stronger than those in Cu .

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Question 561 Mark

Discuss the relative stability in aqueous solutions of +2 oxidation state among the elements: Cr, Mn, Fe and Co. How would you justify this situation?
(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)

Answer

On the basis of electrochemical series, the standard electrode potential shows the following order:
$E^\circ Mn^{2+}/Mn < E^\circ Cr^{2+}/Cr < E^\circ Fe^{2+}/Fe < E^\circ CO^{2+}/CO$
Therefore, $Co^{2+}$ gets easily reduced to metallic cobalt while it is difficult to reduce $Mn^{2+}$. Hence, $Mn^{2+}$ will be the most stable and the increasing stability order will be
$Co^{2+} < Fe^{2+} < Cr^{2+} < Mn^{2+}$

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Question 571 Mark

How would you account for the following:
Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them.

Answer

Due to lanthanoid contraction, they have almost same size (Zr = 160 pm) and (Hf = 159 pm).

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Question 581 Mark

What is the maximum oxidation state shown by actinoids?

Answer

+7.

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Question 591 Mark