Question 11 Mark
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
AnswerThis is due to poor shielding by $5f-$electrons in the actinoids than that by $4f e^{-1}s$ in lanthanoids.
View full question & answer→Question 21 Mark
Write down the electronic configuration of: $Th^{4+}$
Answer$Th^{4+} = 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 4f^{14}, 5s^2, 5p^6, 5d^{10}, 6s^2, 6p^6.$
View full question & answer→Question 31 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.
View full question & answer→Question 41 Mark
Write down the electronic configuration of: $Lu^{2+}$
Answer$Lu^{2+}: 1s^2 \ 2s^2 \ 2p^{6 } \ 3s^{2 } \ 3p^{6 } \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10 } \ 5s^{2 } \ 5p^6 \ 4f ^{14 } \ 5d^1$
Or, $[Xe]^{54} \ 2f^{14 } \ 3d^3$
View full question & answer→Question 51 Mark
Write down the electronic configuration of: $Pm^{3+}$
Answer$Pm^{3+}: 1s^2 \ 2s^{2} \ 2p^{6} \ 3s^{2} \ 3p^{6} \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10} \ 5s^{2 } \ 5p^6 \ 4f^4$
Or, $[Xe]^{54} \ 3d^3$
View full question & answer→Question 61 Mark
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: Iodide
Answer$\ce{K_2Cr_2O_{7 }}$ 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 $\ce{K_2Cr_2O_7}.$
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{I}^-\rightarrow3\text{I}_2+6\text{e}^-$
$\overline{\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+6\text{I}^-\rightarrow3\text{I}_2+2\text{Cr}^{3+}+7\text{H}_2\text{O}}$
View full question & answer→Question 71 Mark
Calculate the ‘spin only’ magnetic moment of $M^{2+}_{(aq)}$ ion $(Z = 27).$
AnswerAtomic 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
$\mu=\sqrt{3(3+2)}$
$\mu=\sqrt{15}$
$\mu=3.8\text{M}$
View full question & answer→Question 81 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?
AnswerThe 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.
View full question & answer→Question 91 Mark
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with$:$ iron$(II)$ solution
AnswerIron $(II)$ solution$:$ Ferrous salts $(Fe^{2+})$ are oxidised ferric $(Fe^{3+})$ salts when they are treated with acidified $\ce{K_2Cr_2O_7}.$
Reaction$:$
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^{+}+6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{Fe}^{2+}\rightarrow6\text{Fe}^{3+}+6\text{e}^-$
$\overline{\text{Cr}_2\text{O}^{2-}_7+6\text{Fe}^2+14\text{H}^+\rightarrow2\text{Cr}^{3+}+6\text{Fe}^{3+}+7\text{H}_2\text{O}}$
View full question & answer→Question 101 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$Ce: 1s^2 \ 2s^{2 } \ 3p^{6 } \ 3s^{2 } \ 3p^{6 } \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10 } \ 5s^{2 } \ 5p^6 \ 4f^{1 } \ 5d^{1 } \ 6s^2$
Magnetic moment can be calculated as$:$
$\mu=\sqrt{\text{n}(\text{n}+2)}$
Where,
$n =$ number of unpaired electrons
The electronic configuration of $Ce^{3+}: 1s^2 \ 2s^2 \ 2p^{6 } \ 3s^{2 } \ 3p^{6 } \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10 } \ 5s^{2 } \ 5p^6 \ 4f^1$
In $Ce^{3+}, n = 1$
Therefore, $\mu=\sqrt{2(2+2)}$
$=\mu=\sqrt{2\times4}$
$=\mu=\sqrt{8}$
$=\mu=2\sqrt{2}$
$= 2.828 \ BM$
View full question & answer→Question 111 Mark
Explain why $Cu^+$ ion is not stable in aqueous solutions?
AnswerIn 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 $Cu.$
$2\text{Cu}^+_{(\text{aq})}\rightarrow\text{Cu}^+_{(\text{aq})}+\text{Cu}_{(\text{s})}$
View full question & answer→Question 121 Mark
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with$:\ce{H_2S}$
Answer$\ce{H_2S: H_2S}$ is oxidised to sulphur.
$\text{Cr}_2\text{O}_7^{2-}+3\text{H}_2\text{S}+8\text{H}^+\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}+3\text{S}$
View full question & answer→Question 131 Mark
What are the different oxidation states exhibited by the lanthanoids?
AnswerIn 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.
View full question & answer→Question 141 Mark
Write down the electronic configuration of: $Ce^{4+}$
Answer$Ce^{4+}: 1s^2 \ 2s^2 \ 2p^{6} \ 3s^{2} \ 3p^{6} \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10} \ 5s^{2} \ 5p^6$
Or, $[Xe]^{54}$
View full question & answer→Question 151 Mark
Write down the electronic configuration of: $CO^{2+}$
Answer$CO^{2+}: 1s^2 \ 2s^2 \ 2p^{6} \ 3s^{2} \ 3p^{6} \ 3d^7$
Or, $[Ar]^{18}d^7$
View full question & answer→Question 161 Mark
Which of the $3d$ series of the transition metals exhibits the largest number of oxidation states and why?
Answer$Mn (Z = 25) = 3d^5 \ 4s^2$
$Mn$ 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 $+7.$
View full question & answer→Question 171 Mark
Silver atom has completely filled d orbitals $(4d^{10})$ in its ground state. How can you say that it is a transition element?
AnswerThe 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.
View full question & answer→Question 181 Mark
Compare the chemistry of actinoids with that of the lanthanoids with special reference to: Electronic configuration
AnswerElectronic 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.
View full question & answer→Question 191 Mark
Write down the electronic configuration of: $Mn^{2+}$
Answer$Mn^{2+}: 1s^2 \ 2s^2 \ 2p^{6 } \ 3s^{2} \ 3p^{6} \ 3d^5$
View full question & answer→Question 201 Mark
Write down the electronic configuration of: $Cr^{3+}$
Answer$Cr^{3+}: 1s^2 \ 2s^2 \ 2p^{6} \ 3s^{2} \ 3p^{6} \ 3d^3$
Or, $[Ar]^{18} \ 3d^3$
View full question & answer→Question 211 Mark
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
AnswerBoth 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.
View full question & answer→Question 221 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 $Fe^{3+}.$
View full question & answer→Question 231 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.
AnswerLawrencium $(Lr)$ 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 $+3.$
View full question & answer→Question 241 Mark
Write down the electronic configuration of: $Cu^+$
Answer$Cu^+: 1s^2 \ 2s^2 \ 2p^{6} \ 3s^{2} \ 3p^{6} \ 3d^{10}$
Or, $[Ar]^{18} \ 3d^{10}$
View full question & answer→Question 251 Mark
Why do transition elements show similarities along the horizontal period?
AnswerAll transition elements contain incompletely filled d-subshell whereas the outer shell electronic configuration remains the same.
View full question & answer→Question 261 Mark
What is the maximum oxidation state shown by actinoids?
Question 271 Mark
Write any one use of pyrophoric alloys.
AnswerPyrophoric alloys emit sparks when struck. Hence, they are used in making flints for lighters.
View full question & answer→MCQ 281 Mark
Electronic configuration of a transition element $X$ in $+3$ oxidation state is $[Ar]3d^5.$ What is its atomic number?
AnswerElectronic configuration of element $X = [Ar] 3d^{5+}$
oxidation state $= 18 + 5 + 3 = 26.$
View full question & answer→Question 291 Mark
Analysis shows that $\ce{FeO}$ has a non$-$stoichiometric composition with formula $Fe_{0.95}O.$ Give reason.
Answer$\ce{FeO}$ has a non$-$stoichiometric composition with the formula $Fe_{0.95}O,$ because $Fe$ is present in both $+2$ and $+3$ oxidation states.
View full question & answer→Question 301 Mark
What are the two important oxidation states of Group 6 elements of the periodic table?
Question 311 Mark
Name one ore each of manganese and chromium.
Answer$\ce{MnO_2} ($pyrolusite$)$ is an ore of $Mn$ whereas $\ce{FeCr_2O_4} ($chromite$)$ is an ore of $Cr.$
View full question & answer→Question 321 Mark
A member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Question 331 Mark
Name a transition element which does not exhibit variable oxidation state
AnswerScandium (Z = 21) does not exhibit variable oxidation states.
View full question & answer→Question 341 Mark
Arrange the following in increasing order of acidic character: $\ce{CrO_3, CrO, Cr_2O_3}.$
Answer$\ce{CrO < Cr_2O_3 < CrO_3}.$ Higher the oxidation state, more will be acidic character.
View full question & answer→Question 351 Mark
On the basis of lanthanoid contraction, explain the following:
Radii of 4d and 5d block elements.
AnswerRadii of 4d and 5d block elements will be almost same.
View full question & answer→Question 361 Mark
Why are transition elements so named?
AnswerTransition elements are so named because their properties are in between those of s-block and p-block.
View full question & answer→Question 371 Mark
What is meant by 'lanthanoid contraction'?
AnswerThe regular decrease in the atomic and ionic radii/(having the same charge) of Lanthanoids with increasing atomic number is known as Lanthanoid contraction.
View full question & answer→Question 381 Mark
Answer the following questions:
Which element of the first transition series has lowest enthalpy ofatomisation?
Answerbecause of the completely filled 3d sub shell no unpaired electron is left for metallic bonding.
View full question & answer→Question 391 Mark
On the basis of Lanthanoid contraction, explain the following:
Stability of the complexes of lanthanoids.
AnswerStability of complexes from La to Lu, increases as the size of the central atom decreases.
View full question & answer→Question 401 Mark
What is the common oxidation state of Cu, Ag and Au?
AnswerThe common oxidation state of Cu, Ag and Au is +1.
View full question & answer→Question 411 Mark
Why are transition elements known as d-block elements?
AnswerThe last electron enters (n - 1) d-orbital, i.e., d-orbital of the penultimate shell. Hence, these are known as d-block elements.
View full question & answer→Question 421 Mark
How many elements are present in the d-block of the periodic table?
Question 431 Mark
On the basis of Lanthanoid contraction, explain the following:
Trends in acidic character of lanthanoid oxides.
AnswerThe acidic nature of lanthanoid oxides increases from La to Lu.
View full question & answer→Question 441 Mark
Give the general electronic configuration of actinoids.
Answer$5f^{1-14} \ 6d^{0-1} \ 7s^2.$
View full question & answer→Question 451 Mark
Which divalent metal ion has maximum paramagnetic character among the first transition metals? Why?
Answer$Mn^{2+}$ has the maximum paramagnetic character because of the maximum number of unpaired electrons, viz. $5.$
View full question & answer→Question 461 Mark
Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number.
AnswerFormula 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.
View full question & answer→Question 471 Mark
$Fe$ has higher melting point than $Cu.$ Why?
AnswerThis is because $Fe (3d^6 \ 4s^1)$ has four unpaired electrons in $3d-$subshell. While $Cu (3d^{10} \ 4s^1)$ has only one electron in the $4s-$subshell.
Hence, metallic bonds in $Fe$ are much stronger than those in $Cu.$
View full question & answer→Question 481 Mark
Discuss the relative stability in aqueous solutions of $+2$ oxidation state among the elements: $\text{Cr, Mn, Fe}$ and Co. How would you justify this situation? $($At. Nos. $Cr = 24, Mn = 25, Fe = 26, Co = 27)$
AnswerOn the basis of electrochemical series, the standard electrode potential shows the following order:
$\ce{E^\circ Mn^{2+}/Mn < E^\circ Cr^{2+}/Cr < E^\circ Fe^{2+}/Fe < E^\circ CO^{2+}/CO}$
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
$\ce{Co^{2+} < Fe^{2+} < Cr^{2+} < Mn^{2+}}$
View full question & answer→Question 491 Mark
Write the formula of an oxo$-$anion of Manganese $(Mn)$ in which it shows the oxidation state equal to its group number.
Answer$\ce{MnO_4^{- } / KMnO_4}$
View full question & answer→Question 501 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.
AnswerDue to lanthanoid contraction, they have almost same size (Zr = 160 pm) and (Hf = 159 pm).
View full question & answer→