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 $+7.$
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.
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.
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.
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 $Fe^{3+}.$
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.
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
$\mu=\sqrt{3(3+2)}$
$\mu=\sqrt{15}$
$\mu=3.8\text{M}$
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 $Cu.$
$2\text{Cu}^+_{(\text{aq})}\rightarrow\text{Cu}^+_{(\text{aq})}+\text{Cu}_{(\text{s})}$
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. $\text{Cr}_2\text{O}_7^{2-}+3\text{H}_2\text{S}+8\text{H}^+\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}+3\text{S}$
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.
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.
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:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{I}^-\rightarrow3\text{I}_2+6\text{e}^-\\ \overline{\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+6\text{I}^-\rightarrow3\text{I}_2+2\text{Cr}^{3+}+7\text{H}_2\text{O}}$
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.
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
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.
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.
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.
The regular decrease in the atomic and ionic radii/(having the same charge) of Lanthanoids with increasing atomic number is known as Lanthanoid contraction.