d and f Block Elements

Identify which metal (M) has the highest second ionization enthalpy.

We are comparing the elements Sc, Ti, V, Cr, Mn, and Fe in terms of their second ionization enthalpy (IE₂). Recall:

The second ionization enthalpy (IE₂) is the energy required to remove one electron from the singly charged ion $ \mathrm{M}^{+} $ to form $ \mathrm{M}^{2+} $.

For most 3d transition metals, the first electron is removed from the 4s orbital.

Particularly large values of IE₂ often occur if, after removal of the first electron, one is forced to remove an electron from a stable configuration (e.g., half-filled d-orbital).

Let us outline the ground-state (neutral atom) electron configurations and the removal of the first and second electrons:

Sc $\bigl[ \mathrm{Ar} \bigr] 3d^1\,4s^2 $

$\mathrm{Sc} \rightarrow \mathrm{Sc}^{+}$: remove 1 electron from 4s

$\mathrm{Sc}^{+} \rightarrow \mathrm{Sc}^{2+}$: remove the second 4s electron

Final: $\mathrm{Sc}^{2+} = [\mathrm{Ar}]\,3d^1$

Ti $\bigl[ \mathrm{Ar} \bigr] 3d^2\,4s^2 $

1st electron from 4s → $\mathrm{Ti}^{+} = [\mathrm{Ar}]\,3d^2\,4s^1$

2nd electron from 4s → $\mathrm{Ti}^{2+} = [\mathrm{Ar}]\,3d^2$

V $\bigl[ \mathrm{Ar} \bigr] 3d^3\,4s^2 $

1st electron from 4s → $\mathrm{V}^{+} = [\mathrm{Ar}]\,3d^3\,4s^1$

2nd electron from 4s → $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$

Cr $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^1 $

1st electron from 4s → $\mathrm{Cr}^{+} = [\mathrm{Ar}]\,3d^5$ (a stable half-filled d$^{5}$ configuration)

2nd electron now must come from the 3d shell → $\mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4$

Because removing an electron from a half-filled d$^{5}$ orbital costs a lot of energy, $\mathrm{Cr}$ generally has a notably high second IE.

Mn $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^2 $

1st electron from 4s → $\mathrm{Mn}^{+} = [\mathrm{Ar}]\,3d^5\,4s^1$

2nd electron from 4s → $\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$ (thus still d$^{5}$ in the end)

The second ionization for Mn is large but less dramatic than for Cr because Mn$^{2+}$ ends with a half-filled d$^{5}$. You are not removing from a half-filled d$^{5}$ to get Mn$^{2+}$.

Fe $\bigl[ \mathrm{Ar} \bigr] 3d^6\,4s^2 $

1st electron from 4s → $\mathrm{Fe}^{+} = [\mathrm{Ar}]\,3d^6\,4s^1$

2nd electron from 4s → $\mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6$

From experimental data (and the arguments above), $\mathrm{Cr}$ indeed has the highest second ionization enthalpy among these six metals.

Determine the electronic configuration of $\mathbf{Cr}^{+}$ and its spin-only magnetic moment.

Neutral $\mathrm{Cr}$: $\bigl[ \mathrm{Ar} \bigr]\,3d^5\,4s^1$

$\mathrm{Cr}^{+}$: Removal of the 4s electron ⇒ $\bigl[ \mathrm{Ar} \bigr]\,3d^5$

The $3d^5$ configuration has 5 unpaired electrons.

The spin-only magnetic moment $\mu$ is given by

$ \mu = \sqrt{n(n+2)} \; \mathrm{BM}, $

where $n$ = number of unpaired electrons. Here $n = 5$, so

$ \mu = \sqrt{5(5+2)} \;=\; \sqrt{35} \;\approx\; 5.92 \;\text{BM}. $

This is often rounded to about 5.9 or 6.0 BM.

The color of lanthanoid ions in solutions is mainly due to the electronic transitions within the 4f subshell. The lanthanoid ions are more likely to be colorless when they have fully filled (with 14 electrons) or completely empty (with 0 electrons) f orbitals because, in these cases, there are no electrons to undergo f-f transitions, and as a result, no absorption of visible light occurs leading to colorlessness.

Let's consider the electronic configurations of the lanthanoid ions provided:

- $\mathrm{Eu}^{3+}$: Europium (Eu) has an atomic number of 63. Neutral europium ([Xe]4f⁷ 6s^2) loses three electrons to form Eu³⁺, leaving it with an electronic configuration equivalent to [Xe]4f⁶. With 6 electrons in the f orbital, it can undergo f-f transitions, thus it is not colorless.

- $\mathrm{Lu}^{3+}$: Lutetium (Lu) has an atomic number of 71. In its 3+ ionic state, lutetium has lost its 6s and 5d electrons and is left with a completely filled 4f orbital ([Xe] 4f¹⁴). This configuration cannot allow for any f-f transitions, as there are no available energy levels within the f orbital for an electron to jump to, making Lu³⁺ colorless.

- $\mathrm{Nd}^{3+}$: Neodymium (Nd) has an atomic number of 60. In its 3+ state ([Xe] 4f³), it clearly has partially filled f orbitals, which can absorb visible light for f-f transitions, so it is not colorless.

- $\mathrm{La}^{3+}$: Lanthanum (La) has an atomic number of 57. In its 3+ ionic state, it has a configuration of [Xe], meaning that its 4f orbital is completely empty. Since there are no electrons in the f orbital to undergo f-f transitions, La³⁺ is colorless.

- $\mathrm{Sm}^{3+}$: Samarium (Sm) has an atomic number of 62. As a 3+ ion ([Xe] 4f⁵), it too has electrons in the f orbital capable of undergoing f-f transitions, so it is not colorless.

From the analysis, the colorless lanthanoid ions among the ones listed are $\mathrm{Lu}^{3+}$ and $\mathrm{La}^{3+}$.

Therefore, the number of colorless lanthanoid ions among the given options is 2.

In order to determine the spin-only magnetic moment value of the species with the least oxidizing ability, we first need to analyze their oxidation states and electron configurations.

1. $\mathrm{VO}_2^{+}$:

For vanadium in $\mathrm{VO}_2^{+}$, the oxidation state is +5. The electronic configuration of V is $[Ar] \, 3d^3 \, 4s^2$. Thus, in +5 oxidation state, vanadium will have zero d-electrons (since it loses 5 electrons) and, consequently, $\mathrm{VO}_2^{+}$ is diamagnetic (since zero unpaired electrons).

2. $\mathrm{MnO}_4^{-}$:

For manganese in $\mathrm{MnO}_4^{-}$, the oxidation state is +7. The electronic configuration of Mn is $[Ar] \, 3d^5 \, 4s^2$. Thus, in +7 oxidation state, manganese will have zero d-electrons (since it loses 7 electrons) and, consequently, $\mathrm{MnO}_4^{-}$ is also diamagnetic (since zero unpaired electrons).

3. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$:

For chromium in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, we have two chromium atoms. Each chromium is in the +6 oxidation state. The electronic configuration of Cr is $[Ar] \, 3d^5 \, 4s^1$. Thus, in +6 oxidation state, each chromium will have zero d-electrons (since it loses 6 electrons) and, consequently, $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is also diamagnetic (since zero unpaired electrons).

From these observations, we note that all the species we analyzed are diamagnetic. However, to identify the species with the least oxidizing ability, we look at their standard reduction potentials (E° values).

Typically, the oxidizing ability increases with increasing positive E° values. Given the nature of these species, we can infer that:

• $\mathrm{MnO}_4^{-}$ is a very strong oxidizing agent (very high E° value).
• $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is also a strong oxidizing agent (high but less than $\mathrm{MnO}_4^{-}$).
• $\mathrm{VO}_2^{+}$ has a moderate oxidizing ability (lesser E° value than both $\mathrm{MnO}_4^{-}$ and $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$).

Therefore, among the given species, $\mathrm{VO}_2^{+}$ has the least oxidizing ability.

As mentioned previously, $\mathrm{VO}_2^{+}$ has zero unpaired electrons, making it diamagnetic. The spin-only magnetic moment is given by the formula:

$\mu = \sqrt{n(n+2)}$

where $n$ is the number of unpaired electrons.

For $\mathrm{VO}_2^{+}$, $n=0$, thus:

$\mu = \sqrt{0(0+2)} = 0 \ \text{BM}$

Given the options, the nearest integer value of the spin-only magnetic moment for the species with the least oxidizing ability (which is $\mathrm{VO}_2^{+}$) is indeed:

0 BM

First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.

1. $\mathrm{CrO}$: In $\mathrm{CrO}$, the oxidation state of chromium is +2. The electronic configuration of chromium (Cr) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$. In the +2 oxidation state, two electrons are removed, typically from the 4s and one of the 3d orbitals, leaving us with the configuration $3d^4$.

To find the spin-only magnetic moment, we use the formula:

$\mu = \sqrt{n(n+2)} \mathrm{BM}$

where n is the number of unpaired electrons. For $\mathrm{Cr}^{2+}$, we have 4 unpaired electrons in the 3d orbitals.

Thus,

$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mathrm{BM}$

2. $\mathrm{Cr}_2 \mathrm{O}_3$: In $\mathrm{Cr}_2 \mathrm{O}_3$, the oxidation state of chromium is +3. The electronic configuration of $\mathrm{Cr}^{3+}$ is $3d^3$ after losing three electrons.

For $\mathrm{Cr}^{3+}$, there are 3 unpaired electrons.

Thus,

$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mathrm{BM}$

3. $\mathrm{CrO}_3$: In $\mathrm{CrO}_3$, the oxidation state of chromium is +6. The electronic configuration of $\mathrm{Cr}^{6+}$ is $3d^0$ after losing six electrons. There are no unpaired electrons for $\mathrm{Cr}^{6+}$.

Since $\mathrm{Cr}^{6+}$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.

The magnetic moment values for each compound are:

• $\mathrm{CrO}$: 4.90 BM (basic oxide)
• $\mathrm{Cr}_2 \mathrm{O}_3$: 3.87 BM (amphoteric oxide)
• $\mathrm{CrO}_3$: 0 BM

The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:

$4.90 + 3.87 = 8.77 \mathrm{BM}$

Expressing in terms of $10^{-2} \mathrm{BM}$,

$8.77 \mathrm{BM} \times 100 = 877 (\times 10^{-2} \mathrm{BM})$

Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $\times 10^{-2} \mathrm{BM}$ (nearest integer).

To determine the spin-only magnetic moments of products A and B formed from the fusion of chromite ore with sodium carbonate in the presence of air, we first need to examine the given reaction and the properties of the products.

The reaction provided is:

$4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 (\mathrm{A}) + 2 \mathrm{Fe}_2 \mathrm{O}_3 (\mathrm{B}) + 8 \mathrm{CO}_2$

• Product A is sodium chromate, $\mathrm{Na}_2 \mathrm{CrO}_4$.


• Product B is ferric oxide, $\mathrm{Fe}_2 \mathrm{O}_3$.

1. For Product A ($\mathrm{Na}_2 \mathrm{CrO}_4$):

• In $\mathrm{Na}_2 \mathrm{CrO}_4$, the chromium is in the +6 oxidation state.


• The electronic configuration of $\mathrm{Cr}^{6+}$ is $[\mathrm{Ar}] 3d^0$.


• Since there are no unpaired electrons in the $3d$ orbitals, the spin-only magnetic moment is zero.

$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{0(0+2)} \text{ BM} = 0 \text{ BM} $

1. For Product B ($\mathrm{Fe}_2 \mathrm{O}_3$):

• In $\mathrm{Fe}_2 \mathrm{O}_3$, the iron is in the +3 oxidation state.


• The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$.


• $\mathrm{Fe}^{3+}$ has 5 unpaired electrons in the $3d$ orbitals.


• The spin-only magnetic moment can be calculated using the formula:

$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{5(5+2)} \text{ BM} = \sqrt{35} \text{ BM} \approx 5.9 \text{ BM} $

• The spin-only magnetic moment of Product A is $0 \text{ BM}$.


• The spin-only magnetic moment of Product B is $5.9 \text{ BM}$.

Therefore, the sum of the spin-only magnetic moments of A and B is:

$ 0 + 5.9 \approx 6 \text{ BM}$

The sum of the spin-only magnetic moment values of A and B is approximately 6 B.M. (to the nearest integer).

Green coloured salt bead represents the metal ion taken is $\mathrm{Fe}^{3+}$ so, $\mathrm{Fe}^{3+}:[\mathrm{Ar}] 4 \mathrm{~s}^0 3 d^5$

So, $\mu=\sqrt{5 \times 7}=5.9=6$

Vanadium has highest enthalpy of atomization among first row transition elements.

$\mathrm{V}_2 \mathrm{O}_5$ is amphoteric

In $\mathrm{V}^{5+}$ there are no unpaired electrons.

Thus, $\mu=0$

$\mathrm{A}$ is $\mathrm{K}_2 \mathrm{MnO}_4$

$\mathrm{B}$ and $\mathrm{C}$ are $\mathrm{KMnO}_4$ and $\mathrm{MnO}_2$

$\mathrm{KMnO}_4(\mu=0)$

$\mathrm{MnO}_2(\mathrm{Mn}^{4+})(\mu=3.87)$

Sum $=3.87=4$ (Nearest integer)

To determine which of these ions have a noble gas configuration, we must consider what the phrase "noble gas configuration" means. Atoms or ions with a noble gas configuration have completely filled electron shells, similar to the electron configuration of the noble gases, which are the elements found in Group 18 of the periodic table. Noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

Let's examine each ion given:

• $\mathrm{Sr}^{2+}(z=38)$: Strontium has atomic number 38 and loses 2 electrons to form a $\mathrm{Sr}^{2+}$ ion, giving it the same electron configuration as krypton (Kr), with atomic number 36. Hence, it has a noble gas configuration.

${\left[\mathrm{Sr}^{2+}\right]=[\mathrm{Kr}]}$


• $\mathrm{Cs}^{+}(z=55)$: Cesium has atomic number 55, and by losing 1 electron to form $\mathrm{Cs}^{+}$, it has the electron configuration of xenon (Xe), with atomic number 54. Thus, it also has a noble gas configuration.

${\left[\mathrm{Cs}^{+}\right]=[\mathrm{Xe}]}$


• $\mathrm{La}^{2+}(z=57)$: Lanthanum has atomic number 57. If it lost 2 electrons to form $\mathrm{La}^{2+}$, it would not have a noble gas configuration because it would have one electron more than xenon (Xe), which means it would not fully match any noble gas electron configuration.

${\left[\mathrm{La}^{2+}\right]=[\mathrm{Xe}] 5 \mathrm{~d}^1}$


• $\mathrm{Pb}^{2+}(z=82)$: Lead has an atomic number of 82, so when it loses 2 electrons to form $\mathrm{Pb}^{2+}$, it has 80 electrons, which is the same electron number as mercury (Hg) and not a noble gas. Therefore, $\mathrm{Pb}^{2+}$ does not have a noble gas configuration.


${\left[\mathrm{~Pb}^{2+}\right]=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^2}$


• $\mathrm{Yb}^{2+}(z=70)$: Ytterbium has an atomic number of 70. By losing 2 electrons to form $\mathrm{Yb}^{2+}$, it has 68 electrons, aligning with the electron configuration of Erbium (Er). $\mathrm{Yb}^{2+}$ and not a noble gas.

${\left[\mathrm{Yb}^{2+}\right]=[Xe]4f^{14}}$


• $\mathrm{Fe}^{2+}(z=26)$: Iron has an atomic number of 26. When it becomes $\mathrm{Fe}^{2+}$ by losing 2 electrons, it has 24 electrons. This does not correspond to any noble gas configuration, as the nearest noble gas is argon (Ar) with 18 electrons.

Therefore, the ions $\mathrm{Sr}^{2+}$ and $\mathrm{Cs}^{+}$ have a noble gas configuration. Altogether, there are 2 ions with a noble gas configuration.