Properties of transition metals
In general, the electronic configuration of outer orbitals of transition elements is (n-1)d1– 10 ns1–2. However, the electronic configurations of outer orbitals of Zn, Cd, Hg, and Cn are represented by the general formula (n-1)d1– 10 ns2. These metals are very hard and have low volatility due to high enthalpy of atomization. The more unpaired electrons, the higher the enthalpy of atomization. The stability of the half-filled d subshell in Mn2+ and the completely filled d10 configuration in Zn2+ are related to their E° values, whereas E° for Ni is related to the highest negative ΔhydH°.
Select the Correct statement from the following:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Cr2+ is reducing and Mn3+ is oxidizing in nature
An ion is a good reducing agent if losing an electron gives it a more stable configuration, and a good oxidising agent if gaining an electron does so. For ions in water, a half-filled t2g3 level and a half-filled d5 subshell are especially stable. Both \( \ce{Cr^{2+}} \) and \( \ce{Mn^{3+}} \) are d4 ions, but they reach stability in opposite directions:
Cr2+ (3d4): Cr (Z = 24) is [Ar]3d54s1, and removing the 4s electron and one 3d electron leaves 3d4. Losing one more electron gives \( \ce{Cr^{3+}} \), 3d3 (t2g3).
Mn3+ (3d4): Mn (Z = 25) is [Ar]3d54s2, and removing the two 4s electrons and one 3d electron leaves 3d4. Gaining one electron gives \( \ce{Mn^{2+}} \), 3d5.
The standard electrode potentials confirm this:
\( \displaystyle E^\circ_{\ce{Cr^3+/Cr^2+}}=-0.41\ \mathrm{V}\quad\text{(Cr}^{2+}\text{ is easily oxidised)} \)
\( \displaystyle E^\circ_{\ce{Mn^3+/Mn^2+}}=+1.57\ \mathrm{V}\quad\text{(Mn}^{3+}\text{ is easily reduced)} \)
Why the other options are incorrect
Option (1) Cr2+ is oxidising and Mn3+ is reducing in nature — Incorrect. Both parts are reversed.
Option (3) Cr2+ is reducing and Mn3+ is also reducing in nature — Incorrect. \( \ce{Mn^{3+}} \) (3d4) readily gains an electron to reach the stable half-filled 3d5 configuration of \( \ce{Mn^{2+}} \) (E° = +1.57 V), so it is an oxidising agent, not a reducing agent.
Option (4) Cr2+ is oxidising and Mn3+ is also oxidizing in nature — Incorrect. \( \ce{Cr^{2+}} \) (3d4) readily loses an electron to reach the stable t2g3 configuration of \( \ce{Cr^{3+}} \) (E° for \( \ce{Cr^{3+}/Cr^{2+}} \) = −0.41 V), so it is a reducing agent, not an oxidising agent.
Note: It seems that there is a mistake in the original NTA paper, which prints the outer electronic configuration of Zn, Cd, Hg and Cn as (n−1)d1–10ns2 when it should read (n−1)d10ns2, because these metals have a completely filled d subshell. The next sentence, “These metals are very hard and have low volatility due to high enthalpy of atomization”, reads as if it describes Zn, Cd and Hg, but it is true of transition metals in general. Zn, Cd and Hg are soft and fairly volatile (mercury is a liquid) and have the lowest enthalpies of atomisation, because they have no unpaired d electrons for metallic bonding.
NCERT (2026-27 edition):
Trends in the M2+/M Standard Electrode Potentials – …
