Transition elements (d-block)
Read the following passage and answer the questions given:
The d-block occupies the large middle section of the periodic table flanked between s- and p-blocks in the modern periodic table. These elements have a general electronic configuration (n-1)d1-10 ns2 (except Pd). These elements possess a partially filled penultimate d-subshell in their ground state. Most of the transition element compounds are colored in solid or in solution state which arise due to excitation of electrons from lower energy d-orbitals to higher energy d-orbitals. The transition metal ions which have completely filled d-orbitals are colorless due to lack of excitation within d-orbitals. Most of the transition metals and their compounds exhibit paramagnetic behavior (attracted by a magnetic field) which is attributed to an incomplete d-subshell and the presence of unpaired electrons. An element or ion having no unpaired electron shows diamagnetic behavior (repelled by a magnetic field). The magnetic moment of transition elements arises due to spin and orbital motion in ions or molecules. The magnetic moment is determined by the number of unpaired electrons (n) and is calculated by using the spin-only formula μ= [n (n+2)]1/2 Bohr magneton (BM). The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the transition metal or its ions. Transition elements exhibit certain important characteristic properties such as variable oxidation state, complex formation, colored ion formation, catalytic activity and metallic properties like high tensile strength, ductility, malleability etc. Transition metals are hard and have high melting points (with exceptions of Zn, Cd, Hg).
The maximum spin only magnetic moment is shown by:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → V2+
As the passage says, the spin-only magnetic moment depends on the number of unpaired electrons, n:
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)
To find n for a transition metal ion, remove the 4s electrons first, then d electrons, and fill the five d orbitals singly before pairing (Hund’s rule):
Sc3+: Sc (Z = 21) is [Ar]3d14s2; removing 4s2 and 3d1 gives 3d0, n = 0, μ = 0
V2+: V (Z = 23) is [Ar]3d34s2; removing 4s2 gives 3d3, n = 3, μ = √15 = 3.87 BM
Cu2+: Cu (Z = 29) is [Ar]3d104s1; removing 4s1 and one 3d electron gives 3d9, n = 1, μ = √3 = 1.73 BM
Ni2+: Ni (Z = 28) is [Ar]3d84s2; removing 4s2 gives 3d8, n = 2, μ = √8 = 2.83 BM
\( \ce{V^{2+}} \) has the most unpaired electrons, so it shows the maximum magnetic moment.
Why the other options are incorrect
Option (1) Sc3+ — Incorrect. It has no d electrons, so it is diamagnetic.
Option (3) Cu2+ — Incorrect. It has only one unpaired electron.
Option (4) Ni2+ — Incorrect. It has only two unpaired electrons.
Note: It seems that there is a mistake in the original NTA paper. The passage gives the general electronic configuration of the d-block elements as (n−1)d1–10 ns2 (except Pd). The correct general configuration is (n−1)d1–10 ns1–2, because some elements have only one ns electron, such as Cr (3d54s1) and Cu (3d104s1), and Pd (4d105s0) has none.
NCERT (2026-27 edition):
Magnetic Properties – …
