D & F Block Elements MCQs for NEET — Chemistry Questions with Answers

According to the context (MAGNETISM_AND_MATTER), "Diamagnetic substances are those which have tendency to move from stronger to the weaker part of the external magnetic field. In other words, unlike the way a magnet attracts metals like iron, it would repel a diamagnetic substance." This indicates weak repulsion by the magnetic field.

The text states (The d - and f - Block Elements, page 102), "Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum." and (MAGNETISM_AND_MATTER) "The individual atoms (or ions or molecules) of a paramagnetic material have a net magnetic moment of their own."

The context states: 'The d orbitals of the transition elements protrude to the periphery of an atom more than the other orbitals (i.e., s and p), hence, they are more influenced by the surroundings as well as affect the atoms or molecules surrounding them. In some respects, ions of a given $d^n$ configuration ($n = 1 – 9$) have similar magnetic and electronic properties. With partly filled d orbitals these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour.' Zn and Cd are not considered transition elements because their d-orbitals are completely filled in the ground state and common oxidation states. Sc (Scandium) in its common $+3$ oxidation state ($3d^0$) has no unpaired electrons. $Cu^{2+}$ has a $3d^9$ configuration, meaning it has one unpaired electron, making it paramagnetic.

The 'spin-only' formula for magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons. For $n=3$, $\mu = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ BM. This is also directly given in 'Table 4.7: Calculated and Observed Magnetic Moments (BM)' for $V^{2+}$ (3d^3).

The text (MAGNETISM_AND_MATTER, page 148) states: 'Some diamagnetic materials are bismuth, copper, lead, silicon, nitrogen (at STP), water and sodium chloride.'

The text states (The d - and f - Block Elements, page 102), 'For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula...'

The text (The d - and f - Block Elements, page 102) clearly states, 'attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism.'

The context (The d - and f - Block Elements, page 91) directly addresses this: 'The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula $(n-1)d^{10}ns^2$. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements.' And further, 'On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state ($3d^1$).'

From Table 4.7, $Ti^{3+}$ has a $3d^1$ configuration with 1 unpaired electron, leading to a calculated magnetic moment of $1.73$ BM. $V^{2+}$ has a $3d^3$ configuration with 3 unpaired electrons, leading to a calculated magnetic moment of $3.87$ BM. Therefore, $V^{2+}$ has a higher magnetic moment than $Ti^{3+}$.

The text (MAGNETISM_AND_MATTER, page 148, and page 152 summary point 5) explicitly states for a superconductor: '$\chi = -1$ and $\mu_r = 0$.'