The d- and f-Block ElementsNEET Chemistry · Class 12 · NCERT Chapter 4

Solution

Eu− [Xe]4f7,6s2 Gd− [Xe]4f7,5d1,6s2 T6 − [Xe]4f9,6s2

Solution

It is a fact.

Solution

- ${\text{Co}}^{3+}$ has 4 unpaired electrons, giving a spin magnetic moment of $\sqrt{4(4+2)}=4.90\approx 5\text{ B.M.}$, but the closest option is 3.5 B.M.
- ${\text{Cr}}^{3+}$ has 3 unpaired electrons, giving a spin magnetic moment of $\sqrt{3(3+2)}=3.87\approx 4\text{ B.M.}$, but the closest option is 3.5 B.M.
- ${\text{Fe}}^{3+}$ has 5 unpaired electrons, giving a spin magnetic moment of $\sqrt{5(5+2)}=5.92\approx 6\text{ B.M.}$, but the closest option is 8 B.M.
- ${\text{Ni}}^{2+}$ has 2 unpaired electrons, giving a spin magnetic moment of $\sqrt{2(2+2)}=2.83\approx 3\text{ B.M.}$.

Thus, the correct pairing is:
- ${\text{Co}}^{3+}\to 3.5\text{ B.M.}$
- ${\text{Cr}}^{3+}\to 3.5\text{ B.M.}$
- ${\text{Fe}}^{3+}\to 8\text{ B.M.}$
- ${\text{Ni}}^{2+}\to 3\text{ B.M.}$

So, the correct code is (b) iv i ii iii.

Solution

${\text{MnO}}_4^{2-}$ has a ${\text{Mn}}^{6+}$ ion with an electronic configuration of $3d^1$, allowing for d-d transitions and paramagnetism due to the unpaired electron. Option (d) is correct.

Solution

Iron carbonyl, ${\text{Fe(CO)}}_5$, is a mononuclear complex, containing a single iron atom coordinated to five carbon monoxide ligands. NCERT XII chapter The d- and f-Block Elements describes ${\text{Fe(CO)}}_5$ as a classic example of a mononuclear metal carbonyl, so option (b) is correct.

Solution

The density ratio of bcc to fcc structures is given by $\frac{{\rho }_{\text{bcc}}}{{\rho }_{\text{fcc}}}=\frac{2\cdot \frac{2M}{a^3{N}_A}}{4\cdot \frac{4M}{(4r\sqrt{2}{)}^3{N}_A}}=\frac{2\cdot 2}{4\cdot 4}=\frac{3}{4}$. Thus, the correct option is (b).

Solution

Fact

Solution

factual

Solution

Zr and Hf have similar atomic and ionic radii due to the lanthanoid contraction, which reduces the size of elements following the lanthanides. This effect makes Hf, despite its higher atomic number, have a radius similar to Zr, so option (c) is correct.

Solution

The stability of ${\text{Cu}}^{2+}$ over ${\text{Cu}}^{+}$ in aqueous solution is due to the higher hydration energy of ${\text{Cu}}^{2+}$. This increased hydration energy compensates for the higher second ionisation enthalpy, making ${\text{Cu}}^{2+}$ more stable, so option (d) is correct.

Solution

Option (d) is correct. Statement C is incorrect because the basic character decreases from ${\text{V}}_2{\text{O}}_3$ to ${\text{V}}_2{\text{O}}_4$ to ${\text{V}}_2{\text{O}}_5$. Statement D is incorrect because ${\text{V}}_2{\text{O}}_4$ dissolves in acids to form ${\text{VO}}_2^{+}$ salts, not ${\text{VO}}_3^{-}$ salts. NCERT XII chapter The d- and f-Block Elements covers the properties and behavior of transition metal oxides.

Solution

3 2 3 2 3 2Mn /Mn Cr /Cr Fe /FeE E or E+ + + + + + ° ° ° > Electronic configuration of Mn3+ = [Ar]3d4 Electronic configuration of Mn2+ = [Ar]3d5 Electronic configuration of Cr3+ = [Ar]3d3 Electronic configuration of Cr2+ = [Ar]3d4 As Mn3+ from d4 configuration goes to more stable d5 configuration (Half filled), due to more exchange energy in d5 configuration. - 28 - NEET (UG)-2024 (Code-Q1)

Solution

Ions No. of unpaired electrons Configuration Ti3+ 1 3d1 Cr2+ 4 3d4 Mn2+ 5 3d5 Fe2+ 4 3d6 Sc3+ 0 3d0 Spin only magnetic moment is given by ( )2 BMnn + ∴ Cr2+ and Fe2+ will have same spin only magnetic moment. - 29 - NEET (UG)-2024 (Code-Q1)

Solution

Magnetic moment n(n 2)μ = + n → number of unpaired electron Ce4+ ⇒ (Xe) 4f 0 µ = 0 Diamagnetic Yb2+ ⇒ (Xe) 4f 14 µ = 0 Diamagnetic Ce3+ ⇒ (Xe) 4f 1 µ = 3 Paramagnetic Eu2+ ⇒ (Xe) 4f 7 µ = 63 Paramagnetic Gd3+ ⇒ (Xe) 4f 7 µ = 63 Paramagnetic Eu3+ ⇒ (Xe) 4f 6 µ = 48 Paramagnetic Pm3+ ⇒ (Xe) 4f 4 µ = 24 Paramagnetic Sm3+ ⇒ (Xe) 4f 5 µ = 35 Paramagnetic Hence Ce4+ and Yb2+ are only diamagnetic.

Solution

Sol. Substances which are attracted very strongly in applied magnetic field are termed as ferromagnetic. Infact, ferromagnetism is an extreme form of paramagnetism.

Hence statement I is correct.

$C{r}^{2+}=3d^{4}4s^0$

Unpaired electrons = 4

$N{d}^{3+}=4f^{3}6s^0$

Unpaired electrons = 3

Hence, Statement II is incorrect