d and f Block Elements

Oxidation state of alkaline earth metal (i.e. $\mathrm{Mg}$ ) is fixed and equal to $(+) 2$.

Oxidation state of $d$-block elements can vary, i.e. of $\mathrm{Fe}$ and $\mathrm{Cr}$ as follows:

Oxidation state of $\mathrm{Co}=(+) 2$ to $(+) 4$ and oxidation state of $\mathrm{Fe}=(+) 2$ to $(+) 6$, but (+)6 is less stable.

Oxidation state of $\mathrm{Cr}=(+)2$ to (+)6, where (+)6 state is more stable.

Hence, option (d) is the correct.

Among the given ions.

$\mathrm{Hg}_2^{2+}, \mathrm{Sr}^{+2}$ and $\mathrm{Li}^{+}$ do not have unpaired electrons, while $\mathrm{Bi}^{3+}$ has one unpaired electron.

Thus, carbonate of $\mathrm{Bi}^{3+}$ show coloured precipitate.

In $\mathrm{CrO}_3$, chromium is in (+)6 oxidation state with half-filled $3 d$-orbitals whereas in $\mathrm{Al}_2 \mathrm{O}_3, \mathrm{Al}$ is in (+) 3 oxidation state with nearest noble gas configuration. Therefore, $\mathrm{CrO}_3$ has lower melting point than $\mathrm{Al}_2 \mathrm{O}_3$.

Thus, assertion and reason both are correct but the given reason is not the correct explanation for the given assertion. The correct reason is the noble gas configuration achieved by aluminium is $\mathrm{Al}_2 \mathrm{O}_3$. Hence, option (b) is the correct.

The given species $\mathrm{S}_2 \mathrm{O}_7^{2-}$ and $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ have same structure and same valency

Thus, option (a) is the correct.

The electronic configuration of

$\begin{aligned} & \mathrm{Yb}^{+2}=[\mathrm{Xe}], 4 f^{14}, 5 d^0, 6 s^0 \\ & \mathrm{Gd}^{+2}=[\mathrm{Xe}], 4 f^7, 5 d^1 6 s^0 \end{aligned}$

Since, $\mathrm{Yb}^{2+}$ has full-filled $(f^{14})$ configuration. It is more stable than that of $\mathrm{Gd}^{2+}$. Thus, assertion is correct but reason is incorrect Hence, option (c) is the correct.

$d^5$-configuration has exactly half-filled configuration, so has more exchange energy, while $d^4$ has only four such (unpaired) electrons thus has less exchange energy.

Maximum exchange energy leads to the stabilisation of the atoms.

Therefore, $d^5$ configuration is more stable than that of $d^4$ configuration.

Hence, option (a) is the correct.

Spin only magnetic moment depends upon the number of unpaired electrons.

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

where, $n=$ number of unpaired electrons The electronic configuration for the given ions is as follow:

For $\mathrm{Mn}^{2+}[\mathrm{Z}=25]=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 3 d^2 4 s^0$

$\begin{gathered} \mu=\sqrt{5(5+2)}=\sqrt{35} \\ \text { For } \operatorname{Cr}^{2+}[Z=24]=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 3 d^4, 4 s^0 \end{gathered}$

$\mu=\sqrt{4(4+2)}=\sqrt{24}$

For $\mathrm{Ti}^{2+}[Z=22]=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 3 d^2, 4 s^0$

$\mu=\sqrt{2(2+2)}=\sqrt{8}$

$\therefore$ Correct order of spin only magnetic moment is $\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Ti}^{2+} \text {. }$

Assertion and reason are true but reason is not correct explanation of the assertion. Separation of Zr and Hf is difficult, it is not because of they lie in the same group of the periodic table. This is due to the lanthanide contraction which causes almost similar radii of both of them.