Chemistry - Complete d-block Question with Solution | TestHub

(i) The oxidation state of $\mathrm{Mn}$ in oxide (${\mathrm{Mn}}_2{\mathrm{O}}_7$) is:

Let us consider $\mathrm{Mn}$ as $\mathrm{x}$.

$\Rightarrow 2\mathrm{x} + 7(-2) = 0$

$\Rightarrow 2\mathrm{x} + (-14) = 0$

$\Rightarrow 2\mathrm{x} -14 = 0$

$\Rightarrow 2\mathrm{x} = 14$

$\Rightarrow \mathrm{x} = 7$

(ii) The oxidation state of Ruthenium in oxide (${\mathrm{RuO}}_4$) is:

Consider $\mathrm{Ru}$ as $x$.

$\Rightarrow \mathrm{x} + 4(-2) = 0$

$\Rightarrow \mathrm{x} - 8 =0$

$\Rightarrow \mathrm{x} = 8$

The oxidation state of Osmium in oxide (${\mathrm{OsO}}_4$) is:

Consider $\mathrm{Os}$ as $\mathrm{x}$.

$\Rightarrow \mathrm{x} + 4(-2) = 0$

$\Rightarrow \mathrm{x} - 8 =0$

$\Rightarrow \mathrm{x} = 8$ 

Hence, Ruthenium and Osmium has +7 oxidation state in its oxide.

(iii) $\mathrm{Sc}$ has ${\mathrm{ns}}^2 (\mathrm{n}-1){\mathrm{d}}^1$ electronic configuration.

It utilizes two electrons from its $\mathrm{ns}$ sub-shell then its oxidation state = +2. When it utilizes both the electrons then its oxidation state = +3.

(iv) In the modern periodic table, chromium metal is present in group 6. It is a transition metal and belongs to the first transition series. The atomic number of chromium is 24. Its electronic configuration is $\left[\mathrm{Ar}\right] 3{\mathrm{d}}^5 4{\mathrm{s}}^1$. 

It has five electrons in the 3d sub-shell and one electron in the 4s sub-shell. Thus, chromium metal has six valence electrons. Chromium can lose some or all of its valence electrons to form ions with different oxidation states. Thus, chromium shows the oxidation states of +1, +2, +3, +4, +5, and +6 respectively.
The most common oxidation states of chromium are +2, +3, and +6.

So, the statement (i), (ii) and (iv) are correct.