Question

which of the following forms an oxide with highest valency $?$
$\left( A \right)V$
$\left( B \right)Cr$
$\left( C \right)Mn$
$\left( D \right)Fe$

Answer 1

The atomic number of vanadium is $23$, chromium is $24$, manganese is $25$ and ferrous is $26$.
Now , we need to check the valence shell electrons of each atom and then the possible oxide which they form. The highest valency will have the highest oxidation state .

Complete step by step answer:
Valency is the term which we usually see for the capacity of atoms to combine with other atoms . Rutherford in his experiment shows that electrons are present outside the nucleus and electrons are responsible for making bonds .
Oxidation state is the measure of total number of electrons loss or gain while forming a bond . It can be positive or negative both .
Now , let us check the oxides of each options
Vanadium has $5$ electrons in its valence shell . so it can lose those $5$ electrons while forming bonds with oxygen .So here the highest oxidation state of vanadium is $+ 5$
The oxide of chromium is $C{r_2}{O_3}$, here the oxidation state of chromium is $+ 3$. In the case of $Cr{O_4}^{2 - }$, here the oxidation state is $+ 6$.
In case of manganese the highest oxidation state while forming an oxide is $+ 7$. When it forms oxide $M{n_2}{O_7}$ which is called manganese heptoxide.
Ferrous oxide has the highest oxidation state of $+ 3$ when it forms a compound of oxide$F{e_2}{O_3}$ which is brown colour.
So option $\left( C \right)Mn$ has the highest valency.

Hence, option C is correct.

Note:
Oxidation state of pure metal is always zero .When we add the oxidation states of all the elements in the compound it must be equal to zero . Generally the oxidation state of hydrogen is$+ 1$ and of oxygen is $- 2$.