An element in compound ABD has oxidation number $ {{A}^{n-}}. $ It is oxidised by $ C{{r}_{2}}O_{7}^{2-} $ in acid medium, In the experiment $ 1.68\times {{10}^{-3}} $ moles of $ {{K}_{2}}C{{r}_{2}}{{O}_{7}} $ were used for $ 3.26\times {{10}^{-3}} $ moles of ABD. The new oxidation number of A after oxidation is:
(A) $ 3 $
(B) $ 3-n $
(C) $ n-3 $
(D) $ +n $

Hint :We know that the oxidation state, or oxidation number, of an element is defined as the charge that appears on an atom of the element when all other atoms attached to it are removed in the form of their ions.

Complete Step By Step Answer:
Here in question we will use the method of finding the oxidation number of an element in a compound. We have to remember that the algebraic sum of the oxidation numbers of all the atoms present in a compound must be equal to zero. Then we will equate the algebraic sum of the oxidation number of all atoms equal to zero. Then we will find out the oxidation state of the required element. Also in the process of oxidation, there occurs a loss of electrons during a reaction by a molecule, atom or ion. It occurs when the oxidation state of a molecule, atom or ion of an element/compound is increased. The process opposite to this is called reduction, which occurs when there takes place a gain of electrons or the oxidation state of an atom of an element/compound decreases.
 $ Cr $ changes its oxidation state from $ +6 $ to $ +3 $ in acidic medium on oxidizing a compound. So, the net change in oxidation per molecule of $ {{K}_{2}}C{{r}_{2}}{{O}_{7}} $ ​ is $ +6 $ (Two $ Cr $ atoms per molecule). As one equivalent of $ {{K}_{2}}C{{r}_{2}}{{O}_{7}} $ ​ oxidizes two equivalents of ABD, then each A atom changes its oxidation state by three. So, the new oxidation state is $ ~-n+3\Rightarrow 3-n. $
Therefore, the correct answer is option B.

Note :
Remember that whenever we are asked such types of questions we have to follow some basic steps. First we have to find out the charge on the other elements and the oxidation number of the asked element be x. After that we will multiply the charge of each element with the number of atoms present in the compound.