When $KMnO_4$ acts as an oxidizing agent and ultimately forms $MnO_4^{2-}$,$MnO_2$,$Mn_2O_3$,and $Mn^{2+}$,what will be the number of electrons transferred in each case,respectively?

How many moles of electrons are required to reduce $2.5 \ mol$ of $Cr_2O_7^{2-}$ to $Cr^{3+}$?

In the following reaction,$I_2 + 5Cl_2 + 6H_2O \longrightarrow X + Y$,the products $X$ and $Y$ respectively are:

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$(a)$ $P_{4(s)} + OH_{(aq)}^{-} \to PH_{3(g)} + HPO_{2(aq)}^{-}$
$(b)$ $N_{2}H_{4(l)} + ClO_{3(aq)}^{-} \to NO_{(g)} + Cl_{(g)}^{-}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \to ClO_{2(aq)}^{-} + O_{2(g)} + H^{+}$

On balancing the given redox reaction,$aCr_2O_7^{2-} + bSO_3^{2-}{(aq)} + cH^{+}{(aq)}$ $\rightarrow 2aCr^{3+}{(aq)} + bSO_4^{2-}{(aq)} + \frac{c}{2}H_2O_{(l)}$,the coefficients $a$,$b$,and $c$ are found to be,respectively-

Assertion : When a metal is treated with conc. $HNO_3$,it generally yields a nitrate,$NO_2$,and $H_2O$.
Reason : Conc. $HNO_3$ reacts with a metal and first produces a metal nitrate and nascent hydrogen. The nascent hydrogen then further reduces $HNO_3$ to $NO_2$.