Chemistry MCQs for NEET — Practice Questions with Answers

In this  complex  iron  is a central metal atom showing  +1 oxidation state.becoz NO is showing +1 charge here.

$\stackrel{\ast }{M{g}_2{P}_2{O}_7}$

$4+2x-2\times 7=0$; $2x=14-4=10$

2x = 10; $x=\frac{10}{2}=+5$.

$Ba{\left(H_2\text{}\stackrel{\ast }{P}\text{}{O}_2\right)}_2$; $Ba{H}_4\text{}\stackrel{\ast }{P_2}\text{}{O}_4$

$2+4+2x-8=0$; 2x = 2

$x=\frac{2}{2}=+1$.

$N{a}_2{\stackrel{\ast }{S}}_4\text{}{O}_6$

$2+4x-12=0$

$4x=10$

$\Rightarrow x=\frac{10}{4}$

$\Rightarrow x=\frac{5}{2}$

$\left[\stackrel{\ast }{Co}{\left(N{H}_3\right)}_{4}ClN{O}_2\right]$

$x+4\left(0\right)+1(-1)+1(-1)=0$

$x+0-1-1=0$

x – 2 = 0; x = +2.

Number of e^– transferred in each case is 1, 3, 4, 5.

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O\times 2$

$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}\times 5$

$\overline{2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O}$

Thus the coefficient of $Mn{O}_4^{-}$, $C_2{O}_4^{2-}$ and H⁺ in the above balanced equation respectively are 2, 5, 16.  

Higher is the reduction potential stronger is the oxidising agent. Hence in the given options. $Mn{O}_4^{-}$ is strongest oxidising agent.  

IO₃^– + aI^– + bH⁺ → cH₂O + dI₂

Step 1 : I^–1 → I₂ (oxidation)

IO₃^– → I₂ (reduction)

Step 2 : 2IO₃^– + 12H⁺ → I₂ + 6H₂O

Step 3 : 2IO₃^– + 12H⁺ + 10e → I₂ + 6H₂O

2I^– → I₂ + 2e

Step 4 : 2IO₃^– + 12H⁺ + 10e^– → I₂ + 6H₂O

[2I^– → I₂ + 2e]5

Step 5 : 2IO₃^– + 10I^– + 12H⁺ → 6I₂ + 6H₂O

IO₃^– + 5I^– + 6H⁺ → 3I₂ + 3H₂O

On comparing, a = 5, b = 6, c = 3, d = 3

In alkaline medium 

$2KMn{O}_4+KI+H_{2}O\to 2Mn{O}_2+2KOH+KI{O}_3$.