Redox Reactions Test - 45

Hint

The total number of electrons that an atom acquires or loses in order to form a chemical connection with another atom is known as the oxidation number.

Explanation

Let the central oxidation number is $$x$$.

The oxidation number of $$O$$ is $$2$$.

Step 1: the oxidation state of  $$CrO_{2}^{-}$$

Thus, the oxidation state if $$Cr$$ in $$CrO_{2}^{-}$$.

$$x+2\times \left( -2 \right)=-1$$

$$x-4=-1$$

$$x=-1+4$$

$$x=3$$

Step 2 : the oxidation state of $$CrO_{4}^{2-}$$,

$$x+4\times \left( -2 \right)=-2$$

$$x-8=-2$$

$$x=8-2$$

$$x=6$$

Step 3 : the oxidation state of $$ClO_{3}^{-}$$,

$$x+3\times \left( -2 \right)=-1$$

$$x-6=-1$$

$$x=6-1=5$$

Step 4: the oxidation state of $$MnO_{4}^{-}$$,

$$x+4\times \left( -2 \right)=-1$$

$$x-8=-1$$

$$x=8-1=+7$$

Hence, the increasing order of the central atom is $$CrO_{2}^{-}<ClO_{3}^{-}<CrO_{4}^{2-}<MnO_{4}^{-}$$.

Final answer

The correct answer is option (A).

$$H^+$$

Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.

Adding up the charges on each side. Making them equal by adding enough electrons $$(e^-)$$ to the more positive side.

Rule of thumb: $$e^-$$ and $$H^+$$ are almost always on the same side.

Hint: Count the total number of electrons in the valence shell.

Calculation of oxidation number is the most important step in the balancing of chemical equation.

So, ionic form is $${ { [Co({ C }_{ 2 }{ O }_{ 4 }) }_{ 2 }{ { (NH }_{ 3 }) }_{ 2 }] }^{ - }$$.

$$Br_3O_3$$ is a tribromooctaoxide.  The  two Br at the terminal, have 3 oxygen atom attached, hence it's oxidation number is +6. The Br atom in the center have 2 oxygen atoms attached ,so it's oxidation state is +4. Hence, +6,4+6 is the sequence.

In the given options, 

The medium is basic clearly seen in the reaction with $$OH^-$$

$$OH^-$$

$$OH^-$$ ions are added in to the reaction in basic medium.

This ions reacts with H+ ions and then gets converted into water.