Redox Reactions Test

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Redox Reactions Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

$$CrO_5$$ has structure as shown, The oxidation number of chromium in the compound is?

A
$$+4$$

B
$$+5$$

C
$$+6$$

D
$$+10$$

E
$$0$$

Solution

By looking the above structure we can observe that 4 oxygen atoms are linked by peroxide linkage. So there oxidation state is -1 as in peroxide.
One oxygen atom is attached normally so its oxidation state is -2.
So oxidation state of $$Cr$$ is $$x+4(-1)+(-2)=0,x=+6$$
Question 2
1 / -0

Oxidation number of Cl in $$CaOCl_2$$ is ____________.

A
$$-1$$ and $$+1$$

B
$$+2$$

C
$$-2$$

D
none of these

Solution

In $$Ca(OCl)Cl$$, two $$Cl$$ atoms are in different oxidation state i.e., one $$Cl^{-}$$ in -1 oxidation state and other as $$OCl^{-}$$ in +1 oxidation state.
Question 3
1 / -0

The oxidation state of iodine in $$IPO_4$$ is:

A
$$+1$$

B
$$+3$$

C
$$+5$$

D
$$+7$$

Solution

Let oxidation state of iodine be x.
$$x-3=0, x=+3$$,
$$\because PO^{3-}_4$$ has combined oxidation number $$-3$$.
$$\therefore x-3=0$$
$$\therefore x=+3$$
Thus oxidation state of iodine is +3.
Question 4
1 / -0

Find the oxidation number of V in $$Rb_4Na[HV_{10}O_{28}]$$.

A
+5

B
+2

C
-5

D
none of these

Solution

Let oxidation state of $$V$$ is $$x$$ in $$Rb_4Na[HV_{10}O_{23}]$$
$$H(1) + 1 + 1 + 10x - 2 (28) = 0$$
$$6 + 10x - 56 = 0 $$
$$10x = 50$$
$$x = +5$$
Question 5
1 / -0

The oxidation state of the underlined element in the given compound is:
$$\underline{C_{12}}H_{22}O_{11}$$

A
+5

B
+4

C
0

D
none of these

Solution

$$C_{12}H_{22}O_{11}$$= Sucrose
$$\Rightarrow 12x+22 \times 1- 11 \times 2=0$$
$$\Rightarrow 12x+22-22=0$$
$$\Rightarrow x=0$$ .
Question 6
1 / -0

The oxidation state of the underlined element in the given compound is:
$$Na_2[\ \underline{Fe}(CN)_5NO\ ]$$

A
-2

B
+2

C
+3

D
none of these

Solution

Given compound $$Na_2[Fe(CN)_5NO]$$
On dividing into ions we get
$$[Fe(CN)_5NO]^{-3}$$ and $$3Na^{+1}$$
$$\Rightarrow x+(-5)+0=-3$$
$$\Rightarrow x= -3+5$$
$$\Rightarrow x=+2$$ .
Question 7
1 / -0

The oxidation state of the underlined element in the given compound is:
$$Ba_2\underline{Xe}O_2$$

A
+2

B
+3

C
+4

D
0

Solution

$$Ba_2 Xe O_2$$
$$\Rightarrow (+2)2+x+2(-2)=0$$
$$\Rightarrow x=0$$
Question 8
1 / -0

The oxidation state of the underlined element in the given compound is:
$$K_2\underline{Ta}F_7$$

A
+7

B
+4

C
+5

D
+1

Solution

$$K_2T_aF_1$$ Let oxidation state be $$x$$
$$ 2 (+1) + x + T(-1) = 0$$
$$x = +5$$
Question 9
1 / -0

In which compound among the following, carbon has a zero oxidation state?

A
$$CH_4$$

B
$$ CH_3Cl$$

C
$$CH_2Cl_2$$

D
$$CCl_4$$.

Solution

$$\text{Sum of total oxidation state of all atoms } = \text{ Overall charge on the compound}.$$
Oxidation state of $$Cl$$ is $$-1$$.Let oxidation state of $$C$$ be $$x$$.
For $$CH_4$$, applying above formula, $$x+4(1)=0,\\x=-4.$$
For $$CH_3Cl$$, applying above formula, $$x+3(1)+1(-1)=0,\\x=-2.$$
For $$CH_2Cl_2$$, applying above formula, $$x+2(1)+2(-1)=0,\\x=0.$$
For $$CCl_4$$, applying above formula, $$x+4(-1)=0,\\x=+4.$$
$$\therefore$$ oxidation state is $$0$$ for $$C$$ in $$CH_2Cl_2$$
Question 10
1 / -0

The oxidation state of the underlined element in the given compound is:
$$Na_2\underline{Mo}O_4$$

A
+4

B
+5

C
+6

D
none of these

Solution

$$Na_2MnO_4$$

Let oxidation state of Mo be $$x$$

$$ 2 (+1) + x + 4 (-2) = 0 $$

$$x = +6$$

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Redox Reactions Test - 36

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Redox Reactions Test - 36

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Question 1

$$+4$$

$$+5$$

$$+6$$

$$+10$$

$$0$$

Solution

Question 2

$$-1$$ and $$+1$$

$$+2$$

$$-2$$

none of these

Solution

Question 3

$$+1$$

$$+3$$

$$+5$$

$$+7$$

Solution

Question 4

+5

+2

-5

none of these

Solution

Question 5

+5

+4

0

none of these

Solution

Question 6

-2

+2

+3

none of these

Solution

Question 7

+2

+3

+4

0

Solution

Question 8

+7

+4

+5

+1

Solution

Question 9

$$CH_4$$

$$ CH_3Cl$$

$$CH_2Cl_2$$

$$CCl_4$$.

Solution

Question 10

+4

+5

+6

none of these

Solution

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