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what is oxidation no. of c in ch3

11 years ago

**Minimum oxidation number: -4 Maximum oxidation number: 4**

Oxidation numbers are hypothetical numbers assigned to an individual atom or ion present in a substance using a set of rules. Oxidation numbers (or oxidation states as they are also called) can be positive, negative, or zero. It is VERY IMPORTANT to remember that oxidation numbers are always reported for one individual atom or ion and not for groups of atoms or ions.

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**DETERMINING OXIDATION NUMBER**

For all compounds, whether covalent, polar covalent, or ionic, we treat as *ionic* for counting electrons and for oxidation-reduction reactions.

**Rule 1:** Sum of the oxidation numbers of all the atoms in the chemical species equals the charge on the species.

Neutral compounds: Sum of oxidation numbers = 0

Ionic species: Sum of oxidation numbers = charge of the ion

**Rule 2:** In Binary Compounds, the more Electronegative (EN) element is assigned to have a negative oxidation number. (See EN trends.)

**Rule 3:** Atoms may have only certain oxidation numbers. The range is:

Maximum oxidation number possible = + Group number.

Minimum oxidation number possible = (Group number - 8) (this number will be negative)

**Atoms which will have known oxidation numbers are:**

**Atoms as****Elements:**Ex. H_{2}, O_{2}, P_{4}, Fe

Oxidation number = 0**Monoatomic Ions:**Cations: Ex. Na

^{+}, Al^{3+}(main group metals)

Oxidation number = + Group Number

Anions: Cl^{-}, O^{2-}

Oxidation number = Group Number - 8**Hydrogen**Combined with Nonmetals: Ex. NH

_{3}, H_{2}O, HCl

Oxidation number = +1

Combined with Metals: Ex. NaH, CaH_{2}(hydrides)

Oxidation number = -1**Oxygen**(Unless O_{2}^{2-}, peroxide)

Oxidation number = -2

Examples:

**CO**: (Sum will equal 0 since it is a neutral molecule)

O will have a -2 ox. number.

1 C + 1 O = 0

(C?) + (-2) = 0

C? = +2

Oxidation number of C in CO is +2

Oxidation number of O in CO is -2 (known)

Check ox. number to see if it falls within range:

+2 is in between the maximum value of C, +4, (Gr#) and the minimum value of C, - 4, (Gr# - 8).

So okay.

**Cr _{2}O_{7}^{2-}**: (Sum of all oxidation numbers will equal -2 since it is an ion.)

2 Cr + 7 O = -2

2(Cr?) + 7(-2) = -2

2(Cr?) + (-14) = -2

2(Cr?) = +12

Cr? = +6

Oxidation number of each Cr in Cr_{2}O_{7}^{2-} is +6

Oxidation number of each O in Cr_{2}O_{7}^{2-} is -2 (known)

Check ox. number to see if it falls within range:

+6 is the maximum value that Cr can have (Gr#). So okay.

**CS _{2}**: (Sum will equal 0 since it is a neutral molecule)

C will have the positive oxidation number since it is less EN than S

S will have a -2 charge since it is Gr # 6, (6 - 8 = -2)

C + 2 S = 0

(C?) + 2 (-2) = 0

(C?) + (-4) = 0

C? = +4

Oxidation number of C in CS_{2} is +4

Oxidation number of each S in CS_{2} is -2 (known)

Check ox. number to see if it falls within range:

+4 is the maximum value that C can have, (Gr#). So okay.

**NH _{4}^{+}**: (Sum will equal +1 since it is an ion)

H will have a +1 ox. number since it is bonded to N, a nonmetal.

N + 4 H = +1

(N?) + 4(+1) = +1

N? = -3

Oxidation number of N in NH_{4}^{+} is -3

Oxidation number of each H in NH_{4}^{+} is +1 (known)

Check ox. number to see if it falls within range:

-3 is the minimum value that N can have, (Gr# - 8). So okay.

**H _{5}IO_{6}**: (Sum will equal 0 since it is neutral species.)

H will have a +1 ox. number since it is combined w/ nonmetals

Iodine will have a + charge since it is less EN than Oxygen

5 H + I + 6 O = 0

5(+1) + (I?) + 6(-2) = 0

(+5) + (I?) + (-12) = 0

(I?) + (-7) = 0

I? = +7

Oxidation number of I in H_{5}IO_{6} is +7

Oxidation number of each H in H_{5}IO_{6} is +1 (known)

Oxidation number of each O in H_{5}IO_{6} is -2 (known)

Check ox. number to see if it falls within range:

+7 is the maximum value that I can have, (Gr#). So okay.

**NaBH _{4}**: (Sum will equal 0 since it is neutral species.)

H will have a -1 ox. number since it is combined w/ metals

Na will have a +1 charge (+ Gr# = +1)

Na + B + 4 H = 0

(+1) + (B?) + 4(-1) = 0

(+1) + (B?) + (-4) = 0

(B?) + (-3) = 0

B? = +3

Oxidation number of B in NaBH_{4} is +3

Oxidation number of each Na in NaBH_{4} is +1 (known)

Oxidation number of each H in NaBH_{4} is -1 (known)

Check ox. number to see if it falls within range:

+3 is the maximum value that B can have, (Gr#). So okay.

**H _{2}MnO_{4}**: (Sum will equal 0 since it is neutral species.)

H is a +1 ox. number since it is combined w/ nonmetals (ignore metal)

Mn will have a + charge since it is less EN than Oxygen

2 H + Mn + 4 O = 0

2(+1) + (Mn?) + 4(-2) = 0

(+2) + (Mn?) + (-8) = 0

(Mn?) + (-6) = 0

Mn? = +6

Oxidation number of Mn in H_{2}MnO_{4} is +6

Check ox. number to see if it falls within range:

+6 is less than the maximum value of Mn, +7, (Gr#). So okay.

**** Note: **If assumed -1 for H, then Mn would be a +10. This is greater than the maximum value allowed for Mn, +7, (Gr#). So +10 for Mn is not allowed as a possible oxidation number.

11 years ago

CH3 isnot any existing compound. As the valency of carbon has not been completed. Bu still if it existed then

let oxdn no of C be x

x+3(+1)=0 (as oxidn no of hydrogen is +1)

x+3=0

x=-3

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