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sid shuk Grade: 11
        

what is oxidation no. of c in ch3

6 years ago

Answers : (2)

prashant madhukar
33 Points
										

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.


 



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:



  1. Atoms as Elements: Ex. H2, O2, P4, Fe
    Oxidation number = 0

  2. Monoatomic Ions:
    Cations: Ex. Na+, Al3+ (main group metals)
    Oxidation number = + Group Number
    Anions: Cl-, O2-
    Oxidation number = Group Number - 8

  3. Hydrogen
    Combined with Nonmetals: Ex. NH3, H2O, HCl
    Oxidation number = +1
    Combined with Metals: Ex. NaH, CaH2 (hydrides)
    Oxidation number = -1

  4. Oxygen (Unless O22-, 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.


Cr2O72-: (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 Cr2O72- is +6
Oxidation number of each O in Cr2O72- 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.


CS2: (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 CS2 is +4
Oxidation number of each S in CS2 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.


NH4+: (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 NH4+ is -3
Oxidation number of each H in NH4+ 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.


H5IO6: (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 H5IO6 is +7
Oxidation number of each H in H5IO6 is +1 (known)
Oxidation number of each O in H5IO6 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.


NaBH4: (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 NaBH4 is +3
Oxidation number of each Na in NaBH4 is +1 (known)
Oxidation number of each H in NaBH4 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.


H2MnO4: (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 H2MnO4 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.

6 years ago
deeksha sharma
40 Points
										

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

6 years ago
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