Answer:
The oxidation state of O in OF₂ is -2
The oxidaiton state of C in CO is +2
The oxidation state of N in K₃N is -3
Explanation:
In the OF₂ molecule, fluorine acts with -1 in this compound as the molecule has 2 atoms of F, the oxygen must act with -2.
Oxygen usually has -2 as oxidation state, except for the peroxides.
In CO, the oxygen has -2 as oxidation state so C works with +2
C usually has +4 as oxidation state but it can act with +2 and -4
For K₃N: oxidation state of K is +1,but we have 3 atoms (+3). Therefore the N must work with -3 as oxidation state. Nitrogen has many oxidation states, as the halogens (except for the F). It can work with 1, 2, +-3, 4 and 5.
Remember that the global charge of the molecule is 0
Answer :
The oxidation state of oxygen (O) in [tex]OF_2[/tex] is, (+2)
The oxidation state of carbon (C) in [tex]CO[/tex] is, (+2)
The oxidation state of nitrogen (N) in [tex]K_3N[/tex] is, (-3)
Explanation :
Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
When the atoms are present in their elemental state then the oxidation number will be zero.
Rules for Oxidation Numbers :
The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.
The oxidation number of a Group 1 element in a compound is +1.
The oxidation number of a Group 2 element in a compound is +2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
(a) The given compound is, [tex]OF_2[/tex]
Let the oxidation state of 'O' be, 'x'
[tex]x+2(-1)=0\\\\x-2=0\\\\x=+2[/tex]
The oxidation state of oxygen (O) in [tex]OF_2[/tex] is, (+2)
(b) The given compound is, [tex]CO[/tex]
Let the oxidation state of 'C' be, 'x'
[tex]x+(-2)=0\\\\x-2=0\\\\x=+2[/tex]
The oxidation state of carbon (C) in [tex]CO[/tex] is, (+2)
(c) The given compound is, [tex]K_3N[/tex]
Let the oxidation state of 'N' be, 'x'
[tex]3(+1)+x=0\\\\3+x=0\\\\x=-3[/tex]
The oxidation state of nitrogen (N) in [tex]K_3N[/tex] is, (-3)
