Courses
Courses for Kids
Free study material
Free LIVE classes
More LIVE
Join Vedantu’s FREE Mastercalss

# Arrange the following in increasing order of oxidation state of Ni.${K_2}[Ni{(CN)_4}]$, ${K_2}[Ni{(F)_6}]$, $Ni{(CO)_4}$ Verified
333.9k+ views
Hint: We can find the oxidation state of any atom by the fact that the sum of oxidation states of all atoms or groups of atoms in a compound is equal to the overall charge on the compound. Cyanide ions have overall charge of -1.

Let’s see what is meant by the oxidation state of a metal.
Oxidation state of any metal is the count of the number of electrons that it has lost in comparison with its ground state electron configuration.
- We can easily find the oxidation state of the metal in any compound if we know the overall charge on the compound and the respective oxidation states of other atoms or charge on a group of atoms in the compound. Let’s find the oxidation state of Nickel atom in all the given compounds first.
i) ${K_2}[Ni{(CN)_4}]$
We know that Cyanide ion ($C{N^ - }$) has an overall charge of -1 as the carbon atom here has +2 oxidation state and nitrogen is in -3 oxidation state. We also know that potassium ions will have +1 oxidation state. SO, we can easily find the oxidation state of nickel atoms in the given compound by following the formula.
Overall charge on ${K_2}[Ni{(CN)_4}]$ = 2(Oxidation state of K) + Oxidation state of Ni + 4(Overall charge on CN)
Overall charge on ${K_2}[Ni{(CN)_4}]$ is zero, so we can write that,
0 = 2(+1) + Oxidation state of Ni + 4(-1)
0 = 2 + Oxidation state of Ni – 4
Oxidation state of Ni = +4 – 2
Oxidation state of Ni in ${K_2}[Ni{(CN)_2}]$ = +2

ii) ${K_2}[Ni{(F)_6}]$
We know that fluoride ions have an oxidation number of -1 and overall charge on this compound is also zero. So, we can write that,
Overall charge on ${K_2}[Ni{(F)_6}]$ = 2(Oxidation state of K) + Oxidation state of Ni + 6(Oxidation state of F)
0 = 2(+1) + Oxidation state of Ni + 6(-1)
0 = 2 + Oxidation state of Ni – 6
Oxidation state of Ni = 6 – 2
Oxidation state of Ni in ${K_2}[Ni{(F)_6}]$ = +4

iii) $Ni{(CO)_4}$
We know that carbon monoxide is a neutral species and does not have any charge on it. Carbon monoxide is neutral because carbon has oxidation number of +2 and oxygen has oxidation number of -2. Let’s find the oxidation state of Ni in this compound.
Overall charge on $Ni{(CO)_4}$ = Oxidation state of Ni + 4(Overall charge on CO)
0 = Oxidation state of Ni + 4(0)
0 = Oxidation state of Ni + 0
Oxidation state of Ni in $Ni{(CO)_4}$ = 0
So, we obtained that oxidation state of Nickel atom in $Ni{(CO)_4}$, ${K_2}[Ni{(CN)_4}]$ and ${K_2}[Ni{(F)_6}]$ is 0, +2 and +4 respectively.

So, we can arrange them according to oxidation state of Nickel atom in an increasing order as:
$Ni{(CO)_4}$ < ${K_2}[Ni{(CN)_4}]$ < ${K_2}[Ni{(F)_6}]$

Note:

Remember that carbon monoxide in its complexes with metal is not carrying any charge means it is electrically neutral. It is easier for avoiding confusion that in some cases, you put the overall charge of some species in the formula for finding oxidation states because some of its constitutional atoms can also be found in other oxidation states in other compounds.

Last updated date: 22nd Sep 2023
Total views: 333.9k
Views today: 3.33k