The correct answer is +3.
An ionic compound is made up of cation and anion. The given ionic compound is T1. It is made up of Пcation and I anion, Oxidation state of an element represents the charge acquired by an element when it is bonded. It is equal to the number of electrons gained or lost in the formation of compound. The total charges of all the atoms in a compound is zero.
Let the oxidation state of TI be x Calculate the oxidation state of Tin T
Oxidation state of '11+3(Oxidation state of 1)=0
x+3(-1)=0
X=+3
So, the apparent oxidation state of T1 is +3,
It is given that the anion of TII, is I. Calculate the oxidation state of T1 in TII,.
Oxidation state of TI+Oxidation state of I=0
x+(-1)=0
X=+1
Thus, the actual oxidation state of T1 is +1.
Electrons in this structure are indicated by dots.
The I atom has 7 valence electrons. Calculate the total number of valence electrons.
Total number of valence electrons = [(3×7)+1] 22
The total number of valence electrons in 1, is 22. Therefore, the number of valence electron pairs in l," is 11, which is 22/2. Draw the skeletal structure of 1, by assigning 2 valence electron pairs between 1 and I atoms (two on each side of I).
In this way, two valence electron pairs are used. Out of the remaining 9 (11-2) valence electron pairs, six valence electron pairs are distributed as lone pairs around the two I atoms to complete the octet of 1. The remaining three valence electron pair is assigned to the central I atom as a lone pair.
Draw the complete Lewis dot structure of 1,
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