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Question

Determine the following regarding ICl4

(i) Number of valance electrons in central atom
(ii) Number of Ligands
(iii) Anionic charge
(iv) Cationic charge


  1. (i) - 6 ; (ii) - 5 ; (iii) - 0 ; (iv) - 1

  2. (i) - 7 ; (ii) - 7 ; (iii) - 1 ; (iv) - 0

  3. (i) - 6 ; (ii) - 5 ; (iii) - 1 ; (iv) - 0

  4. (i) - 7 ; (ii) - 4 ; (iii) - 1 ; (iv) - 0


Solution

The correct option is D

(i) - 7 ; (ii) - 4 ; (iii) - 1 ; (iv) - 0


As we know,

The central atom is taken as which is least in number

Therefore, central atom is Iodine

Iodine is a halogen

The valance electrons will be 7

Now, we know the ligands are the atoms which are attached to the central atom.

Since, 4 chlorine atoms are attached to iodine (which is the central atom)

Therefore, number of Ligands = 4

Since, there is a negative charge on the compound.

Then we can say that

Anionic charge = 1 (negative)

Cationic charge = 0 (Because there is no positive charge)

That's why the correct option is (d)

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