Question

Determine the following regarding $IC{l}_4^{-}$

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

• A. (i) - 6 ; (ii) - 5 ; (iii) - 1 ; (iv) - 0
• B. (i) - 7 ; (ii) - 7 ; (iii) - 1 ; (iv) - 0
• C. (i) - 6 ; (ii) - 5 ; (iii) - 0 ; (iv) - 1
• D. (i) - 7 ; (ii) - 4 ; (iii) - 1 ; (iv) - 0

Answer 1

The correct option is D

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

As we know,

$\Rightarrow$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)