In $[F e ({N H}_{3})_{6}] {C l}_{3}$ the number of covalent bonds is:

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Solution

The correct option is C 18
In $[F e (N H_{3})_{6}] C l_{3}$ the $N H_{3}$ molecules are bonded to $F e$ with the aid of $6$ coordinate bond. Each $N H_{3}$ molecule has $3$ covalent $N - H$ bonds. Since there are $6 N H_{3}$ molecules, the total number of pure covalent bonds are $6 \times 3 = 18$ . The chlorine atoms exist as $C l^{-}$ ions and hence does not contribute to covalent bond

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