MCQ
Calculate the steric numbers for iodine in $(\text{IF}_4^−) $and for bromine in $(\text{BrO}_4^−).$
- ✓$6$ and $4$
- B$5$ and $6$
- C$4$ and $6$
- D$3$ and $4$
✓
Answer
Correct option: A.
$6$ and $4$
The central $I^-$ has eight valence electrons
Each F atom has seven valence electrons of its own and needs to one of the electrons from $I^-$ to attain noble $-$ gas configuration $(s^2p^6)$. Thus, four, of the $I^-$ valence electrons take part in covalent bonds, leaving the remaining four to form two lone pairs.
Thus, $\ce{SN} = 4 ($bonded atoms$) +\ 2 ($lone pairs$) = 6$
The central $\ce{Br}^-$ has also eight electrons. In $(\text{BrO}_4^−).$, each oxygen atom needs to share two of electrons from $Br^-$ to attain noble $-$ gas configurations thus all the electrons are used in bonding and no lone pair exists.
Thus, $\ce{SN} = 4 ($bonded pair$) \ + 0 ($lone pairs$) = 4.$
Each F atom has seven valence electrons of its own and needs to one of the electrons from $I^-$ to attain noble $-$ gas configuration $(s^2p^6)$. Thus, four, of the $I^-$ valence electrons take part in covalent bonds, leaving the remaining four to form two lone pairs.
Thus, $\ce{SN} = 4 ($bonded atoms$) +\ 2 ($lone pairs$) = 6$
The central $\ce{Br}^-$ has also eight electrons. In $(\text{BrO}_4^−).$, each oxygen atom needs to share two of electrons from $Br^-$ to attain noble $-$ gas configurations thus all the electrons are used in bonding and no lone pair exists.
Thus, $\ce{SN} = 4 ($bonded pair$) \ + 0 ($lone pairs$) = 4.$
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