• Option 3 is correct, No, some atoms have wrong number of electrons around them. The symbol of atom are Cl, O, O, O. In ClO₃^{- TOTAL NUMBER OF VALENCE ELECTRON = 7+3×6+1=7+18+1=26electrons}(Cl has 7valence shell electron, O has 6 valence shell electron and here O atom are 3 in number so 6×3=18electrons , as it is an anion so one negative charge present on anion will be added in total number of valence electron. In the given structure total number of valence electrons is 26 but the number of electrons around the atoms are wrong. Out of three oxygen atom, 2 oxygen atoms should form double bonds with Cl and one oxygen should form single bond. This single bonded oxygen will have 3 lone pair and a negative charge and other 2 oxygen atoms will form double bond and have 2 lone pair of electrons.
• IF ₄^- Option 1 is correct yes, its a reasonable structure. In this structure total number of valence electron =7+7 × 4 +1 =7+28+1 =36 electrons ( 7 valence shell electron in I, 7in F and in the structure there are 4 F atoms so 7×4=28 electrons and one negative charge is added to the total valence shell electron). The electrons around each atom are correctly arranged Iodine atom has 4 bonds and 2 lone pair of electrons. 4 F bonds form single bond with I and has 3 lone pair of electrons.
• BeH₂ option 2is correct. No, the total number of valence electron is wrong. Total number of valence electron =2+1×2 = 2+2 =4 electrons. ( 2 from Beryllium and one from each hydrogen. However, in the structure total no. Of valence electron is 8. Thus, in actual structure there should be no lone pair on Be.  
• The correct structures are given below :

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